1. The mean of a data set is 50, and the standard deviation is 10. What is the z-score
for a value of 60?
A. 1
B. 0
C. 2
D. -1
Answer: a) 1
Rationale: The formula for the z-score is Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ,
where XXX is the data point, μ\muμ is the mean, and σ\sigmaσ is the standard
deviation. Here, Z=60−5010=1Z = \frac{60 - 50}{10} = 1Z=1060−50=1.
2. In a Poisson distribution with λ=3\lambda = 3λ=3, what is the probability of
observing exactly 1 event?
A. 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3
B. 32e−32!\frac{3^2 e^{-3}}{2!}2!32e−3
C. 33e−33!\frac{3^3 e^{-3}}{3!}3!33e−3
D. 30e−30!\frac{3^0 e^{-3}}{0!}0!30e−3
Answer: a) 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3
Rationale: The Poisson probability mass function is P(X=k)=λke−λk!P(X=k) =
\frac{\lambda^k e^{-\lambda}}{k!}P(X=k)=k!λke−λ. Here λ=3\lambda = 3λ=3 and
k=1k = 1k=1, so the probability is 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3.
3. What is the probability of drawing a king or a queen from a standard deck of 52
cards?
A. 1/13
B. 2/52
C. 4/52
D. 8/52
Answer: c) 4/52
Rationale: There are 4 kings and 4 queens in a deck, so the probability of drawing a
king or queen is 852=213\frac{8}{52} = \frac{2}{13}528=132.
4. The variance of a set of data is 16. What is the standard deviation?
A. 4
B. 16
C. 32
D. 8
Answer: a) 4
Rationale: The standard deviation is the square root of the variance. 16=4\sqrt{16} =
416=4.
5. What is the probability of drawing two red cards in a row from a standard deck of 52
cards without replacement?
A. 126\frac{1}{26}261
B. 1352⋅1251\frac{13}{52} \cdot \frac{12}{51}5213⋅5112
C. 152⋅1351\frac{1}{52} \cdot \frac{13}{51}521⋅5113
D. 1352⋅1352\frac{13}{52} \cdot \frac{13}{52}5213⋅5213
Answer: b) 1352⋅1251\frac{13}{52} \cdot \frac{12}{51}5213⋅5112
, Rationale: The probability of drawing a red card on the first draw is
2652\frac{26}{52}5226. After drawing one red card, there are 25 red cards left, so the
probability of drawing another red card is 2551\frac{25}{51}5125.
6. In a normal distribution, what percentage of data values lie between the mean and 2
standard deviations above the mean?
A. 68%
B. 13.5%
C. 95%
D. 99.7%
Answer: b) 13.5%
Rationale: In a normal distribution, approximately 13.5% of the data lies between the
mean and 2 standard deviations above the mean.
7. What is the variance of a binomial distribution with n=12n = 12n=12 and p=0.5p =
0.5p=0.5?
A. 6
B. 3
C. 2.4
D. 4
Answer: d) 4
Rationale: The variance of a binomial distribution is σ2=n⋅p⋅(1−p)\sigma^2 = n \cdot p
\cdot (1 - p)σ2=n⋅p⋅(1−p). For n=12n = 12n=12 and p=0.5p = 0.5p=0.5, the variance
is σ2=12⋅0.5⋅0.5=4\sigma^2 = 12 \cdot 0.5 \cdot 0.5 = 4σ2=12⋅0.5⋅0.5=4.
8. Which of the following is the formula for the binomial distribution?
A. P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-
k}P(X=k)=(kn)pk(1−p)n−k
B. P(X=k)=1k!e−1P(X=k) = \frac{1}{k!} e^{-1}P(X=k)=k!1e−1
C. P(X=k)=1k2P(X=k) = \frac{1}{k^2}P(X=k)=k21
D. P(X=k)=knP(X=k) = \frac{k}{n}P(X=k)=nk
Answer: a) P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-
k}P(X=k)=(kn)pk(1−p)n−k
Rationale: This is the standard formula for the probability mass function of a binomial
distribution, where nnn is the number of trials, kkk is the number of successes, and ppp
is the probability of success on a single trial.
9. Which of the following describes the shape of the distribution for a large number of
independent trials of a binomial experiment?
A. Uniform distribution
B. Normal distribution
C. Poisson distribution
D. Exponential distribution
Answer: b) Normal distribution
Rationale: According to the Central Limit Theorem, for a large number of trials, a
binomial distribution approaches a normal distribution.
10. A box contains 10 balls: 4 red, 3 blue, and 3 green. What is the probability of
selecting a red ball at random?
