Category: Probability & Statistics Objective Questions

250+ TOP MCQs on Baye’s Theorem and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Baye’s Theorem”.

1. Three companies A, B and C supply 25%, 35% and 40% of the notebooks to a school. Past experience shows that 5%, 4% and 2% of the notebooks produced by these companies are defective. If a notebook was found to be defective, what is the probability that the notebook was supplied by A?
a) 44⁄69
b) 25⁄69
c) 13⁄24
d) 11⁄24
Answer: b
Clarification: Let A, B and C be the events that notebooks are provided by A, B and C respectively.
Let D be the event that notebooks are defective
Then,
P(A) = 0.25, P(B) = 0.35, P(C) = 0.4
P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02
P(A│D) = (P(D│A)*P(A))/(P(D│A) * P(A) + P(D│B) * P(B) + P(D│C) * P(C) )
= (0.05*0.25)/((0.05*0.25)+(0.04*0.35)+(0.02*0.4)) = 2000/(80*69)
= 25⁄69.

2. A box of cartridges contains 30 cartridges, of which 6 are defective. If 3 of the cartridges are removed from the box in succession without replacement, what is the probability that all the 3 cartridges are defective?
a) (frac{(6*5*4)}{(30*30*30)})
b) (frac{(6*5*4)}{(30*29*28)})
c) (frac{(6*5*3)}{(30*29*28)})
d) (frac{(6*6*6)}{(30*30*30)})
Answer: b
Clarification: Let A be the event that the first cartridge is defective. Let B be the event that the second cartridge is defective. Let C be the event that the third cartridge is defective. Then probability that all 3 cartridges are defective is P(A ∩ B ∩ C)
Hence,
P(A ∩ B ∩ C) = P(A) * P(B|A) * P(C | A ∩ B)
= (6⁄30) * (5⁄29) * (4⁄28)
= (6 * 5 * 4)⁄(30 * 29 * 28).

3. Two boxes containing candies are placed on a table. The boxes are labelled B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 1⁄3 and the probability of selecting box B2 is 2⁄3. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. What is the probability that Suresh will win the TV (that is, she will select a cinnamon candy)?
a) 7⁄33
b) 6⁄33
c) 13⁄33
d) 20⁄33
Answer: c
Clarification: Let A be the event of drawing a cinnamon candy.
Let B1 be the event of selecting box B1.
Let B2 be the event of selecting box B2.
Then, P(B1) =1⁄3 and P(B2) = 2⁄3
P(A) = P(A ∩ B1) + P(A ∩ B2)
= P(A|B1) * P(B1) + P(A|B2)*P(B2)
= (7⁄11) * (1⁄3) + (3⁄11) * (2⁄3)
= 13⁄33.

4. Two boxes containing candies are placed on a table. The boxes are labelled B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 1⁄3 and the probability of selecting box B2 is 2⁄3. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. If he wins a colour TV, what is the probability that the marble was from the first box?
a) 7⁄13
b) 13⁄7
c) 7⁄33
d) 6⁄33
Answer: a
Clarification: Let A be the event of drawing a cinnamon candy.
Let B1 be the event of selecting box B1.
Let B2 be the event of selecting box B2.

Then, P(B1) = 1⁄3 and P(B2) = 2⁄3
Given that Suresh won the TV, the probability that the cinnamon candy was selected from B1 is
P(B1|A) = (P(A|B1) * P( B1))/( P(A│B1) * P( B1) + P(A│B1) * P(B2))
(frac{(frac{7}{11})*(frac{1}{3})}{(frac{7}{11})*(frac{1}{3})+(frac{3}{11})*(frac{2}{3})})
= 7⁄13.

