Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.2

Question 1.
Define Index Number.
Solution:
Index Numbers are the indicators which reflect the changes over a specified period of time in price of different commodities, production, sales, cost of living etc.

“An Index Number is a device which shows by its variations the change in a magnitude which is not capable of accurate measurements in it-self or of direct valuation in practice”. – Whel-don

“An Index number is a statistical measure of fluctuations in a variable arranged in the form of a series and using a base. Period for maxing H comparisons” – Lawrence J Kalpan.

Question 2.
State the uses of Index Number.
Solution:
(1) Index number is an important tool for formulating decision and management p policies.
(2) It helps in studying the trends and tendencies.
(3) It determines the inflation and deflation in an Economy.

Question 3.
Mention the classification of Index Number.
Solution:
Index number can be classified as follows,
(i) Price Index Number:
It measures the general changes in the retail or wholesale price level of a particular or group of commodities.

(ii) Quantity Index Number:
These are indices of measure the changes in the quantity of goods manufactured in a factory.

(iii) Cost of living Index Number:
These are intended to study the effect of change in the price level on the cost of living of different classes of people.

Question 4.
Define Laspeyre’s Price index number.
Solution:
Laspeyre’s Price index number
P\(_{ 01 }^{L}\) ⥪ \(\frac { Σp_1q_0 }{Σp_0q_0}\) × 100
where P₁ = Current year price
p₀ = base year price
q₀ = base year quantity

Question 5.
Explain Paasche’s price index number.
Solution:
Paasches price index number
P\(_{ 01 }^{L}\) = \(\frac { Σp_1q_1 }{Σp_0q_1}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 6.
Write note on Fisher’s price index number.
Solution:
Fishers price index number
P^F = \(\sqrt { P^L×P^P}\)
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\) × 100
where P₁ = current year price
q₁ = current year quantity
p₀ = base year price
q₀ = base year quantity

Question 7.
State the test of adequacy of index number.
Solution:
There are two tests which are used to test the adequacy for an index number. The two tests are as follows
(i) Time Reversal Test
(ii) Factory Reversal Test

Question 8.
Define Time Reversal Test.
Solution:
It is an important test for testing the consistency of a good index number. This test maintains time consistency by working both forward and backward with respect to time (here time refers to base year and current year). Symbolically the following relationship should be satisfied. P₀₁ × p₁₀ = 1

Fisher’s index number formula satisfies the above relationship
P\(_{ 01 }^{F}\) = \(\sqrt { \frac{Σp_1p_0×Σp_1q_1}{Σp_0q_0×Σp_0q_1}}\)

when the base year and current year are interchanged, we get
P\(_{ 10 }^{F}\) = \(\sqrt { \frac{Σp_0q_1×Σp_0q_0}{Σp_1q_1×Σp_1q_0}}\)
P\(_{ 01 }^{F}\) × P\(_{ 10 }^{F}\) = 1

Question 9.
Explain factor reversal test.
Solution:
This is another test for testing the consistency of a good index number. The product of price index number and quantity index number from the base year to the current year should be equal to the true value ratio. That is, the ratio between the total value of current period and total
value of the base period is known as true value ratio. Factor Reversal Test is given by,

where P₀₁ is the relative change in price
Q₀₁ is the relative change in quantity.

Question 10.
Define true value ratio.
Solution:
\(\frac { Σp_1q_1 }{Σp_0q_0}\) is the ratio of the total value in the current period to the total value in the base period and this ratio is called the true value ratio.

Question 11.
Discuss about cost of Living Index Number.
Solution:
Cost of living index numbers are generally designed to represent the average change over time in the prices paid by the ultimate consumer for a specified quantity of goods and services cost of living index number is also known as consumer price index number.

It is well known that a given change in the level of prices (retail) affects the cost of living of different classes of people in different manners. The general index number fails to reveal this. Therefore it is essential to construct a cost of living index number which helps us in determining the effect of rise and fall in prices on different classes of consumers living in different areas.

Question 12.
Define family budget method.
Solution:
In this method, the weights are calculated by multiplying prices and quantity of the base year.
(i.e.) V = Σp₀q₀. The formula is given by,
Cost of Living Index Number = \(\frac { Σpv }{Σv}\)
where P = \(\frac { p_1 }{p_0}\) × 100 is the price relative.
v = Σp₀q₀ is the value relative.

Question 13.
State the uses of cost of Living Index Number.
Solution:
(i) It indicates whether the real wages of workers are rising or falling for a given time.
(ii) It is used by the administrators for regulating dearness allowance or grant of bonus to the workers.

Question 14.
Calculate by a suitable method, the index number of price from the following data:

Solution:

Question 15.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 16.
Calculate price index number for 2005 by (a) Laspeyre’s (b) Paasche’s method

Solution:

Question 17.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 18.
Using the following data, construct Fisher’s Ideal index and show how it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Question 19.
Calculate Fisher’s index number to the following data. Also show that it satisfies Time Reversal Test.

Solution:

Question 20.
Th following are the group index numbers and the group weights of an average working class family’s budget. Construct the cost of living index number:

Solution:

Question 21.
Construct the cost of living Index number for 2015 on the basis of 2012 from the following data using family budget method.

Solution:

Question 22.
Calculate the cost of living index by aggregate expenditure method:

Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.1

Question 1.
Define Time series.
Solution:
A Time Series consists of data arranged chronologically – Croxton & Cowden
When quantitative data are arranged in the order of their occurrence, the resulting series is called the Time Series – Wessel & Wallet.

A time series consists of a set of observations arranged in chronological order (either ascending or descending). Times Series has an important objective to identify the variations and try to eliminate the variations and also helps us to estimate or predict the future values.

Question 2.
What is the need for studying time series?
Solution:
Time series analysis is one of the statistical methods used to determine the patterns in data collected for a period of time. Generally, each of us should know about the past data to observe and understand the changes that have taken place in the past and current time. One can also identify the regular or irregular occurrence of any specific feature over a time period in a time series data.

Question 3.
State the uses of time series.
Solution:
It helps in the analysis of the past behavior.
It helps in forecasting and for future plans.
It helps in the evaluation of current achievements.
It helps in making comparative studies between one time period and others.
Therefore time series helps us to study and analyze the time related data which involves in business fields, economics, industries, etc.

Question 4.
Mention the components of the time series.
Solution:
There are four types of components in a time series. They are as follows
(i) Secular Trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

Question 5.
Define secular trend.
Solution:
It is a general tendency of time series to increase or decrease or stagnates during a long period of time, an upward tendency is usually observed in population of a country, production, sales, prices in industries,income or individuals etc., A downward tendency is observed in deaths, epidemics, prices of electronic gadgets, water sources, mortality rate etc. It is not necessarily that the increase or decrease should be in the same direction throughout the given period of time.

Question 6.
Write a brief note on seasonal variations
Solution:
As the name suggests, tendency movements are due to nature which repeat themselves periodically in every seasons. These variations repeat themselves in less than one year time. It is measured in an interval of time. Seasonal variations may be influenced by natural force, social customs and traditions. These variations are the results of such factors which uniformly and regularly rise and fall in the magnitude. For example, selling of umbrellas’ and raincoat in the rainy season, sales of cool drinks in summer season, crackers in Deepawali season, purchase of dresses in a festival season, sugarcane in Pongal season.

Question 7.
Explain cyclic variations.
Solution:
These variations are not necessarily uniformly periodic in nature. That is, they may or may not follow exactly similar patterns after equal intervals of time. Generally one cyclic period ranges from 7 to 9 years and there is no hard and fast rule in the fixation of year for a cyclic period. For example, every business cycle has a Start-Boom – Depression. Recover, maintenance during booms and depressions, changes in government monetary policies, changes in interest rates.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
These variations do not have particular pattern and there is no regular period of time of their occurrences. These are accidently changes which are purely random or unpredictable. Normally they are short-term variations, but its occurrence sometimes has its effect so intense that they may give rise to new cyclic or other movements of variations. For example floods, wars, earthquakes, Tsunami, strikes, lockouts etc.

Question 9.
Define seasonal index.
Solution:
Seasonal Index for every season (monthly or quarterly) is calculated as follows
Seasonal Index (S.I) = \(\frac { Seasonal Average }{Grand Average}\) × 100
If the data is given monthwise
Seasonal Index = \(\frac { Monthly Average }{Grand Average}\) × 100
If quarterly data is given
Seasonal Index = \(\frac { Quarterly Average }{Grand Average}\) × 100

Question 10.
Explain the method of fitting a straight line.
Solution:
(i) The straight line trend is represented by the equation Y = a + bX …………. (1)
Where Y is the actual value, X is time, a, b are constants.
(ii) The constants ‘a and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………… (3)
Where ‘n’ = number of years given in the data.
(iii) By taking the mid-point of the time as the ori-gin, we get ΣX = 0
(iv) When ΣX = 0, the two normal equations reduces to
ΣY = na + b(0) ; a = \(\frac { ΣY }{n}\) = \(\bar { Y}\)
ΣXY = a(0) + bΣX² ; b = \(\frac { ΣXY }{ΣX^2}\)
The constant ‘a’ gives the mean of Y and ‘b gives the rate of change (slope),
(v) By substituting the values of ‘a and ‘b’ in the trend equation (1), we get the Line of Best Fit.