A. 1/10
for a value of 60?
A. 1
B. 0
C. 2
D. -1
Answer: a) 1
Rationale: The formula for the z-score is Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ,
where XXX is the data point, μ\muμ is the mean, and σ\sigmaσ is the standard
deviation. Here, Z=60−5010=1Z = \frac{60 - 50}{10} = 1Z=1060−50=1.
2. In a Poisson distribution with λ=3\lambda = 3λ=3, what is the probability of
observing exactly 1 event?
A. 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3
B. 32e−32!\frac{3^2 e^{-3}}{2!}2!32e−3
C. 33e−33!\frac{3^3 e^{-3}}{3!}3!33e−3
D. 30e−30!\frac{3^0 e^{-3}}{0!}0!30e−3
Answer: a) 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3
Rationale: The Poisson probability mass function is P(X=k)=λke−λk!P(X=k) =
\frac{\lambda^k e^{-\lambda}}{k!}P(X=k)=k!λke−λ. Here λ=3\lambda = 3λ=3 and
k=1k = 1k=1, so the probability is 31e−31!\frac{3^1 e^{-3}}{1!}1!31e−3.
3. What is the probability of drawing a king or a queen from a standard deck of 52
cards?
A. 1/13
B. 2/52
C. 4/52
D. 8/52
Answer: c) 4/52
Rationale: There are 4 kings and 4 queens in a deck, so the probability of drawing a
king or queen is 852=213\frac{8}{52} = \frac{2}{13}528=132.
4. The variance of a set of data is 16. What is the standard deviation?
A. 4
B. 16
C. 32
D. 8
Answer: a) 4
Rationale: The standard deviation is the square root of the variance. 16=4\sqrt{16} =
416=4.
5. What is the probability of drawing two red cards in a row from a standard deck of 52
cards without replacement?
A. 126\frac{1}{26}261
B. 1352⋅1251\frac{13}{52} \cdot \frac{12}{51}5213⋅5112
C. 152⋅1351\frac{1}{52} \cdot \frac{13}{51}521⋅5113
D. 1352⋅1352\frac{13}{52} \cdot \frac{13}{52}5213⋅5213
Answer: b) 1352⋅1251\frac{13}{52} \cdot \frac{12}{51}5213⋅5112
, Rationale: The probability of drawing a red card on the first draw is
2652\frac{26}{52}5226. After drawing one red card, there are 25 red cards left, so the
probability of drawing another red card is 2551\frac{25}{51}5125.
6. In a normal distribution, what percentage of data values lie between the mean and 2
standard deviations above the mean?
A. 68%
B. 13.5%
C. 95%
D. 99.7%
Answer: b) 13.5%
Rationale: In a normal distribution, approximately 13.5% of the data lies between the
mean and 2 standard deviations above the mean.
7. What is the variance of a binomial distribution with n=12n = 12n=12 and p=0.5p =
0.5p=0.5?
A. 6
B. 3
C. 2.4
D. 4
Answer: d) 4
Rationale: The variance of a binomial distribution is σ2=n⋅p⋅(1−p)\sigma^2 = n \cdot p
\cdot (1 - p)σ2=n⋅p⋅(1−p). For n=12n = 12n=12 and p=0.5p = 0.5p=0.5, the variance
is σ2=12⋅0.5⋅0.5=4\sigma^2 = 12 \cdot 0.5 \cdot 0.5 = 4σ2=12⋅0.5⋅0.5=4.
8. Which of the following is the formula for the binomial distribution?
A. P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-
k}P(X=k)=(kn)pk(1−p)n−k
B. P(X=k)=1k!e−1P(X=k) = \frac{1}{k!} e^{-1}P(X=k)=k!1e−1
C. P(X=k)=1k2P(X=k) = \frac{1}{k^2}P(X=k)=k21
D. P(X=k)=knP(X=k) = \frac{k}{n}P(X=k)=nk
Answer: a) P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-
k}P(X=k)=(kn)pk(1−p)n−k
Rationale: This is the standard formula for the probability mass function of a binomial
distribution, where nnn is the number of trials, kkk is the number of successes, and ppp
is the probability of success on a single trial.
9. Which of the following describes the shape of the distribution for a large number of
independent trials of a binomial experiment?
A. Uniform distribution
B. Normal distribution
C. Poisson distribution
D. Exponential distribution
Answer: b) Normal distribution
Rationale: According to the Central Limit Theorem, for a large number of trials, a
binomial distribution approaches a normal distribution.
10. A box contains 10 balls: 4 red, 3 blue, and 3 green. What is the probability of
selecting a red ball at random?
A. 1/10