5. Suppose box A contains 4 red and 5 blue coins and box B contains 6 red and 3 blue coins. A coin is chosen at random from the box A and placed in box B. Finally, a coin is chosen at random from among those now in box B. What is the probability a blue coin was transferred from box A to box B given that the coin chosen from box B is red?
a) 15⁄29
b) 14⁄29
c) 1⁄2
d) 7⁄10
Answer: a
Clarification: Let E represent the event of moving a blue coin from box A to box B. We want to find the probability of a blue coin which was moved from box A to box B given that the coin chosen from B was red. The probability of choosing a red coin from box A is P(R) = 7⁄9 and the probability of choosing a blue coin from box A is P(B) = 5⁄9. If a red coin was moved from box A to box B, then box B has 7 red coins and 3 blue coins. Thus the probability of choosing a red coin from box B is 7⁄10 . Similarly, if a blue coin was moved from box A to box B, then the probability of choosing a red coin from box B is 6⁄10.
Hence, the probability that a blue coin was transferred from box A to box B given that the coin chosen from box B is red is given by
(P(E|R)=frac{P(R|E)*P(E)}{P(R)})
=(frac{(frac{6}{10})*(frac{5}{9})}{(frac{7}{10})*(frac{4}{9})+(frac{6}{10})*(frac{5}{9})})
= 15⁄29.

6. An urn B1 contains 2 white and 3 black chips and another urn B2 contains 3 white and 4 black chips. One urn is selected at random and a chip is drawn from it. If the chip drawn is found black, find the probability that the urn chosen was B1.
a) 4⁄7
b) 3⁄7
c) 20⁄41
d) 21⁄41
Answer: d
Clarification: Let E1, E2 denote the vents of selecting urns B1 and B2 respectively.
Then P(E1) = P(E2) = 1⁄2
Let B denote the event that the chip chosen from the selected urn is black .
Then we have to find P(E1 /B).
By hypothesis P(B /E1) = 3⁄5
and P(B /E2) = 4⁄7
By Bayes theorem P(E1 /B) = (P(E1)*P(B│E1))/((P(E1) * P(B│E1)+P(E2) * P(B│E2)) )
= ((1/2) * (3/5))/((1/2) * (3/5)+(1/2)*(4/7) ) = 21/41.

7. At a certain university, 4% of men are over 6 feet tall and 1% of women are over 6 feet tall. The total student population is divided in the ratio 3:2 in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?
a) 2⁄5
b) 3⁄5
c) 3⁄11
d) 1⁄100
Answer: c
Clarification: Let M be the event that student is male and F be the event that the student is female. Let T be the event that student is taller than 6 ft.
P(M) = 2⁄5 P(F) = 3⁄5 P(T|M) = 4⁄100 P(T|F) = 1⁄100
P(F│T) = (P(T│F) * P(F))/(P(T│F) * P(F) + P(T│M) * P(M))
= ((1/100) * (3/5))/((1/100) * (3/5) + (4/100) * (2/5) )
= 3⁄11.

8. Previous probabilities in Bayes Theorem that are changed with help of new available information are classified as _________________
a) independent probabilities
b) posterior probabilities
c) interior probabilities
d) dependent probabilities
Answer: b
Clarification: None.

250+ TOP MCQs on Sampling Distribution of Means and Answers

Probability and Statistics Questions on “Sampling Distribution of Means”.

1. If the mean of population is 29 then the mean of sampling distribution is __________
a) 29
b) 30
c) 21
d) 31
Answer: a
Clarification: In a sampling distribution the mean of the population is equal to the mean of the sampling distribution. Hence mean of population=29. Hence mean of sampling distribution=29.

2. In systematic sampling, population is 240 and selected sample size is 60 then sampling interval is ________
a) 240
b) 60
c) 4
d) 0.25
Answer: c
Clarification: Sampling interval is defined as the interval in which the population is divided. The sampling interval is given as the population/sample size = 240/60 = 4.

3. The method of selecting a desirable portion from a population which describes the characteristics of whole population is called as ________
a) sampling
b) segregating
c) dividing
d) implanting
Answer: a
Clarification: The method of selecting a desirable portion from a population which describes the characteristics of whole population is called as Sampling. It is useful in combining the related samples and hence making the distribution easy to manipulate.