Question 11.
State the two normal equations used in fitting a straight line.
Solution:
The constants ‘a’ and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………. (3)
Where ‘n’ = number of years given in the data.

Question 12.
State the different methods of measuring trend.
Solution:
Following are the methods by which we can measure the trend.
(i) Freehand or Graphic Method
(ii) Method of Semi-Averages
(iii) Method of Moving Averages

Question 13.
Compute the average seasonal movement for the following series

Solution:
Computation of seasonal. Index by the method of simple averages.

Question 14.
The following figures relates to the profits of a commercial concern for 8 years

Find the trend of profits by the method of three yearly moving averages.
Solution:

Question 15.
Find the trend of production by the method of a five-yearly period of moving average for the following data:

Solution:
I Computation of five – yearly moving averages

Question 16.
The following table gives the number of small- scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.

Solution:

Question 17.
The annual production of a commodity is given as follows:

Fit a straight line trend by the method of least squares
Solution:

Therefore the required equation of the straight line
Y = a + bx
Y = 169.428 + 3.285 X
⇒ Y = 169.428 + 3.285 (x – 1998)

Question 18.
Determine the equation of a straight line which best fits the following data

Compute the trend values for all years from 2000 to 2004
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 54 + 5.4X
Y = 54 + 5.4 (x – 2002)
The trend value can be obtained by when x = 2000
Yt = 54 + 5.4 (2000 – 2002)
Y = 54+ 5.4 (-2)
= 54 – 10.8
= 43.2
When x = 2001
Yt = 54 + 5.4 (2001 – 2002)
Y = 54 + 5.4 (-1)
= 54 – 5.4
= 48.6
When x = 2002
Yt = 54 + 5.4 (2002 – 2002)
V = 54 + 5.4 (0)
= 54
When x = 2003
Yt = 54 + 5.4 (2003 – 2002)
Y = 54 + 5.4 (1)
= 54 + 5.4
= 59.4
When x = 2004
Yt = 54 + 5.4 (2004 – 2002)
Y = 54 + 5.4(2)
= 54 + 10.8
= 64.8

Question 19.
The sales of a commodity in tones varied from January 2010 to December 2010 as follows: in year 2010 Sales (in tones)

Fit a trend line by the method of semi-average.
Solution:
Since the number of years is even (twelve). We can equally divide the given data it two equal parts and obtain averages of first six months and last six.

Question 20.
Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003 and 2004.

Solution:

Question 21.
Use the method of monthly averages to find T the monthly indices for the following data of production of a commodity for the years 2002, 2003 and 2004.

Solution:
Computation of Seasonal Index by the method of simple averages

Question 22.
The following table shows the number of salesmen working for a certain concern:

Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 48.8 + 2X
(i.e) Y = 48.8 + 2 (x – 1994)
When x = 1992
Yt = 48.8 + 2 (1992 – 1994)
Y = 48.8 + 2 (-2)
= 48.8 – 4
= 44.8
When x = 1993
Yt = 48.8 + 2 (1993 – 1994)
Y = 48.8 + 2 (-1)
= 48.8 – 2
= 46.8
When x = 1994
Yt = 48.8 + 2 (1994 – 1994)
Y = 48.8 + 2(0)
= 48.8
When x = 1995
Yt = 48.8 + 2 (1995 – 1994)
Y = 48.8 + 2(1)
= 50.8
When x = 1996
Yt = 48.8 + 2 (1996 – 1994)
Y = 48.8 + 2 (2)
= 48.8 + 4
= 52.8
When x = 1997
Yt = 48.8 + 2 (1997 – 1994)
Y = 48.8 + 2 (3)
= 48.8 + 6
= 54.8

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.
Explain the types of sampling.
Solution:
There are two types of sampling. They are
(1) Non-Random sampling or Non-probability sampling
(2) Random sampling or probability sampling Random sampling refers to selection of samples from the population in a random manner. A random sample is one where each and every item in the population has an equal chance of being selected.

“Every member of a parent population has had equal chances of being included “Dr. Yates.

“A random sample is a sample selected in such a way that every item in the population has an equal chance of being included”-Harver.

The different types of probability sampling are
(1) sampling random sampling
(2) stratified random sampling
(3) systematic sampling

Question 2.
Write short note on sampling distribution and standard error.
Solution:
sampling distribution:
Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.

For instance if we draw a sample a size n from a given finite population of N, then the total number of possible samples is
Nc_n = \(\frac { N! }{n!(N-n)!}\) = k(say)

Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its standard Error abbreviated as S.E. The standard Error (S.E) of some of the well-known statistics, for large samples, are given below, where n is the samples size, σ² is the population variance.

Question 3.
Explain the procedures of testing of hypothesis
Solution:
The following are the steps involved in hypothesis testing problems:
1. Null hypothesis: Set up the null hypothesis H₀

2. Alternative hypothesis: Set up the alternative hypothesis. This will enable us to decide whether we have to use two tailed test or single tailed test (right or left tailed)

3. Level of significance: Choose the appropriate level of significant (a) depending on the reliability of the estimates and permissible risk. This is to be fixed before sample is drawn, i.e., a is fixed in advance.

4. Test statistic : Compute the test statistic
Z = \(\frac { t-E(t) }{\sqrt{var(t)}}\) = \(\frac { t-E(t) }{S.E(t)}\) N(0, 1) as n → ∞

5. Conclusion: We compare the computed value of Z in step 4 with the significant value or critical value or table value Zα at the given level of significance.
(i) If |Z | < Zα i.e., if the calculated value of is less than critical value we say it is not significant. This may due to fluctuations of sampling and sample data do not provide us sufficient evidence against the null hypothesis which may therefore be accepted.

(ii) If |Z |> Zα i.e., if the calculated value of Z is greater than critical value Zα then we say it is significant and the null hypothesis is rejected at level of significance α.

Question 4.
Explain in detail about the test of significance for single mean.
Solution:
Let xi , (i = 1, 2, 3, …, n) is a random sample of size from a normal population with mean µ and variance σ² then the sample mean is distributed normally with mean and variance
\(\frac { σ^2 }{n}\), i.e \(\bar { x }\) N(µ, \(\frac { σ^2 }{n}\))

Thus for large samples, the standard normal variate corresponding to \(\bar { x }\) is
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\) N (0, 1)

Under the null hypothesis that the sample has been drawn from a population with mean and variance σ², i.e., there is no significant difference between the sample mean (\(\bar { x }\)) and the population mean (α), the test statistic (for large samples) is:
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\)

Question 5.
Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad?
Solution:
sample size n = 500
No. of bad pine apples = 65
sample proportion = P = \(\frac { 65 }{500}\) = 0.13
Q = 1 – p ⇒ Q = 1 – 0.13
∴ Q = 0.87
The S.E for sample proportion is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.13)(0.87) }{500}}\)
= \(\sqrt{\frac { 0.1131 }{500}}\) = \(\sqrt{0.0002262}\)
= 0.01504
∴ S.E = 0.015
Hence the standard error for sample proportion is S.E = 0.015

Question 6.
A sample of 100 students are drawn from a school The mean weight and variance of the sample are 67.45 kg and 9 kg respectively find (a) 95% and (b) 66% confidence intervals for estimating the mean weight of the students.
Solution:
sample size n = 100
The sample mean = \(\bar { x }\) = 67.45
The sample variance S² = 9
The sample standard deviation S = 3
S.E = \(\frac { S }{√n}\) = \(\frac { 3 }{\sqrt{100}}\) = \(\frac { 3 }{10}\) = 0.3

(a) The 95% confidence limits for µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E < µ < \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (1.96 × 0.3) ≤ µ ≤ 67.45 + (1.96 × 0.3) 67.45 – 0.588 ≤ µ ≤ 67.45 + 0.588
66.862 ≤ µ ≤ 68.038
The confidential limits is (66.86, 68.04)

(b) The 99% confidence limits for estimating µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E ≤ µ ≤ \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (2.58 × 0.3) ≤ µ ≤ 67.45 + (2.58 × 0.3)
67.45 – 0.774 ≤ µ ≤ 67.45 + 0.774
66.676 ≤ µ ≤ 68.224
∴ The 99% confidence limits is (66.68, 68.22)

Question 7.
The mean I.Q of a sample of 1600 children was 99. it is likely that this was a random sample from a population with mean I.Q 100 and standard deviation 15? (Test at 5% level of significance)
Solution:
sample size n = 1600
\(\bar { x }\) = 99
sample mean
Population mean µ = 100
population S.D σ = 15
under the Null hypothesis H₀ : µ = 100
Alternative hypothesis H₁ : µ = 100 (two tails)
Level of significance µ = 0.05

z = -2.666
z = -2.67
Calculated value |z| = 2.67
critical value at 5% level of significance is
z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is greater than table value i.e z ⇒ z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is rejected. Therefore we conclude that the sample mean differs, significantly from the population mean.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3

Choose the correct Answer:

Question 1.
A ……………… may be finite or infinite according as the number of observation or items in it is finite or infinite
(a) Population
(b) census
(c) parameter
(d) none of these
Solution:
(a) population

