QMB 3200 ACTUAL EXAM PAPER 2026
QUESTIONS WITH SOLUTIONS GRADED A+
◉ A sample statistic, such as x bar , that estimates the value of the
corresponding population parameter is known as a:
a. population parameter
b. point estimator
c. Both a parameter and a population parameter are correct.
d. parameter. Answer: b. point estimator
◉ The standard deviation of xbar is referred to as the
a. standard error of the mean
b. standard x
c. sample standard mean
d. sample mean deviation. Answer: a. standard error of the mean
◉ Which of the following is(are) point estimator(s)?
a. σ
b. μ
c. s
d. All of these answers are correct.. Answer: c. s
,◉ A probability distribution for all possible values of a sample statistic
is known as a
a. parameter
b. sample statistic
c. sampling distribution
d. simple random sample. Answer: c. sampling distribution
◉ A population has a mean of 180 and a standard deviation of 24. A
sample of 64 observations will be taken. The probability that the mean
from that sample will be between 183 and 186 is
a. 0.8185
b. 0.1359
c. 0.4772
d. 0.3413. Answer: b. 0.1359
1. st. error: 24/sqrt(64) = 3
2. z value (upper): (186-180)/3 = 2
3. Norm.s.dist(1, true) = .9772
4. z value (lower): (183-180)/3= 1
5.Norm.s.dist(-1,true)= .8413
6. 0.9772-.8413=0.1359
,◉ The basis for using a normal probability distribution to approximate
the sampling distribution of xbar and pbar is:
a. the empirical rule
b. the central limit theorem
c. Chebyshev's theorem
d. Bayes' theorem. Answer: b. the central limit theorem
◉ A finite population correction factor is needed in computing the
standard deviation of the sampling distribution of sample means:
a. whenever the sample size is less than 5% of the population size
b. whenever the population is infinite
c. whenever the sample size is more than 5% of the population size
d. The correction factor is not necessary if the population has a normal
distribution. Answer: c. whenever the sample size is more than 5% of the
population size
◉ A sample of 66 observations will be taken from a process (an infinite
population). The population proportion equals 0.12. The probability that
the sample proportion will be less than 0.1768 is
a. 0.4222
b. 0.0778
c. 0.9222
d. 0.0568. Answer: c. 0.9222
, 1. Mean of the sample proportions = 0.12
2. St. dev. of pbar = sqrt [ p(1-p) / n] = sqrt [ (0.12)(1-.12) / 66] = .04
μ = 0.12
σ = 0.04
3. standardize x to z = (x - μ) / σ = (0.1768-0.12) / 0.04 = 1.42
4. norm.s.dist(1.42,true) = 0.9222
◉ A population has a mean of 53 and a standard deviation of 21. A
sample of 49 observations will be taken. The probability that the sample
mean will be greater than 57.95 is
a. 0
b. .0495
c. .4505
d. None of the alternative answers is correct.. Answer: b. .0495
1. 21/sqrt(49)=3
2. 57.93-53=4.95
3. 4.95/3 = 1.65
4.norm.s.dist(1.65,true) = .9505
5. 1-.9505= .0495
◉ A numerical measure from a sample, such as a sample mean, is known
as
QUESTIONS WITH SOLUTIONS GRADED A+
◉ A sample statistic, such as x bar , that estimates the value of the
corresponding population parameter is known as a:
a. population parameter
b. point estimator
c. Both a parameter and a population parameter are correct.
d. parameter. Answer: b. point estimator
◉ The standard deviation of xbar is referred to as the
a. standard error of the mean
b. standard x
c. sample standard mean
d. sample mean deviation. Answer: a. standard error of the mean
◉ Which of the following is(are) point estimator(s)?
a. σ
b. μ
c. s
d. All of these answers are correct.. Answer: c. s
,◉ A probability distribution for all possible values of a sample statistic
is known as a
a. parameter
b. sample statistic
c. sampling distribution
d. simple random sample. Answer: c. sampling distribution
◉ A population has a mean of 180 and a standard deviation of 24. A
sample of 64 observations will be taken. The probability that the mean
from that sample will be between 183 and 186 is
a. 0.8185
b. 0.1359
c. 0.4772
d. 0.3413. Answer: b. 0.1359
1. st. error: 24/sqrt(64) = 3
2. z value (upper): (186-180)/3 = 2
3. Norm.s.dist(1, true) = .9772
4. z value (lower): (183-180)/3= 1
5.Norm.s.dist(-1,true)= .8413
6. 0.9772-.8413=0.1359
,◉ The basis for using a normal probability distribution to approximate
the sampling distribution of xbar and pbar is:
a. the empirical rule
b. the central limit theorem
c. Chebyshev's theorem
d. Bayes' theorem. Answer: b. the central limit theorem
◉ A finite population correction factor is needed in computing the
standard deviation of the sampling distribution of sample means:
a. whenever the sample size is less than 5% of the population size
b. whenever the population is infinite
c. whenever the sample size is more than 5% of the population size
d. The correction factor is not necessary if the population has a normal
distribution. Answer: c. whenever the sample size is more than 5% of the
population size
◉ A sample of 66 observations will be taken from a process (an infinite
population). The population proportion equals 0.12. The probability that
the sample proportion will be less than 0.1768 is
a. 0.4222
b. 0.0778
c. 0.9222
d. 0.0568. Answer: c. 0.9222
, 1. Mean of the sample proportions = 0.12
2. St. dev. of pbar = sqrt [ p(1-p) / n] = sqrt [ (0.12)(1-.12) / 66] = .04
μ = 0.12
σ = 0.04
3. standardize x to z = (x - μ) / σ = (0.1768-0.12) / 0.04 = 1.42
4. norm.s.dist(1.42,true) = 0.9222
◉ A population has a mean of 53 and a standard deviation of 21. A
sample of 49 observations will be taken. The probability that the sample
mean will be greater than 57.95 is
a. 0
b. .0495
c. .4505
d. None of the alternative answers is correct.. Answer: b. .0495
1. 21/sqrt(49)=3
2. 57.93-53=4.95
3. 4.95/3 = 1.65
4.norm.s.dist(1.65,true) = .9505
5. 1-.9505= .0495
◉ A numerical measure from a sample, such as a sample mean, is known
as