4. If the standard deviation of a population is 50 and the sample size is 16 then the standard deviation of the sampling distribution is ________
a) 11.25
b) 12.25
c) 13.25
d) 14.25
Answer: b
Clarification: The standard deviation of a population ϕ is given as σ/(n)1/2 where n is the sample size and σ is the standard deviation of population.
Substituting the values of n=16 and σ=50.
σ/(n)1/2
50/(16)1/2
we get ϕ=12.25.

5. In sampling distribution what does the parameter k represents ________
a) Sub stage interval
b) Secondary interval
c) Multi stage interval
d) Sampling interval
Answer: d
Clarification: In sampling distribution the parameter k represents Sampling interval. It represents the distance between which data is taken.

6. If the distribution of sample and population changes then the mean of Sampling distribution must be equal to ________
a) standard deviation of population
b) variance of population
c) sample of population
d) mean of population
Answer: d
Clarification: In a sampling distribution irrespective of the variation in sample and population the mean of the population is equal to the mean of the sampling distribution. If repeated random samples of a given size n are taken from a population of values for a quantitative variable, where the population mean is μ and the population standard deviation is σ (sigma) then the mean of all sample means (x-bars) is population mean μ.

7. The cluster sampling, stratified sampling or systematic samplings are types of ________
a) direct sampling
b) indirect sampling
c) random sampling
d) non random sampling
Answer: c
Clarification: The cluster sampling, stratified sampling or systematic samplings are types of random sampling. The advantage of probability sampling methods is that they ensure that the sample chosen is representative of the population.

8. Which of the following is classified as unknown or exact value that represents the whole population?
a) predictor
b) guider
c) parameter
d) estimator
Answer: c
Clarification: The unknown or exact value that represents the whole population is called as parameter. Generally parameters are defined by small Roman symbols.

9. A sample size is considered large in which of the following cases?
a) n > or = 30
b) n > or = 50
c) n < or = 30
d) n < or = 50
Answer: a
Clarification: Generally a sample having 30 or more sample values is called a large sample. By the Central Limit Theorem such a sample follows a Normal Distribution.

10. The selected clusters in a clustering sampling are known as ________
a) elementary units
b) primary units
c) secondary units
d) proportional units
Answer: a
Clarification: In Cluster the population is divided into various groups called as clusters. The selected clusters in a sample are called as elementary units.

250+ TOP MCQs on Permutations and Combinations and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Permutations and Combinations”.

1. If 16Pr-1 : 15Pr-1 = 16 : 7 then find r.
a) 10
b) 12
c) 7
d) 8
Answer: a
Clarification: We know that (^nP_r = frac{n!}{(n-r)!} )
Hence (^{16}P_{r-1} : , ^{15}P_{r-1} = 16 : 7 )
([frac{16!}{16-(r-1)!}] ÷ [frac{15!}{15-(r-1)}] = 16 ÷ 7 )
16 ÷ (17 – r) = 16 ÷ 7
17 – r = 7
Hence r = 10.

2. In a colony, there are 55 members. Every member posts a greeting card to all the members. How many greeting cards were posted by them?
a) 990
b) 890
c) 2970
d) 1980
Answer: c
Clarification: First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are
54 + 54 + 54 …
54 (55times) = 54 x 55 = 2970.

3. Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.
a) 36
b) 48
c) 144
d) 96
Answer: c
Clarification: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

4. Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition.
a) 666700
b) 666600
c) 678860
d) 665500
Answer: b
Clarification: The given digits are 1, 3, 5, 7, 9
Sum of r digit number= n-1Pr-1
(Sum of all n digits)×(1111… r times)
N is the number of non zero digits.
Here n=5, r=4
The sum of 4 digit numbers
4P3 (1+3+5+7+9)(1111)=666600.