Question 2.
A …………….. of statistical individuals in a population is called a sample.
(a) Infinite set
(b) finite subset
(c) finite set
(d) entire set
Solution:
(b) finite subset

Question 3.
A finite subset of statistical individuals in a population is called ………………
a) a sample
(b) a population
(c) universe
(d) census
Solution:
(a) a sample

Question 4.
Any statistical measure computed from sample data is known as ……………..
(a) Parameter
(b) random sample
(c) Infinite measure
(d) uncountable
Solution:
(b) random sample

Question 5.
A ………………. is one where each item in the universe has an equal chance of known opportu¬nity of being selected
(a) Parameter
(b) random sample
(c) statistic
(d) entire data
Solution:
(b) random sample

Question 6.
A random sample is a sample selected in such a way that every item in the population has an equal chance of being included
(a) Harper
(b) fisher
(c) karl pearson
(d) Dr. yates
Solution:
(a) Harper

Question 7.
Which one of the following is probability sampling
(a) Purposive sampling
(b) judgement sampling
(c) sample random sampling
(d) Convenience sampling
Solution:
(c) sample random sampling

Question 8.
In simple random sampling of drawing any unit, the probability of drawing any unit at the draw is ?
(a) \(\frac { n }{N}\)
(b) \(\frac { 1 }{N}\)
(c) \(\frac { N }{n}\)
(d) n
Solution:
(b) \(\frac { 1 }{N}\)

Question 9.
In ……………. the heterogeneous groups divided into homogeneous groups
(a) Non-probability sample
(b) a sample random sample
(c) a stratified random sample
(d) Systematic sample
Solution:
(c) a stratified random sample

Question 10.
Errors in sampling are of
(a) Two types
(b) three types
(c) four types
(d) five types
Solution:
(a) Two types

Question 11.
The method of obtaining the most likely value of the population parameter using statistic is called
(a) estimate
(b) estimate
(c) biased estimate
(d) standard error
Solution:
(d) standard error

Question 12.
An estimator is a sample statistic used to estimate a
(a) population parameter
(b) biased estimate
(c) sample size
(d) census
Solution:
(a) population parameter

Question 13.
…………… is a relative property, which states that one estimate is efficient relative to another.
(a) efficiency
(b) sufficiency
(c) unbiased
(d) consistency.
Solution:
(a) efficiency

Question 14.
If probability p[|\(\bar { θ }\) – θ|< ∈|< ∈|] 1 → µ as n → α for any positive then \(\bar { θ }\) is said to estimator of θ
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(d) Consistent

Question 15.
An estimator is said to be ………….. if it contains all the information in the data about the parameter it estimates.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(b) sufficient

Question 16.
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie called an ………….. interval estimate of the parameter
(a) point estimate
(b) interval estimate
(c) standard error
(d) confidence
Solution:
(b) interval estimate

Question 17.
A ……………… is a statement or an assertion about the population parameter
(a) hypothesis
(b) statistic
(c) sample
(d) census
Solution:
(a) hypothesis

Question 18.
Type I error is
(a) Accept H₀ when it is true
(b) Accept H₀ when it is false
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(c) Reject H₀ when it is true

Question 19.
Type II error is?
(a) Accept H₀ when it is wrong
(b) Accept H₀ when it is when it is true
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(a) Accept H₀ when it is wrong

Question 20.
The standard error of sample mean is?
(a) \(\frac { σ }{\sqrt{2n}}\)
(b) \(\frac { σ }{n}\)
(c) \(\frac { σ }{√n}\)
(d) \(\frac { σ^2 }{√n}\)
Solution:
(c) \(\frac { σ }{√n}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2

Question 1.
Mention two branches of statistical inference?
Solution:
(i) Estimation (ii) Testing of Hypothesis

Question 2.
What is an estimator?
Solution:
Any sample statistic which is used to estimate an unknown population parameter is called an estimator ie., an estimator is a sample statistic used to estimate a population parameter.

Question 3.
What is an estimate?
Solution:
When we observe a specific numerical value of our estimator, we call that value is an estimate. In other words, an estimate is a specific value of a statistic.

Question 4.
What is point estimation?
Solution:
When a single value as an estimate, the estimate is called a point estimate of the population parameter. In other words, an estimate of a population parameter given by a single number is called as point estimation.

Question 5.
What is interval estimation?
Solution:
Generally, there are situation where point estimation is not desirable and we are interested in finding limits within which the parameter would be expected to lie is called an interval estimation.

Question 6.
What is confidence interval?
Solution:
Let us choose a small value of a which is known as level of significance (1% or 5%) and determine two constants says c₁ and c₂ such that p(c₁ < θ < c₂|t) = 1 – α

The quantities c₁ and c₂ so determined are known as the confidence Limits and the interval [c₁, c₂] with in which the unknown value of the population parameter is expected to lie is known as confidence interval.(1 – α)is called as confidence coefficient.

Question 7.
What is null hypothesis? Give an example.
Solution:
According to prof R. A. fisher, “Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true”, and it is denoted by H₀.

For example: If we want to find the population mean has a specified value µ₀, then the null hypothesis H₀ is set as follows H₀ : µ = µ₀

Question 8.
Define alternative hypothesis.
Solution:
Any hypothesis which is complementry to the null hypothesis is called as the alternative hypothesis and is usually denoted by H₁.

For example: If we want to test the null hypothesis that the population has specified mean µ i.e., H₀ : µ = µ₀ then the alternative hypothesis could be any one among the following:
(i) H₁ : µ ≠ µ₀ (µ > or µ < µ₀)
(ii) H₁ : µ > µ₀
(iii) H₁ : µ < µ₀

Question 9.
Define critical region.
Solution:
A region corresponding to a test statistic in the sample space which tends to rejection of H₀ is called critical region or region of rejection.

Question 10.
Define critical value.
Solution:
The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depend upon.
(i) The level of significance
(ii) The alternative hypothesis whether it is two-tailed or single tailed

Question 11.
Define level of significance.
Solution:
The probability of type 1 error is known as level. of significance and it is denoted by The level of significance is usually employed in testing of hypothesis are 5% and 1%. The level of significance is always fixed in advanced before collecting the sample information.

Question 12.
What is type I error
Solution:
There is every chance that a decision regarding a null hypothesis may be correct or may not be correct. The error of rejecting H₀ when it is true is called type I error.

Question 13.
What is single tailed test.
Solution:
When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one-tailed test. Here H₁ : µ > µ₀ and H₁ : µ < µ₀ are known as one tailed alternative.

Question 14.
A sample of 100 items, draw from a universe with mean value 64 and S.D 3, has a mean value 63.5. Is the difference in the mean significant?
Solution:
sample size n = 100 ; sample mean \(\bar { x}\) = 63.5
sample SD S = 3;
population mean µ = 64 population SD σ = 3
Null Hypothesis H₀ : µ = 64 (the sample has been drawn from the population mean µ = 64 and SD σ = 3)
Alternative Hypothesis H₁ : µ ≠ 64 (two tail) i.e the sample has not been drawn from the population mean µ = 64 and SD σ = 3
The level of significance α = 5% = 0.05
Test statistic
z = \(\frac { 63.5-64}{\frac{3}{\sqrt{100}}}\) = \(\frac { -0.5 }{(\frac{3}{10})}\) = \(\frac { -0.5 }{0.3}\) = -1.667
|z| = 1.667
∴ calculated z = 1.667
critical value at 5% level of
significance is z_{\(\frac { α }{2}\)} = 1.96
Inference:
At 5% level of significance Z < Z_{\(\frac { α }{2}\)} since the calculated value is less than the table value the null hypothesis is accepted.

Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height of 67.39 inches and standard deviation 1.30 inches?
Solution:
sample size n = 400; sample mean \(\bar { x}\) = 67.47 inches
sample SD S = 1.30 inches population mean
µ = 67.39 inches
population SD σ = 1.30 inches
Null Hypothesis H₀ : µ = 67.39 inches (the sample has been drawn from the population mean µ = 67.39 inches; population SD σ = 1.30 inches)
Alternative Hypothesis H₁ = µ ≠ 67.39 inches(two tail)
i.e the sample has not been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches
The level of significance α = 5% = 0.05
Test static:

Thus the calculated and the significant value or Z_{\(\frac { α }{2}\)} = 1.96
table value comparing the calculated and table values
Z_{\(\frac { α }{2}\)} (i.e.,) 1.2308 < 1.96
Inference: since the calculated value is less than value i.e Z > Z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is accepted Hence we conclude that the data doesn’t provide us any evidence against the null hypothesis. Therefore, the sample has been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches.

Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
sample size n = 100
sample mean \(\bar { x}\) = 72
sample SD S = 8
population mean µ = 76
under the Null hypothesis H₀ : p = 76
Against the alternative hypothesis H₀ : µ ≠ 76 (two mail)
Level of significance µ = 0.05
Test statistic:

since alternative hypothesis is of two tailed test we can take |Z| = 5.
∴ critical value 5% level of significance is z > z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is less than table value i.e z > z_{\(\frac { α }{2}\)} at 5% level of significance the null hypothesis H0 is rejected Therefore, we conclude that there is significant difference between the sample mean and population mean µ = 76 and SD σ = 8.

Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance.
Solution:
Sample size n = 50
Sample mean \(\bar { x}\) = 1800
Sample SD S = 100
population mean µ = 1850
Null hypothesis H₀ : µ = 1850
Level of significance µ = 0.01

= -3.5355
∴ z = -3.536
calculated value |z| = 3.536
Critical value at 1% level of significance is z_{\(\frac { α }{2}\)} = 2.58
Inference:
Since the calculated value is greater than table (ie) z > z_{\(\frac { α }{2}\)} at 1% level of significance, the null hypothesis is rejected, therefore we conclude that we is rejected, Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1

Question 1.
What is population?
Solution:
The group of individuals considered under study is called as population. The word population here refers not only to people but to all items that have been chosen for the study.

Question 2.
What is sample?
Solution:
A selection of a group of observation/individuals/population in such a way that is represents the population is called as sample.

Question 3.
What is statistic?
Solution:
Statistic: Any statistical measure computed from sample is known as statistic.

Question 4.
Define parameter.
Solution:
Parameter: The statistical constants of the population like mean (µ),variance (σ²) are referred as population parameters.

Question 5.
What is sampling distribution of a statistic?
Solution:
Sampling distribution of a statistic is the frequency distribution which is formed with various of a statistic computed from different samples of the same size drawn from the same population.

Question 6.
What is standard error?
Solution:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error abbreviated as S.E.

Question 7.
Explain in detail about simple random sampling with a suitable example.
Solution:
In this technique the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. Simple random sampling may be done, with or without replacement of a samples selected. In a simple random sampling with replacement there is a possibility of selecting the same sample any number of times. So, simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units ,the probability of drawing any unit at the first draw is \(\frac { 1 }{N}\), the probability of drawing any unit in the second draw from among the available (N – 1) units is \(\frac { 1 }{(N-1)}\), and so on.

For example, if we want to select 10 students, out of 100 students, then we must write the names/roll number of all the 100 students on slips of the same size and mix them, then we make a blindfold selection of 10 students. This method is called unrestricted random sampling, because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population or universe is infinite, this method is inapplicable.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
In stratified random sampling, first divide the population into sub-populations, which are called strata. Then,the samples are selected from each of the strata through random techniques. The collection of all the samples from all strata gives the stratified random samples.

When the population is heterogeneous or different segments or groups with respect to the variable or characteristic under study, then stratified Random sampling methos is studied. First, the population is divided into homogeneous number of sun-groups of strata before the sample is drawn. A sample from each stratum at random. Following steps are involved for selecting random sample in a stratified random sampling method.

(a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.

(b) After the population is stratified,a sample of a specified size is drawn at random from each stratum using Lottery Method or table of random number method.

Question 9.
Explain in detail about systematic random sampling with example.
Solution:
In a systematic sampling, randomly select the first sample from the first k units. Then every k th members, starting with the first selected sample, is included in the sample.

Systematic sampling is a commonly used technique,if the complete and up-to-date list of the sampling units is available. We can arrange the items in numerical, alphabetical, selecting the first at random, the rest being automatically selected according to some pre-determined pattern. A systematic is formed by selecting every item from the population, where k refers to the sample interval. The sampling interval can be determined by divided the size of the population by the size of the sample to be chosen.

That is K = \(\frac { N }{n}\), where k is an integer.
k = sampling interval, size of the population, sample size
Procedure for selection of samples by systematic sampling method

(i) If we want to select a sample of 10 students from a class of 100 students,the sampling interval is Calculated as k = \(\frac { N }{n}\) = \(\frac { 100 }{10}\) = 10
Thus sampling interval = 10 denotes that for every 10 samples one sample

(ii) The first sample is selected from the first 10(sampling interval) samples through selection procedures.

(iii) If the selected first random sample is 5, then the rest of the samples are automatically selected by incriminating the value of the sampling interval 9k = 10. i.e, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

Question 10.
Explain in detail about sampling error.
Solution:
Sampling Error:
Error, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors, sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice. Sampling Errors arise primarily due to the following reasons:
(a) Faulty selection of the sample instead of correct sample by defective sampling technique.
(b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
(c) In area surveys,while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as faulty demarcation of sampling units.

Question 11.
Explain in detail about non-sampling error.
Solution:
Non-sampling Errors:
The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments (tape, scale) are called Non¬sampling errors. It may arise in the following ways:
(a) Due to neglience and carelessness of the part of either investigator or respondents
(b) Due to lack of trained and qualified investigators.
(c) Due to framing of a wrong questionnaire.
(d) Due to apply wrong statistical measure.
(e) Due to incomplete investigation and sample survey.

Question 12.
State any two merits of simple random sampling.
Solution:
Merits:
1. Personal bias is completely eliminated.
2. This method is economical as it saves time, money and labour.
3. The method requires minimum knowledge about the population in advance.

Question 13.
State any three merits of stratified random sampling.
Solution:
Merits:
(a) A random stratified sample is superior to a sample random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy
(c) it is easy to administer, if the population under study is sub divided
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had the geographical areas.

Question 14.
State any two demerits of systematic random sampling.
Solution:
Demerits:
1. Systematic samples are not random samples.
2. If N is not multiple of n-then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

Question 15.
State any two merits for systematic random sampling.
Solution:
Merits:
1. This is simple and convenient method.
2. This method distributes the sample more evenly over the entire listed population.
3. The time and work is reduced much.

Question 16.
Using the following Tippet’s random number table

Draw a sample of 10 three digit numbers which are even numbers.
Solution:
There are many ways to select 10 random samples from the given Tippets random number number table since the population size is three digit numbers, Here the door numbers must be even (ie) the unit digit must be even. Here we consider column wise selection of random numbers starting from first column.

So the first sample is 416 and other 9 samples are 056, 664, 952, 748, 524, 914, 154, 340 and 140.
Tippets random number Table

Question 17.
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning of good apples.
Solution:
sample size = 600; Number of success = 600 – 36
= 564
sample proportion p = \(\frac { 564 }{600}\) = 0.94
600
population proportion (p) = probability of getting good apple
= 96%
= \(\frac { 96 }{100}\) {∵ 4% of the apples 100 are defective}
P = 0.96
Q = 1 – p = 1 – 0.96
Q = 0.04
The S.E for a sample proporation is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.96)(0.04) }{600}}\)
\(\sqrt{\frac { 0.0384 }{600}}\) = \(\sqrt{0.000064}\)
∴ S.E = 0.008
Hence the standard error foe sample proportion is S.E = 0.008

Question 18.
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate standard error of mean.
Solution:
Given n = 1000; \(\bar{x}\) = 119 lbs (pounds)
s = 30 lbs is known in this problem.
since σ is unknown, so we consider \(\bar{σ}\) = s and µ = 120 lbs
S.E = \(\frac { \bar{σ} }{√n}\) = \(\frac { s }{√n}\) = \(\frac { 30 }{\sqrt{1000}}\)
= \(\frac { 30 }{31.623}\) = 0.9487
Therefore the standard error for the average weight of large group of students of 120 lbs is 0.9487

Question 19.
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Solution:
Sample size n = 60
Sample S.D S = 2.5
population S.D a = 3
The standard error for sample S.D is given by
\(\sqrt{\frac { σ^2 }{2n}}\) = \(\sqrt{\frac { (3)^2 }{2(60)}}\) = \(\frac { 3 }{\sqrt{120}}\)
= \(\frac { 3 }{10.954}\) = 0.27387
= 0.2739
Thus standard error for sample S.D = 0.2739

Question 20.
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village?
Solution:
sample size n = 400
case (i):
sample proporation of vegetarian p = \(\frac { 3 }{10.954}\) = \(\frac { 230 }{400}\)
p = 0.575
q = 1 – p
= 1 – 0.575
q = 0.425
Sample error S.E= \(\sqrt{\frac { pq }{n}}\)
= \(\sqrt{\frac { 0.575×0.425 }{400}}\) = \(\sqrt{\frac { 0.223125 }{400}}\)
\(\sqrt{0.0005578125}\)
S.E = 0.2361

Case(ii):
sample size n = 400
since both vegetarian and non- vegetarian foods are equally popular in that village
sample proparation of vegetarian p = \(\frac { 1 }{2}\) = 0.5
q = 1 -p ⇒ q = 1 – 0.5
q = 0.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.4

Choose the correct answer

Question 1.
Normal distribution was invented by
(a) Laplace
(b) De-Moivre
(c) Gauss
(d) all the above
Solution:
(b) Demoivre

Question 2.
If X ~ N(9, 81) the standard normal variate Z will be
(a) Z = \(\frac { X-81 }{9}\)
(b) Z = \(\frac { X-9 }{81}\)
(c) Z = \(\frac { X-9 }{9}\)
(d) Z = \(\frac { 9-X }{9}\)
Solution:
(b) Z = \(\frac { X-9 }{81}\)
Hint:
Here µ = 9
σ² = 81
∴ σ = 9
Z = \(\frac { X-µ }{σ}\) = \(\frac { X-9 }{9}\)

