1. In a normal distribution, approximately what percentage of data values lie within one
standard deviation of the mean?
A. 68%
B. 95%
C. 99.7%
D. 50%
Answer: a) 68%
Rationale: In a normal distribution, approximately 68% of data values lie within one
standard deviation of the mean.
2. The central limit theorem states that:
A. The distribution of the sample means is normally distributed.
B. The population must be normally distributed.
C. The variance of a sample increases as the sample size increases.
D. The sample mean is equal to the population mean.
Answer: a) The distribution of the sample means is normally distributed.
Rationale: The central limit theorem states that the distribution of sample means will
approach a normal distribution as the sample size increases, regardless of the
population distribution.
3. 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.
4. 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.
5. 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).
6. What is the standard deviation of a binomial distribution with n=8n = 8n=8 and
p=0.6p = 0.6p=0.6?
A. 1.6
B. 1.4
C. 2.4
D. 2.8
Answer: a) 1.6
Rationale: The standard deviation of a binomial distribution is given by
σ=n⋅p⋅(1−p)\sigma = \sqrt{n \cdot p \cdot (1 - p)}σ=n⋅p⋅(1−p). For n=8n = 8n=8 and
p=0.6p = 0.6p=0.6, the standard deviation is 8⋅0.6⋅0.4=1.6\sqrt{8 \cdot 0.6 \cdot
0.4} = 1.68⋅0.6⋅0.4=1.6.
7. A dataset has a mean of 30 and a standard deviation of 5. What is the z-score of the
value 40?
A. 2
B. 1
C. 0
D. 3
Answer: a) 2
Rationale: The z-score is Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ, where X=40X
= 40X=40, μ=30\mu = 30μ=30, and σ=5\sigma = 5σ=5. Therefore, Z=40−305=2Z =
\frac{40 - 30}{5} = 2Z=540−30=2.
8. A sample of 50 students has a mean score of 75 and a standard deviation of 10.
What is the standard error of the mean?
A. 1
B. 10
C. 1.41
D. 2
Answer: c) 1.41
Rationale: The standard error of the mean is given by SE=σn\text{SE} =
\frac{\sigma}{\sqrt{n}}SE=nσ, where σ=10\sigma = 10σ=10 and n=50n = 50n=50.
Therefore, SE=1050≈1.41\text{SE} = \frac{10}{\sqrt{50}} \approx
1.41SE=5010≈1.41.
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.
standard deviation of the mean?
A. 68%
B. 95%
C. 99.7%
D. 50%
Answer: a) 68%
Rationale: In a normal distribution, approximately 68% of data values lie within one
standard deviation of the mean.
2. The central limit theorem states that:
A. The distribution of the sample means is normally distributed.
B. The population must be normally distributed.
C. The variance of a sample increases as the sample size increases.
D. The sample mean is equal to the population mean.
Answer: a) The distribution of the sample means is normally distributed.
Rationale: The central limit theorem states that the distribution of sample means will
approach a normal distribution as the sample size increases, regardless of the
population distribution.
3. 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.
4. 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.
5. 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).
6. What is the standard deviation of a binomial distribution with n=8n = 8n=8 and
p=0.6p = 0.6p=0.6?
A. 1.6
B. 1.4
C. 2.4
D. 2.8
Answer: a) 1.6
Rationale: The standard deviation of a binomial distribution is given by
σ=n⋅p⋅(1−p)\sigma = \sqrt{n \cdot p \cdot (1 - p)}σ=n⋅p⋅(1−p). For n=8n = 8n=8 and
p=0.6p = 0.6p=0.6, the standard deviation is 8⋅0.6⋅0.4=1.6\sqrt{8 \cdot 0.6 \cdot
0.4} = 1.68⋅0.6⋅0.4=1.6.
7. A dataset has a mean of 30 and a standard deviation of 5. What is the z-score of the
value 40?
A. 2
B. 1
C. 0
D. 3
Answer: a) 2
Rationale: The z-score is Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ, where X=40X
= 40X=40, μ=30\mu = 30μ=30, and σ=5\sigma = 5σ=5. Therefore, Z=40−305=2Z =
\frac{40 - 30}{5} = 2Z=540−30=2.
8. A sample of 50 students has a mean score of 75 and a standard deviation of 10.
What is the standard error of the mean?
A. 1
B. 10
C. 1.41
D. 2
Answer: c) 1.41
Rationale: The standard error of the mean is given by SE=σn\text{SE} =
\frac{\sigma}{\sqrt{n}}SE=nσ, where σ=10\sigma = 10σ=10 and n=50n = 50n=50.
Therefore, SE=1050≈1.41\text{SE} = \frac{10}{\sqrt{50}} \approx
1.41SE=5010≈1.41.
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.