5. If nPr = 3024 and nCr = 126 then find n and r.
a) 9, 4
b) 10, 3
c) 12, 4
d) 11, 4
Answer: a
Clarification: (frac{^nP_r}{^nC_r} = frac{3024}{126} )
(^nP_r = frac{n!}{(n-r)!} )
(^nC_r = frac{n!}{(n-r)!×r!} )
Hence ( [frac{n!}{(n-r)!}]÷[frac{n!}{(n-r)!×r!}] ) = 24
24 = r!
Hence r = 4
Now nP4 = 3024
(frac{n!}{(n-4)!} = 3024 )
n(n-1)(n-2)(n-3) = 9.8.7.6
n = 9.

6. Find the number of rectangles and squares in an 8 by 8 chess board respectively.
a) 296, 204
b) 1092, 204
c) 204, 1092
d) 204, 1296
Answer: b
Clarification: Chess board consists of 9 horizontal 9 vertical lines. A rectangle can be formed by any two horizontal and two vertical lines. Number of rectangles = 9C2 × 9C2 = 1296. For squares there is one 8 by 8 square four 7 by 7 squares, nine 6 by 6 squares and like this
Number of squares on chess board = 12+22…..82 = 204
Only rectangles = 1296-204 = 1092.

7. There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear?
a) 1130
b) 550
c) 1129
d) 1140
Answer: a
Clarification: Number of points in plane n = 20.
Number of colinear points m = 5.
Number of triangles from by joining n points of which m are colinear = nC3 – mC3
Therefore the number of triangles = 20C3 – 5C3 = 1140-10 = 1130.

8. In how many ways can we select 6 people out of 10, of which a particular person is not included?
a) 10C3
b) 9C5
c) 9C6
d) 9C4
Answer: c
Clarification: One particular person is not included we have to select 6 persons out of 9 which can be done in 9C6 ways.

9. Number of circular permutations of different things taken all at a time is n!.
a) True
b) False
Answer: b
Clarification: The number of circular permutations of different things taken all at a time is n-1! and the number of linear permutations of different things taken all at a time is n!.

10. Is the given statement true or false?
nCr= nCn-r
a) True
b) False
Answer: a
Clarification: The property of combination states nCr= nCn-r
As (^nC_r = frac{n!}{(n-r)!×r!} )
(^nC_{n-r} = frac{n!}{r!×(n-r)!} = ^nC_r. )

250+ TOP MCQs on Sampling Distribution of Proportions and Answers

Probability and Statistics Questions and Answers for Campus interviews on “Sampling Distribution of Proportions”.

1. The probability of selecting a sample containing n items from a population with N items without replacement in a Sampling Distribution is?
a) 1/NCn
b) 1/nCN
c) 1/2n
d) 1/2N
Answer: a
Clarification: The number of ways of selecting and samples of size n from a population containing N atoms is NCn. The probability of selecting of each sample is 1/NCn.

2. Find the number of all possible samples from a population containing 18 items from which 6 items are selected at random without replacement.
a) 18564
b) 15864
c) 20264
d) 21564
Answer: a
Clarification: The number of ways of selecting n samples from a population containing n items is NCn. The population is N = 18 and sample size is n = 6. Therefore the number of possible samples are 18C6 = 18564.

3. A pack of cards contains 52 cards. A player selects 4 cards at random without replacement. Find all possible combinations of the cards selected.
a) 207752
b) 270752
c) 270725
d) 207725
Answer: c
Clarification: Considering the experiment to be a sampling distribution where the population contains 52 cards and each sample contains 4 cards. The number of possible samples without replacement are NCn that is 52C4 = 207725 samples.

4. A population contains N items out of which n items are selected with replacement. Find the probability of the sample being selected.
a) 1/N
b) 1/nN
c) 1/NCn
d) 1/Nn
Answer: d
Clarification: The number of samples containing n items selected from a population of N items is Nn. The probability of selection of each sample is 1/Nn.