Question 3.
If Z is a standard normal variate, the proportion of items lying between Z = -0.5 and Z = -3.0 is
(a) 0.4987
(b) 0.1915
(c) 0.3072
(d) 0.3098
Solution:
(c) 0.3072
Hint:

p(-3.0 < z < -0.5)
= p(0.5 < z < 3.0) – p(0 < z < o.5)
= 0.4987 – 0.1915 = 0.3072

Question 4.
If X ~ N(µ, σ2), the maximum probability at the point of inflexion of normal distribution
(a) (\(\frac { 1 }{\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}
(b) (\(\frac { 1 }{\sqrt{2π}}\))e^{\(\frac { 1 }{2}\)}
(c) (\(\frac { 1 }{σ\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}
(d) (\(\frac { 1 }{\sqrt{2π}}\))
Solution:
(c) (\(\frac { 1 }{σ\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}

Question 5.
In a parametric distribution the mean is equal to variance is
(a) binomial
(b) normal
(c) poisson
(d) all the above
Solution:
(c) poisson

Question 6.
In turning out certain toys in a manufacturing company, the average number of defectives is 1%. The probability that the sample of 100 toys there will be 3 defectives is
(a) 0.0613
(b) 0.613
(c) 0.00613
(d) 0.3913
Solution:
(a) 0.0613
Hint:
Given
p = 0.01 and n = 100
λ = np = 0.01 × 100 = 1
p(x = x) = \(\frac { e^{-λ}λ^x }{x!}\)
p(x = x) = \(\frac { e^{-1}(1)^3 }{3!}\) = \(\frac { 0.3678 }{6}\) = 0.0613

Question 7.
The parameters of the normal distribution
f(x) (\(\frac { 1 }{\sqrt{72π}}\)) \(\frac { e^{-(x-10)^2} }{72}\) – ∞ < x < ∞
(a) (10, 6)
(b) (10, 36)
(c) (6, 10)
(d) (36, 10)
Solution:
(b) (10, 36)
Hint:

Here σ = 6 and µ = 10
∴ σ = (6)² = 36

Question 8.
A manufacturer produces switches and experiences that 2 per cent switches are defective. The probability that in a box of 50 switches, there are at most two defective is:
(a) 2.5 e^-1
(b) e^-1
(c) 2 e^-1
(d) none of the above
Solution:
(a) 2.5 e^-1
Hint:

Question 9.
An experiment succeeds twice as often as it fails. Die chance that in the next six trials, there shall be at least four successes is
(a) 240/729
(b) 489/729
(c) 496/729
(d) 251/729
Solution:
(c) 496/729
Hint:
p = 2 q ⇒ p = 2(1 – p)
p (2 – 2p) ⇒ 3p = 2
p = 2/3 and q = 1 – p ⇒ q = 1 – 2/3
q = 1/3 and n = 6
p(X = x) = nc_xp^xq^n-x
p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

Question 10.
If for a binomial distribution b(n, p) mean = 4 and variance = 4/3, the probability, P(X ≥ 5) is equal to:
(a) (2/3)⁶
(b) (2/3)⁵( 1/3)
(c) (1/3)⁶
(d) 4(2/3)⁶
Solution:
(d) 4(2/3)⁶
Hint:
In a binomial distribution
mean np = 4 → (1)
variance npq = 4/3 → (2)
(2) ÷ (1) ⇒ npq = 4/3 ⇒ q = 1/3
p = (1 – q) ⇒ p = 1 – 1/3
∴ p = 2/3
p(X = x) = nc_rp^xq^n-x
p(x ≥ 5) = p(x = 5) + p(x = 6)

Question 11.
The average percentage of failure in a certain examination is 40. The probability that out of a group of 6 candidates atleast 4 passed in the examination are:
(a) 0.5443
(b) 0.4543
(c) 0.5543
(d) 0.4573
Solution:
(b) 0.4543
Hint:
given:
n = 6
q = 40/100 = 2/5
p = 1 – q = 1 – 2/5 = 3/5
p(X = x) = nc_xp^xq^n-x
p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

= \(\frac { 8505 }{5^6}\) = \(\frac { 1701 }{3125}\)
= 0.54432

Question 12.
Forty percent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage?
(a) 6.00
(b) 6.45
(c) 7.20
(d) 7.50
Solution:
(a) 6.00
Hint:
Given
P = \(\frac { 40 }{100}\) and n = 15
Mean np = \(\frac { 40 }{100}\) × 15
= 0.4 × 15
= 6.00

Question 13.
Which of the following statements is/are true regarding the normal distribution curve?
(a) it is symmetrical and bell shaped curve
(b) it is asymptotic in that each end approaches the horizontal axis but never reaches it
(c) its mean, median and mode are located at the same point
(d) all of the above statements are true.
Solution:
(d) all of these above statements are true

Question 14.
Which of the following cannot generate a Poisson distribution?
(a) The number of telephone calls received in a ten-minute interval
(b) The number of customers arriving at a petrol station
(c) The number of bacteria found in a cubic feet of soil
(d) The number of misprints per page
Solution:
(b) The number of customers arriving a petrol station

Question 15.
The random variable X is normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that X is between 72 and 84?
(a) 0.683
(b) 0.954
(c) 0.271
(d) 0.340
Solution:
(d) 0.340
Hint:
In a normal distribution
S.D(σ) = 10
mean(µ) = 70 and

= p(0.2 < z < 1.4) – p(0 < z < 0.2)
= 0.4192 – 0.0793
= 0.3399

Question 16.
The starting annual salaries of newly qualified chartered accountants (CA’s) in South Africa follow a normal distribution with a mean of Rs 180,000 and a standard deviation of Rs 10,000. What is the probability that a randomly selected newly qualified CA will earn between Rs 165,000 and Rs 175,000 per annum?
(a) 0.819
(b) 0.242
(c) 0.286
(d) 0.533
Solution:
(b) 0.242
Hint:
In a normal distribution
µ = 180,000 and σ = 10,000
z = \(\frac { X-µ }{σ}\) = \(\frac { X-180000 }{10000}\)
p(165,000 < x < 1,75,000) = ?
when x = 165,000
z = \(\frac { 165000-180000 }{10000}\) = \(\frac { 15000 }{10000}\)
z = -1.5

when x = 175,000
z = \(\frac { 175000-180000 }{10000}\) = \(\frac { -5000 }{10000}\) = \(\frac { -1 }{2}\)
z = -0.5
p(165,000 < x < 175,000)
= p(-1.5 < z < -0.5)
= p(0.5 < z < 1.5)
= p(0 < z < 1.5) – p(0 < z < 0.5)
= 0.4332 – 0.1915 = 0.2417

Question 17.
In a large statistics class the heights of the students are normally distributed with a mean of 172 cm and a variance of 25 cm. What proportion of students are between 165 cm and 181 cm in height?
(a) 0.954
(b) 0.601
(c) 0.718
(d) 0.883
Solution:
(d) 0.883
Hint:

In a normal distribution
µ = 172; σ² = 25 then σ = 5
z = \(\frac { x-µ }{σ}\) = \(\frac { x-172 }{5}\)
p(165 < x < 181) = ?
when x = 165 z = \(\frac { 165-172 }{5}\) = \(\frac { -7 }{5}\) = -1.4
when x = 181 z = \(\frac { 181-172 }{5}\) = \(\frac { 9}{5}\) = 1.8
p(165 < x < 181) = p(-1.4 < z < 1.8)
= p(-1.4 < z < 0) + p(0 < z < 1.8)
= p(0 < z < 1.4) + p(0 < z < 1.8)
= 0.4192 + 0.4641
= 0.8833

Question 18.
A statistical analysis of long-distance” telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What proportion of calls lasts less than 180 seconds?
(a) 0.214
(b) 0.094
(c) 0933
(d) 0.067
Solution:
(d) 0.067
Hint:
In a normal distribution

µ = 240 and σ = 40

Question 19.
Cape town is estimated to have 21% of homes whose owners subscribe to the satelite service, DSTV. If a random sample of your home in taken, what is the probability that all four home subscribe to DSTV?
(a) 0.2100
(b) 0.5000
(c) 0.8791
(d) 0.0019
Solution:
(d) 0.0019
Hint:
p = \(\frac { 21 }{100}\) = 0.21
p(x = 4) = (0.21) × (0.21) × (0.21) × (0.21)
= 0.00194481

Question 20.
Using the standard normal table, the sum of the probabilities to the right of z = 2.18 and to the left of z = -1.75 is:
(a) 0.4854
(b) 0.4599
(c) 0.0146
(d) 0.0547
Solution:
(d) 0.0547
Hint:

p(z < – 1.75) + p(z > 2.18)
= p(-∞ < z < 0) – p(-1.75 < z < 0) + p(0 < z < ∞) – p(0 < z < 2.19)
= 0.5 – p(0 < z < 1.75) + 0.5 – p(0 < z < 2.18)
= (0.5 – 0.4599) + (0.5 – 0.4854)
= 0.0401 + 0.0146
= 0.0547

Question 21.
The time until first failure of a brand of inkjet printers is normally distributed with a mean of 1,500 hours and a standard deviation of 200 hours. What proportion of printers fails before 1000 hours?
(a) 0.0062
(b) 0.0668
(c) 0.8413
(d) 0.0228
Solution:
(a) 0.0062
Hint:

