1. If a random variable X follows a Poisson distribution with a mean of 4, what is the
probability of observing exactly 2 events?
A. 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
B. 24e−24!\frac{2^4 e^{-2}}{4!}4!24e−2
C. 43e−43!\frac{4^3 e^{-4}}{3!}3!43e−4
D. 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
Answer: a) 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
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, where λ=4\lambda = 4λ=4 and
k=2k = 2k=2. Thus, P(X=2)=42e−42!P(X=2) = \frac{4^2 e^{-
4}}{2!}P(X=2)=2!42e−4.
2. What is the mode of a normal distribution?
A. The value with the highest frequency.
B. The value at the upper quartile.
C. The mean value.
D. The median value.
Answer: a) The value with the highest frequency.
Rationale: In a normal distribution, the mode is the value that occurs most frequently
and is located at the peak of the distribution, which is also the mean and median.
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. What is the variance of a binomial distribution with parameters n=10n = 10n=10
and p=0.4p = 0.4p=0.4?
A. 4
B. 6
C. 2.4
D. 1.6
Answer: c) 2.4
Rationale: The variance of a binomial distribution is given by σ2=n⋅p⋅(1−p)\sigma^2 =
n \cdot p \cdot (1 - p)σ2=n⋅p⋅(1−p). Therefore, σ2=10⋅0.4⋅0.6=2.4\sigma^2 = 10
\cdot 0.4 \cdot 0.6 = 2.4σ2=10⋅0.4⋅0.6=2.4.
5. In a binomial distribution, if n=4n = 4n=4 and p=0.5p = 0.5p=0.5, what is the
probability of getting exactly 2 successes?
A. 6/166/166/16
B. 1/21/21/2
C. 4/164/164/16
D. 1/41/41/4
, Answer: a) 6/166/166/16
Rationale: Using the binomial probability formula, 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, with n=4n = 4n=4, p=0.5p
= 0.5p=0.5, and k=2k = 2k=2, the probability is (42)(0.5)2(0.5)2=6/16\binom{4}{2}
(0.5)^2 (0.5)^2 = 6/16(24)(0.5)2(0.5)2=6/16.
6. 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.
7. The value of the correlation coefficient lies between:
A. -1 and 1
B. 0 and 1
C. -1 and 0
D. -∞ and ∞
Answer: a) -1 and 1
Rationale: The correlation coefficient ranges from -1 (perfect negative correlation) to 1
(perfect positive correlation).
8. 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
B. 2/5
C. 3/10
D. 4/10
Answer: b) 2/5
Rationale: There are 4 red balls and 10 balls in total. The probability is
410=25\frac{4}{10} = \frac{2}{5}104=52.
9. In a simple random sample, what is the probability of selecting a specific individual if
the sample size is 20 and the population size is 100?
A. 1/20
B. 1/100
C. 20/100
D. 5/100
Answer: b) 1/100
Rationale: In a simple random sample, the probability of selecting a specific individual
is the same for each individual and is 1population size\frac{1}{\text{population
size}}population size1. Therefore, 1100\frac{1}{100}1001.
10. What is the variance of a binomial distribution with n=12n = 12n=12 and p=0.5p =
0.5p=0.5?
probability of observing exactly 2 events?
A. 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
B. 24e−24!\frac{2^4 e^{-2}}{4!}4!24e−2
C. 43e−43!\frac{4^3 e^{-4}}{3!}3!43e−4
D. 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
Answer: a) 42e−42!\frac{4^2 e^{-4}}{2!}2!42e−4
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, where λ=4\lambda = 4λ=4 and
k=2k = 2k=2. Thus, P(X=2)=42e−42!P(X=2) = \frac{4^2 e^{-
4}}{2!}P(X=2)=2!42e−4.
2. What is the mode of a normal distribution?
A. The value with the highest frequency.
B. The value at the upper quartile.
C. The mean value.
D. The median value.
Answer: a) The value with the highest frequency.
Rationale: In a normal distribution, the mode is the value that occurs most frequently
and is located at the peak of the distribution, which is also the mean and median.
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. What is the variance of a binomial distribution with parameters n=10n = 10n=10
and p=0.4p = 0.4p=0.4?
A. 4
B. 6
C. 2.4
D. 1.6
Answer: c) 2.4
Rationale: The variance of a binomial distribution is given by σ2=n⋅p⋅(1−p)\sigma^2 =
n \cdot p \cdot (1 - p)σ2=n⋅p⋅(1−p). Therefore, σ2=10⋅0.4⋅0.6=2.4\sigma^2 = 10
\cdot 0.4 \cdot 0.6 = 2.4σ2=10⋅0.4⋅0.6=2.4.
5. In a binomial distribution, if n=4n = 4n=4 and p=0.5p = 0.5p=0.5, what is the
probability of getting exactly 2 successes?
A. 6/166/166/16
B. 1/21/21/2
C. 4/164/164/16
D. 1/41/41/4
, Answer: a) 6/166/166/16
Rationale: Using the binomial probability formula, 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, with n=4n = 4n=4, p=0.5p
= 0.5p=0.5, and k=2k = 2k=2, the probability is (42)(0.5)2(0.5)2=6/16\binom{4}{2}
(0.5)^2 (0.5)^2 = 6/16(24)(0.5)2(0.5)2=6/16.
6. 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.
7. The value of the correlation coefficient lies between:
A. -1 and 1
B. 0 and 1
C. -1 and 0
D. -∞ and ∞
Answer: a) -1 and 1
Rationale: The correlation coefficient ranges from -1 (perfect negative correlation) to 1
(perfect positive correlation).
8. 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
B. 2/5
C. 3/10
D. 4/10
Answer: b) 2/5
Rationale: There are 4 red balls and 10 balls in total. The probability is
410=25\frac{4}{10} = \frac{2}{5}104=52.
9. In a simple random sample, what is the probability of selecting a specific individual if
the sample size is 20 and the population size is 100?
A. 1/20
B. 1/100
C. 20/100
D. 5/100
Answer: b) 1/100
Rationale: In a simple random sample, the probability of selecting a specific individual
is the same for each individual and is 1population size\frac{1}{\text{population
size}}population size1. Therefore, 1100\frac{1}{100}1001.
10. What is the variance of a binomial distribution with n=12n = 12n=12 and p=0.5p =
0.5p=0.5?