5. A box contains 26 pairs of napkins. If 3 pairs of napkins are selected at random with a replacement then the number of possible samples is _______
a) 17675
b) 17566
c) 17576
d) 17556
Answer: c
Clarification: The number of samples formed with n items from a population containing N items is Nn.
Here N = 26 and n = 3.
Hence samples are Nn = 263 = 17576.

6. A sample was formed consisting of 8 students from a total of 56 students for certain task. Find the sampling fraction of the population of students.
a) 1/7
b) 7
c) 49
d) 1/49
Answer: a
Clarification: In a sampling distribution if N is the population size, n is the sample size then number of sampling fractions is n/N. Hence N=56 and n=8 which gives n/N as 8/56 = 1/7.

7. Find the population proportion p for an IPL team having total 30 players with 10 overseas players.
a) 1/2
b) 1/3
c) 2/3
d) 1/4
Answer: b
Clarification: The population proportion p for a population consisting of N items with X specialized items is given as X/N. Now N=30, X=10. Hence p = X/N = 10/30 = 1/3.

8. It is provided that for a sampling distribution E(X)=11 and ϕ=13. Find the bias in the sampling.
a) 2
b) 4
c) 6
d) 3
Answer: a
Clarification: The bias for sampling distribution is given as |E(X) – ϕ|.
|11 – 13| = 2
Hence bias is 2.

9. Find the standard error of population proportion p for sampling with replacement. The population proportion is 0.5 and size of sample is 4.
a) 0.5
b) 0.25
c) 0.225
d) 0.375
Answer: b
Clarification: The standard error of population proportion in sampling distribution with replacement is given as [p*(1-p)/n]1/2. Here p=0.5 and n=4. Hence substituting the values we get
[0.5*(1-0.5)/4]1/2
= 0.25.

10. Find the value of standard error Ẋ in a sampling distribution without replacement. Given that the standard deviation of the population of 100 items is 25.
a) 3
b) 4
c) 2
d) 5
Answer: c
Clarification: Standard error in a sampling distribution with replacement is given by Ẋ = σ/(n)1/2. The standard error with the replacement of items remains the same that is Ẋ = σ/(n)1/2.
Hence n = 100 and σ = 25
Ẋ = σ/(n)1/2
Ẋ = 25/(100)1/2
Ẋ = 2.5
The value of Ẋ = 2.5.

Probability and Statistics for Campus Interviews,

250+ TOP MCQs on Random Variables and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Random Variables”.

1. Consider a dice with the property that that probability of a face with n dots showing up is proportional to n. The probability of face showing 4 dots is?
a) (frac{1}{7} )
b) (frac{5}{42} )
c) (frac{1}{21} )
d) (frac{4}{21} )
Answer: d
Clarification: P (n) is proportional to n where n=
1,2,3,…6 is random variable.
P(n) = kn
P(1)+P(2)….P (6) = 1
K(1+2+3+4+5+6) = 1
(K=frac{1}{21} )
Hence P(4) = 4K = (frac{4}{21}. )

2. Let X be a random variable with probability distribution function f (x)=0.2 for |x|<1
= 0.1 for 1 < |x| < 4
= 0 otherwise
The probability P (0.5 < x < 5) is _____
a) 0.3
b) 0.5
c) 0.4
d) 0.8
Answer: c
Clarification: P (0.5 < x < 5) = Integrating f (x) from
0.5 to 5 by splitting in 3 parts that is from 0.5 to 1
and from 1 to 4 and 4 to 5 we get
P (0.5 < x < 5) = 0.1 + 0.3 + 0
P (0.5 < x < 5) = 0.4.

3. Runs scored by batsman in 5 one day matches are 50, 70, 82, 93, and 20. The standard deviation is ______
a) 25.79
b) 25.49
c) 25.29
d) 25.69
Answer: a
Clarification: The mean of 5 innings is
(50+70+82+93+20)÷5 = 63
S.D = [1⁄n (x(n)-mean)2]0.5
S.D = 25.79.