Question 22.
The weights of newborn human babies are normally distributed with a mean of 3.2 kg and a standard deviation of 1.1 kg. What is the probability that a randomly selected newborn baby weighs less than 2.0 kg?
(a) 0.138
(b) 0.428
(c) 0.766
(d) 0.262
Solution:
(a) 0.138
Hint:

Question 23.
Monthly expenditure on their credit cards, by credit card holders from a certain bank, follows a normal distribution with a mean of Rs 1,295.00 and a standard deviation of Rs 750.00. What proportion of credit card holders spend more than Rs 1,500.00 on their credit cards per month?
(a) 0.487
(b) 0.392
(c) 0.500
(d) 0.791
Solution:
(b) 0.392
Hint:
In a normal distribution

mean m = 1295 and S.D σ = 7.50
p(x > 1500) = ?
when x = 1500
z = \(\frac { 1500-1295 }{750}\) = \(\frac { 205 }{750}\) = 0.273
(ie) p(x > 1500) = p(z > 0.273)
= 0.5 – p(0 < z < 0.273)
= 0.5 – 0.1064
= 0.3936

Question 24.
Let z be a standard normal variable. If the area to the right of z is 0.8413, then the value of z must be:
(a) 1.00
(b) -1.00
(c) 0.00
(d) -0.41
Solution:
(b) -1.00
Hint:

p(-c < z < ∞) = 0.8413
p(-c < z < 0) + 0.5 = 0.8413
p(-c < z < 0) = 0.813 – 0.5 = 0.3413
p(0 < z < c) = 0.3413 ⇒ c = 1.00
∴ -c = -1.00

Question 25.
If the area to the left of a value of z (z has a standard normal distribution) is 0.0793, what is the value of z?
(a) -1.41
(b) 1.41
(c) -2.25
(d) 2.25
Solution:
(a) -1.41
Hint:

p(z < -c) = 0.0793 ie p(z > c) = 0.0793
p(0 < z < ∞) – p(0 < z < c) = 0.0793
0. 5 – p(0 < z < c) = 0.0793
p(0 < z < c) = 0.5 – 0.0793
p(0 < z < c) = 0.4207
c = 1.41
then -c = -1.41

Question 26.
If P(Z > z) = 0.8508 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) -1.04
(d) 1.04
Solution:
(c) -1.04
Hint:
since the area is greater than 0.5 then the z must be negative

p(z < Z) = 0.8508
p(-z < Z < 0) + p(0 < z < ∞) = 0.8508
p(-z < Z < 0) + 0.5 = 0.8508
p(-z < Z < 0) = 0.8508 – 0.5 = 0.3508
p(0 < Z < z) = 0.3508
z = 1.04 then -z = -1.04

Question 27.
If P(Z > z) = 0.5832 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) 1.04
(d) -0.21
Solution:
(d) -0.21
Hint:
p(z > Z) = 0.5832
since area >0.5 then then z must be negative

p(- z < Z < ∞) = 0.5832
p(-z < Z < 0) + p(0 < Z < ∞) = 0.5832
p(-z < Z < 0) + 0.5 = 0.5832
p(-z < Z < 0) = 0.0832
z = 0.21
then -z = -0.21

Question 28.
In a binomial distribution, the probability of success is twice as that of failure. Then out of 4 trials, the probability of no success is
(a) 16/81
(b) 1/16
(c) 2/27
(d) 1/81
Solution:
(d) 1/81
Hint:
In a binomial distribution
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ 3p = 2 ⇒ p = 2/ 3
q = 1 – p = 1 – 2/3
∴ q = 1/3
p(X = x) = nc_xp^xq^n-x = 4c_x (\(\frac { 2 }{3}\))^x (\(\frac { 1 }{3}\))^4-x
p(X = 0) = 4c₀ (\(\frac { 2 }{3}\))⁰ (\(\frac { 1 }{3}\))^4-0
= (1)(1)(\(\frac { 1 }{3}\))⁴ = \(\frac { 1 }{81}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Miscellaneous Problems

Question 1.
A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain
(a) no more than 2 rejects?
(b) at least 2 rejects?
Solution:
In a binomial distribution
n = 10; p = \(\frac { 12 }{100}\) = \(\frac { 3 }{25}\); q = 1 – p = 1 – \(\frac { 3 }{25}\); q = \(\frac { 22 }{25}\)
p(X = x) = nc_xp^xq^n-x
(a)p(no more than 2 rejects)
p(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

(b) p (at least 2 rejects) = p (x ≥ 2)

Question 2.
Hospital records show that of patients suffering from a certain disease 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?
Solution:
Let x be the random variable of patience suffering from a certain disease
In a binomial distribution
p (X = x) = nc_xp^xq^n-x

Question 3.
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
Solution:
In a poisson distribution
mean(λ) = \(\frac { 3 }{20}\) = 0.15
p(X = x) = \(\frac { e^{-λ}λ^x }{x!}\)
p(not be more than one failure) = p(x ≤ 1)
p(x = 0) + p(x = 1)

= 0.86074 × (1.15)
= 0.98981

Question 4.
Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
1. Find the probability that none passes in a given minute.
2. What is the expected number passing in two minutes?
Solution:
In a poisson distribution
Average per hour = 300 vehicles
mean per minute = \(\frac { 300 }{60}\) = 5
∴ λ = 5

= e^-5(12.5)
= 0.0067379 × 12.5
= 0.08422375
= 0.08422375 × 10²

Question 5.
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Raghul wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Raghul takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x denotes the scores of a national test mean
µ = 500 and standard deviation σ = 100
standard normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-5000 }{100}\)
when x = 585
z = \(\frac { 585-500 }{100}\) = \(\frac { 85 }{100}\) = 0.85
p(x ≤ 585) = p(z ≤ 0.85)
p(z ≤ 0.85) = p(-∞ < z < 0) + p(0 < z < 0.85)
= 0.5 + 0.3023
= 0.8023
for n = 100;
p(z ≤ 0.85) = 100 × 0.8023
= 80.23
∴ Raehul scores 80.23%

We can determine the scores of 70% of the students as follows:
from the table for the area 0.35
We get z₁ = -1.4(as z₁ lies to left of z = 0)
similarly z₂ = 1.4

Now z₁ = \(\frac { x_1-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
-1.4 × 100 = x₁ – 500 ⇒ x₁ 500 – 140
x₁ = 360
Again z₂ = \(\frac { x_2-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
1.4 × 100 = x₂ – 500 ⇒ x₂ = 140 + 500
= x₂ = 640
Hence 70% of students score between 360 and 640
But Raghul scored 585. His score is not better than the score of 70% of the students.
∴ He will not be admitted to the university.

Question 6.
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time.
(i) less than 19.5 hours?
(ii) between 20 and 22 hours?
Solution:
Let x denotes the time taken to assemable cars mean µ = 20 hours and S.D σ = 2 hours
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-20 }{2}\)
(i) p(less than 19.5 hours) = p(x < 19.5)
when x = 19.5
z = \(\frac { 19.5 }{2}\) = \(\frac { -0.5 }{2}\) = 0.25
p(x < 19.5) = p(z < – 0.25)
= p(-∞ < z < 0) – p(-0.25 < z < 0)
= 0.5 – p(0 < z < 0.25)
= 0.5 – 0.0987
= 0.4013

(ii) p(between 20 and 22 hours) = p(20 < x < 22)

when x = 20
z = \(\frac { 20-20 }{2}\) = \(\frac { 0 }{2}\) = 0
when x = 22;
z = \(\frac { 22-20 }{2}\) = \(\frac { 2 }{2}\) = 1
p(20 < x < 22) = p(0 < z < 1)
= 0.3413

Question 7.
The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000.
(a) What percent of people earn less than $40,000?
(b) What percent of people earn between $45,000 and $65,000?
(c) What percent of people earn more than $75,000
Solution:
Let x denotes the annual salaries of employees in a large company
mean µ = 50,000 and S.D σ = 20,000
Standard normal variate z = \(\frac { x-µ }{σ}\)
(a) p(people earn less than $40,000) = p(x < 40,000)
when x = 40,000
z = \(\frac { 40,000-50,000 }{20,000}\) = \(\frac { 10,000 }{20,000}\)
z = -0.5
p(x < 40,000) = p(z < -0.5)
= p(-∞ < z < 0) – p(-0.5 < z < 0)
= 0.5 -p(-0.5 < z <0)
= 0.5 – p(0 < z < 0.5) (due to symmetry)
= 0.5 – 0.01915
= 0.3085
= p(x < 40,000) in percentage = 0.3085 × 100 = 30.85