4. Find median and mode of the messages received on 9 consecutive days 15, 11, 9, 5, 18, 4, 15, 13, 17.
a) 13, 6
b) 13, 18
c) 18, 15
d) 15, 16
Answer: b
Clarification: Arranging the terms in ascending order 4, 5, 9, 11, 13, 14, 15, 18, 18.
Median is (frac{(n+1)}{2} ) term as n = 9 (odd) = 13.
Mode = 18 which is repeated twice.

5. Mode is the value of x where f(x) is a maximum if X is continuous.
a) True
b) False
Answer: a
Clarification: For a continuous variable mode is defined as the value where f(x) is a maximum or it is defined as the quantity repeated maximum number of times.

6. E (XY)=E (X)E (Y) if x and y are independent.
a) True
b) False
Answer: a
Clarification: By the property of Expectation
E (XY) = E (X) E (Y).
That is the Expectation of a composite function XY is the product of the individual expectations of X and Y.

7. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is ______
a) (frac{1}{2} )
b) (frac{1}{3} )
c) (frac{1}{4} )
d) (frac{1}{6} )
Answer: a
Clarification: p=0.5 (Probability of tail)
q=1-0.5=0.5
n=4 and x is binomial variate.
P (X=x) = nCx px qn-x.
P (X=3) = 4C3 (0.5)3 = 1⁄2.

8. If E denotes the expectation the variance of a random variable X is denoted as?
a) (E(X))2
b) E(X2)-(E(X))2
c) E(X2)
d) 2E(X)
Answer: b
Clarification: By property of Expectation
V (X) = E (X2)-(E(X))2.

9. X is a variate between 0 and 3. The value of E(X2) is ______
a) 8
b) 7
c) 27
d) 9
Answer: d
Clarification: Integrating f(x) = x2 from 0 to 3 we get E(X2) = 32 = 9.

10. The random variables X and Y have variances 0.2 and 0.5 respectively. Let Z= 5X-2Y. The variance of Z is?
a) 3
b) 4
c) 5
d) 7
Answer: d
Clarification: Var(X) = 0.2, Var(Y) = 0.5
Z = 5X – 2Y
Var(Z) = Var(5X-2Y)
= Var(5X) + Var(2Y)
= 25Var(X) + 4Var(Y)
Var(Z) = 7.

250+ TOP MCQs on Chi-Squared Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Chi-Squared Distribution”.

1. A dice is tossed 120 times with the following results

No. turned up 1 2 3 4 5 6
Frequency 30 25 18 10 22 15

Test the hypothesis that the dice is unbiased (X2 = 11.7). Calculate the frequency observed for Chi Square distribution.
a) Dice is unbiased, 11.3
b) Dice is biased, 12.9
c) Dice is unbiased, 10.9
d) Dice is biased, 12.3
Answer: b
Clarification: Step 1: Null Hypothesis: dice is unbiased.
Step 2: Calculation of Expected frequency:
Since the dice is unbiased P(r) = 1/6.
r = 1, 2, 3, 4, 5, 6
Expected frequency f(r) = N*r = 120* 1/6 = 20
Step3: Calculation of X2
X2 (=∑frac{[(fe-fo)]^2}{fe} )
Hence X2 = 12.90 > 11.90.
Thus dice is biased.

2. Consider a set of 18 samples from a standard normal distribution. We square each sample and sum all the squares. The number of degrees of freedom for a Chi Square distribution will be?
a) 17
b) 18
c) 19
d) 20
Answer: b
Clarification: In Chi Square Distribution the number of standard normal derivatives or samples equals the number of degrees of freedom.
Here total number of standard normal derivatives = 18.
Hence the number of degrees of freedom for a Chi Square distribution = 18.