(b) p(people ear between $45,000 and $65,000)
p(45000 < x < 65000)
When x = 45,000;
z = \(\frac { 45,000-50,000 }{20,000}\) = \(\frac { -5000 }{20,000}\) = \(\frac { -1 }{4}\)
z = -0.25
when x = 65,000;
z = \(\frac { 65,000-50,000 }{20,000}\) = \(\frac { 15000 }{20,000}\) = \(\frac { 3 }{4}\)
z = 0.75
p(45000 < x < 65000) = p(-0.25 < z < 0.75)
= p(-0.25 < z < 0) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= 0.0987 + 0.2734 = 0.3721
p(45000 < x < 65000) in percentage = 0.3721 × 100
= 37.21

p(people earn more than$75,000) = p(x > 70000)
when x = 75,000;
z = \(\frac { 75,000-50,000 }{20,000}\) = \(\frac { 25000 }{20,000}\) = \(\frac { 5 }{4}\)
z = 1.25
p(x > 75,000) = p(x > 1.25)
= p(0 < z < ∞) – p(0 < z < 1.25) = 0.5 – 0.3944 = 0.1056 p(x > 750,000)in percent = 01056 × 100
= 10.56

Question 8.
X is a normally normally distributed variable with mean µ = 30 and standard deviation σ = 4. Find
(a) P(x < 40) (b) P(x > 21)
(c) P(30 < x < 35)
Solution:
x is a normally distributed variable with mean µ = 30 and standard deviation σ = 4

Then the normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-30 }{4}\)

(a) p(x < 40) = ?
when x = 40;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = 2.5
p(x < 40) = p(z < 2.5)
= p(-∞ < z < 0) + p(0 < z < 2.5) = 0.5 + 0.4938 = 0.9938 (b) p(x > 21) = ?
when x = 21;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = -2.25
p(x > 21) = p(z > -2.25)
= p(-2.25 < z < 0) + p(0 < z < ∞)
= p(0 < z < 2.25) + 0.5
= 0.4878 + 0.5
= 0.9878

(c) p(30 < x < 35) = ?
when x = 30;
z = \(\frac { 30-30 }{4}\) = \(\frac { 0 }{4}\) = 0
when x = 35;
z = \(\frac { 35-30 }{4}\) = \(\frac { 5 }{4}\) = 1.25
p(30 < x < 35) = p(0 < z < 1.25)
= 0.3944

Question 9.
The birth weight of babies is Normally distributed with mean 3,500 g and standard deviation 500 g. What is the probability that a baby is born that weighs less than 3,100 g?
Solution:
Let x be a normally distributed variable with mean 3,500 g and standard deviation 500 g

Here µ = 3500 and σ = 500
The standard normal variate z = \(\frac { x-µ }{σ}\)
p(weight less than variate 3100 g) = p(x < 3100)
when x = 3100;
z = \(\frac { 3100-3500 }{500}\) = \(\frac { -400 }{500}\) = \(\frac { -4 }{5}\)
z = -0.8
∴ p(z < 3100) = p(z < -0.8)
= p(-∞ < z < 0) – p(-0.8 < z < 0)
= 0.5 – p(0 < z < 0.8)
= 0.5 – 0.2881
= 0.2119

Question 10.
People’s monthly electric bills in chennai are normally distributed with a mean of Rs 225 and a standard deviation of Rs 55. Those people spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is Rs 100 or less?
Solution:
Let X be a normally distributed variable with a mean of Rs 225 and a standard deviation of Rs 55

Here µ = 225 and σ = 55
The standard normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-225 }{55}\)
p(a bill have Rs 100 or less) = p(x ≤ 100)
when x = 100;
z = \(\frac { 100-225 }{55}\) = \(\frac { -125 }{55}\) = -2.27
p(x ≤ 100) = p(z < -2.27)
p(z < -2.27) = p(-∞ < z < 0) – p(-2.27 < z < 0)
= 0.5 – p(0 < z < 2.27)
= 0.5 – 04884
= 0.0116

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3

Question 1.
Define normal distribution.
Solution:
A random variable X is said to follow a normal distribution with parameters mean µ and varaince σ², if its probability density function is given by

Question 2.
Define standard normal variate.
Solution:
A random variable Z = (X – µ)/σ follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e Z – N (0, 1). Its Probability density function is given by:
φ(z) = \(\frac { 1 }{\sqrt {2π}}\) e^-x²/2 -∞ < z < ∞

Question 3.
Write down the conditions in which the normal distribution is a limiting case of binomial distribution.
Solution:
The normal distribution of a variable when represented graphically, takes the shape of a symmetrical curve, known as the Normal Curve. The curve is asymptotic to x-axis on its either side.

Question 4. m
Write down any five characteristics of normal probability curve.
Solution:
Chief Characteristics or Properties of Normal Probability distribution and Normal probability Curve.
The normal probability curve with mean µ and standard deviation σ has the following properties:
(i) the curve is bell-shaped and symmetrical about the line x = u.
(ii) Mean, median and mode of the distribution coincide.
(iii) x-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (x) axis)
(iv) No portion of the curve lies below the x-axis as f(x) being the probability function can never be negative.
(v) The points of inflexion of the curve are x = µ ± σ

Question 5.
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for
(i) more than 2,150 hours
(ii) less than 1,950 hours
(iii) more 1,920 hours but less than 2,100 hours.
Solution:
Let x denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.
Here m = 2040; σ = 60 and N = 2000
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-2040 }{60}\)
(i) p(morethan 2,150 hours)
p(x > 2150)
when x = 2150
z = \(\frac { 2150-2040 }{60}\) = \(\frac { 110 }{60}\)= 1.833
p(x > 2150) = p(z > 1.833)
= p(0 < z < ∞) – p(0 < z < 1.833)
= 0.5 – 0.4664
= 0.0336
∴ Number of bulbs whose burning time is more than 2150 hours=0.0336 × 2000
= 67.2 = 67(approximately)

(ii) p(less than 1950 hours)
p(x < 1950)
when x = 1950
z = \(\frac { 1950-3040 }{60}\) = \(\frac { -90 }{60}\)= -1.5
p(x < 1950) = p(z < -1.5) = p(z > 1.5)
= 0.5 – 0.4332
= 0.068

Numbers of bulbs whose burning time is less than
1950 = 0.0668 × 2000 = 133.6
= 134 (approximately)

(iii) p(more 1,920 hours but less than 2,100 hours)
= p(1920 < x < 2100)
when x = 1950
z = \(\frac { 1920-2040 }{60}\) = \(\frac { -120 }{60}\)= -2
when x = 2100
z = \(\frac { 2100-2040 }{60}\) = \(\frac { -60 }{60}\)= -2
∴ p(1920 < x < 2040) = p(-2 < z < 1)
= p(0 < z < 2) + p(0 < z < 1)
= 0.4772 + 0.3413
= 0.8185

∴ Number of bulbs whose burning time more than 1920 hours but less than 2100 hours) = 0.8185 × 2000
= 1637

Question 6.
In a distribution 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.
Solution:

z = \(\frac { x-µ }{σ}\)
Given that
p(x < 50) = 0.3 p(x > 86) = 0.1
p(z < -c) = 0.3
p(-c < z < 0) = 0.5 – 0.3
p(-c < z < 0) = 0.2 {from the table}
p(0 < z < c) = 0.2
c = 0.53
then -c = -0.53
∴ \(\frac { 50-µ }{σ}\) = -0.53
50 – µ = -0.53σ
µ – 0.53σ = 50 → 1
p(x < 50) = 0.1
p(0 < z < ∞) = -p(0 < z < c₁) = 0.1
p(0 < z < ∞) = p(0 < z < c₁) + 0.1
0.5 = p(0 < z < c₁) + 0.1
p(0 < z < c₁) = 0.5 – 0.1
p = (0 < z < c₁) = 0.4
c₁ = 1.29
∴ \(\frac { 86-µ }{σ}\) = 1.29
86 – µ = 1.29 σ
µ + 1.29σ = 86 → 2
solving eqn 1 & 2
eqn 2 ⇒ m + 1.29σ = 86
eqn 1 ⇒ m + 0.53σ = 50
– + – ………….
………… 1.82 ………… σ = 36 ………….
σ = \(\frac { 36 }{1.82}\) ∴ = 19.78
Substitute σ = 19.78 in eqn 1
µ – 0.53(19.78) = 50
µ -10.48 = 50
µ = 50 + 10.48
µ = 60.48
Mean = 60.48 and standard deviation = 19.78

Question 7.
X is normally distributed with mean 12 and sd 4. Find P (x ≤) 20 and P (0 ≤ x ≤ 12)
Solution:
x is normally distribution with mean 12 and sd 4
∴ µ = 12 and σ = 4
Standard normal variable
z = \(\frac { x-µ }{σ}\) = \(\frac { x-12 }{4}\)
(i) p(x ≤ 20)

when x = 20
z = \(\frac { 20-12 }{4}\) = \(\frac { 8 }{4}\) = 2
v(x ≤ 20) = \(\frac { 8 }{4}\) = 2
p(x ≤ 20) = p(z ≤ 2)
= 0.5 + p(0 < z < 2)
= 0.5 + 0.4772
= 0.9772

(ii) p(0 < x < 12 )

when x = 0
z = \(\frac { 0-12 }{4}\) = \(\frac { -12 }{4}\) = -3
when x = 12
z = \(\frac { 12-12 }{4}\) = \(\frac { 0 }{4}\) = 0
p(0 ≤ x ≤ 12) = p(-3 ≤ z ≤ 0)
= p(0 ≤ z ≤ 3)
= 0.4987

Question 8.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height
(a) greater than 2 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
let x denote the height of a student N = 500; m = 68.0 inches and σ = 3.0 inches the standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-68 }{3}\)
a(greater than 72 inches)
p = p(x > 72)
when x = 72
z = \(\frac { 72-68 }{3}\) = \(\frac { 4 }{3}\) = 1.33
p(x > 72) = p(z > 1.33)
= 0.5 – 0.4082
= 0.0918

Number of students whose height are greater than 72 inches
= 0.0918 × 500
= 45.9
= 46 (approximately)

(b) p(less than or equal to 64 inches)
p(x ≤ 64)
when x = 64
z = \(\frac { 64-68 }{3}\) = \(\frac { -4 }{3}\) = -1.33
p(x ≤ 64) = p(z ≤ -1.33)
p(z ≥ -1.33)
= 0.5 – 0.4082
= 0.0918

∴ Number of heights whose ate less than or equal to 64 inches =0.0918 × 500
= 45.9
= 46 (approximately)

(c) p(between 65 and 71 inches)
p(65 ≤ x ≤ 71)
when x = 65

z = \(\frac { 65-68 }{3}\) = \(\frac { -3 }{3}\) = -1
when x = 71
z = \(\frac { 71-68 }{3}\) = \(\frac { 3 }{3}\) = 1
p(65 ≤ x ≤ 71) = p(-1 < z < 1)
= p(-1 < z < 0) + p(0 < z < 1)
= p(0 < z < 1) + p(0 < z < 1)
= 2 × [p(0 < z < 1)]
= 2 × 0.3413
= 0.6826
∴ Number of students whose height between 65 and 7 inches = 0.6826 × 500
= 341.3
= 342 (approximately)

Question 9.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints.
Solution:
let x be the random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second

µ = 16,28 and σ = 0.12
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-16.28 }{0.12}\) = 1
p(less than 16.35 seconds) = p(x < 16.35)
when x = 16.35
z = \(\frac { 16.35-16.28 }{0.12}\) = \(\frac { 0.07 }{0.12}\) = 0.583
p(x< 16.35) = p(z < 0.583)
= p(—∞ < z < 0) + p(0 < z < 0.583)
= 0.5 + 0.2190
= 0.7190

Question 10.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height (a) greater than 2 inches (b) less than or equal to 64 inches (c) between 65 and 71 inches.
Solution:
Let x be a normal variate with mean 400 labour days and standard deviation of 100 labour days
m = 400 and σ = 100
The construction work should be completed within 450 days.
The standard normal variate
\(\frac { x-µ }{6}\) = \(\frac { x-400 }{100}\)
personality for 1 labour day = Rs 10,000
If personality amount is = 2,00,000 than No of excess

days = \(\frac { 200000 }{10000}\) = 20
∴ x = 450 + 20 = 470
when x = 470
z = \(\frac { 470-400 }{100}\) = \(\frac { 70 }{100}\) = 0.7
= p(x ≥ 470) = p(z ≥ 0.7)
= 0.5 – 0.2580
= 0.2420

(ii) p(at most 500 days) = p(x ≤ 500 )
when x = 500

z = \(\frac { 500-400 }{100}\) = \(\frac { 100 }{100}\) = 1
p(x ≤ 500) = p(z ≤ 1)
= p(∞ < z < 0) -r- p(0 < z < 1)
= 0.5 + 0.3415
= 0.8413

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Define possion distribution.
Solution:
Poisson distribution was derived in 1837 by a French Mathematician Simeon D. Poisson.
A random variable X is said to follow a Poisson distribution with parameter X if it assumes only non-negative values and its probability mass function is given by

Question 2.
Write any 2 examples for possion distribution
Solution:
1. The number of alpha particles emitted by a radioactive substance in a fraction of a second.
2. Number of road accidents occurring at a particular interval of time per day.

Question 3.
Write the condition for which the possion distribution is limiting case of binomial distribution
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
(i) n, the number of trials is indefinitely large i.e n → ∞
(ii) p, the constant probability of success in each trial is very small, i.e. p → 0.
(iii) np = λ is finite. Thus p = \(\frac { λ }{n}\) and q = 1 – (\(\frac { λ }{n}\))
where λ, is a positive real number.

Question 4.
Derive the mean and variance of possion distribution.
Solution:
Derivation of Mean and variance of Poisson distribution

Variance (X) = E(X²) – E(X)²

Variance (X) = E(X²) – E(X)²
= λ² + λ – (λ)²
= λ

Question 5.
Mention the properties of possion distribution.
Solution:
Poisson distribution is the only distribution in which the mean and variance are equal.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? Give e^(-2.8) = 0.06
Solution:
Since the mortality rate for a certain disease in 7 in loop
∴ p = \(\frac { 7 }{1000}\) and n = 400
The value of mean A = λp = 400 × \(\frac { 7 }{1000}\)
∴ λ = 2.8
Let x be a random variable following
distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
∴ the distribution is p(x = 2) = \(\frac { e^{-2-8}(2.8)^2 }{2!}\)
\(\frac { 0.06×7.84 }{2}\)
= 0.2352

Question 7.
Mention the properties of possion distribution.
Solution:
p(defective bulbs) = \(\frac { 5 }{100}\)
n = 120
The value of mean λ = np = 120 × \(\frac { 5 }{100}\)
λ = 6
Hence, x follows possion distribution with
P(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(x = 0) = \(\frac { e^{-6}(6)^0 }{0!}\) = e^-6
= 0.0025

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a possion variate, with mean 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused
Solution:
In a possion distribution n=2
mean λ = 1.5
x follows poison distribution
With in p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(neither car is used) = p(x = 0)
\(\frac { e^{-1.5}(1.5)^0 }{0!}\) = e^-1.5
= 0.2231

(ii) p(some demand is refused) = p(x > 2)

= 1 – 0.2231 [1 + 1.5 + 1.125]
= 1 – 0.2231 [3.625]
= 1-0.8087
= 0.1913

Question 9.
The average number of phone calls per minute into the switch board of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be (i) no phone at all (ii) exactly 3 calls (iii) atleast 5 calls
Solution:
The average number of phone cells per minute into the switch board of a company is λ = 2.5
x follows poisson distribution with

(iii) p(atleast 5 calls) = p(x ≥ 5)
= p(x = 5) + p(x = 6) + …………..
= 1 – p(x < 5)

Question 10.
The distribution of the number of road accidents pre day in a city is possion with mean 4. Find the number of days out of 100 days when there will be (i) no accident (ii) atleast 2 accidents and (iii) at most 3 accidents.
Solution:
In a possion distribution
mean λ = 4
n = 100
x follows possion distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) = \(\frac { e^{-4}(4)^x }{x!}\)
(i) p(no accident) = p(x = 0)
= \(\frac { e^{-4}(4)^0 }{0!}\) = e^-4 = 0.0183
out of 100 days there will be no accident
= n × p(x = 0)
= 100 × 0.0183 = 1.83
= 2 days (approximately)

(ii) p(atleast 2 accidents)
= p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 4) + …………
= 1 – p(x < 2)
= 1 – [p(x = 0) + p(x = 1)]
= 1 – [\(\frac { e^{-4}(4)^0 }{0!}\) + \(\frac { e^{-4}(4)^1 }{1!}\)]
= 1 – e^-4 [l + 4]
= 1 – 0.0183(5) = 1 – 0.0915
= 0.9085
= Out of 100 days there will be atleast 2 accidents = n × p(x ≥ 2)
= 100 × 0.9085
= 90.85
= 91 days (approximately)

(iii) p(atmost 3 accident) = p(x ≤ 3)
= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

out of 100 days there will be at most 3 acccident = n × p(x ≤ 3)
= 100 × 0.4331
= 43.31
= 43 days(approximately)

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200/ calculate the probability that in a factory employing 300 workers there will be atleast two fatal accidents in year, (given e^-0.25 = 0.7788 Solution:
Let p be the probability of a fatal accident in a factory during the yeart
p = \(\frac { 1 }{1200}\) and n = 300 1200
λ = np = 300 × \(\frac { 1 }{1200}\) = \(\frac { 1 }{4}\)
λ = 0.25
x follows poison distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) + \(\frac { e^{-0.25}(0.25) }{x!}\)
p(atleasttwo fatal accidents) = p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 3) + p(x = 4) + ……….
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}
= 1 – 0.7788 [1 + 0.25]
= 1 – 0.7788(1.25)
= 1 – 0.9735 = 0.0265
∴ p(x ≥ 2) = 0.0265

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute (i) No customer appears (ii) three or more customers appear.
Solution:
The average number of customers ,who appear in a counter of a certain bank per minute = 2
∴ λ = 2
x follows poisson distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
x follows poisson distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(no customber appears) = p(x = 0)
= \(\frac { e^{-2}(2)^0 }{0!}\) = e^-2
= 0.1353

(ii) p(three or more customers appears) = p(x ≥ 3)
= p(x = 3) + p{x = 4) + p(x = 5) + ……
= 1 – p(x < 3)
= 1 – {p(x = 0) + p(x = 1) + p(x = 2)}

= 1 – 2^-2 [1 + 2 + 2]
= 1 – 0.1353(5)
= 1 – 0.6765
= 0.3235