3. What is the mean of a Chi Square distribution with 6 degrees of freedom?
a) 4
b) 12
c) 6
d) 8
Answer: c
Clarification: By the property of Chi Square distribution, the mean corresponds to the number of degrees of freedom.
Degrees of freedom = 6.
Hence mean = 6.

4. Which Chi Square distribution looks the most like a normal distribution?
a) A Chi Square distribution with 4 degrees of freedom
b) A Chi Square distribution with 5 degrees of freedom
c) A Chi Square distribution with 6 degrees of freedom
d) A Chi Square distribution with 16 degrees of freedom
Answer: d
Clarification: When the number of degrees of freedom in Chi Square distribution increases it tends to correspond to normal distribution. The option with a maximum number of degrees of freedom is 16.

5. A bag contains 80 chocolates. This bag has 4 different colors of chocolates in it. If all four colors of chocolates were equally likely to be put in the bag, what would be the expected number of chocolates of each color?
a) 12
b) 11
c) 20
d) 9
Answer: c
Clarification: If all four colors were equally likely to be put in the bag, then the expected frequency for a given color would be 1/4th of the chocolates.
N = 80, r = 1/4
So, the expected frequency = N*r = (1/4)*(80) = 20.

6. Suppose a person has 8 red, 5 green, 12 orange, and 15 blue balls. Test the null hypothesis that the colors of the balls occur with equal frequency. What is the Chi Square value you get?
a) 5.6
b) 5.68
c) 5.86
d) 5.8
Answer: d
Clarification: By Chi Square Test we get,
Observed frequency f0= (8+5+12+15)/4 = 10
X2 =(∑frac{[(fe-fo)]^2}{fe} )
The sum of each (expected – observed)2/expected = (10-8)2/10 + (10-5)2/10 + (10-12)2/10 + (10-15)2/10 = 5.8.

7. A faculty is interested in whether there is a relationship between gender and subject at his college. He tabulated some men and women on campus and asked them if their subject was Mathematics (M), Geography (G), and Science (S). What would be the expected frequency of women in Geography based on this table?

M G S Total
Women 10 14 10 34
Men 11 22 14 23
Total 21 36 24 57

a) 31.12
b) 11.32
c) 12.13
d) 13.12
Answer: d
Clarification: The expected value of women in social sciences is the product of the total number of women and the total number of social science majors divided by the total number of participants. (22*34)/57 = 13.12.

8. In a sample survey of public opinion answer to the question:
1) Do you drink?
2) Are you in favor of local option sale of Liquor

Yes No Total
Yes 56 31 87
No 18 6 24
Total 74 37 111

Infer or not the local option on the sale of liquor is dependent on individual drinker? Find the value of X2 for degrees of freedom at level of significance 3.841.
a) 0.957
b) 0.975
c) 0.759
d) 0.795
Answer: a
Clarification: Step 1: Null hypothesis: The option on the sale of liquor is not dependent with the individual drinking.
Step 2: Calculation of theoretical frequencies (Expected)
Expected frequency of (1,1) cell
fe11 = 87*74/111 = 58
fe12 = 87*37/111 = 29
fe21 = 24*74/111 = 16
fe22 = 24*32/111 = 8

Step 3: calculation of X2 distribution we know that
X2 =( ∑frac{[(fe-fo)]^2}{fe} )
X2 = 0.957 < 3.841.
Hence the null hypothesis is accepted
Thus the sale of liquor does not depend on the individual drinker.

9. The Variance of Chi Squared distribution is given as k.
a) True
b) False
Answer: b
Clarification: The Mean of Chi Squared distribution is given as k. The Variance of Chi Squared distribution is given as 2k.

10. Which of these distributions is used for a testing hypothesis?
a) Normal Distribution
b) Chi-Squared Distribution
c) Gamma Distribution
d) Poisson Distribution
Answer: b
Clarification: Chi-Squared Distribution is used for testing hypothesis. The value of X2 decides whether the hypothesis is accepted or not.