This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Mean and Variance of Distribution”.
1. The expectation of a random variable X (E(X)) can be written as _________
a) \(\frac{d}{dt} [M_X (t)](t=0) \)
b) \(\frac{d}{dx} [M_X (t)](t=0) \)
c) \(\frac{d^2}{dt^2} [M_X (t)](t=0) \)
d) \(\frac{d^2}{dx^2} [M_X (t)](t=0) \)
View Answer
Explanation: Expectation of a random variable X can be written as the first differentiation of Moment generating function, which can be written as \(\frac{d}{dt} [M_X (t)](t=0). \)
2. If the probability of hitting the target is 0.4, find mean and variance.
a) 0.4, 0.24
b) 0.6, 0.24
c) 0.4, 0.16
d) 0.6, 0.16
View Answer
Explanation: p = 0.4
q = 1-p
= 1-0.4 = 0.6
Therefore, mean = p = 0.4 and
Variance = pq = (0.4) (0.6) = 0.24.
3. If the probability that a bomb dropped from a place will strike the target is 60% and if 10 bombs are dropped, find mean and variance?
a) 0.6, 0.24
b) 6, 2.4
c) 0.4, 0.16
d) 4, 1.6
View Answer
Explanation: Here, p = 60% = 0.6 and q = 1-p = 40% = 0.4 and n = 10
Therefore, mean = np = 6
Variance = npq = (10)(0.6)(0.4)
= 2.4.
4. If P(1) = P(3) in Poisson’s distribution, what is the mean?
a) \(\sqrt{2} \)
b) \(\sqrt{3} \)
c) \(\sqrt{6} \)
d) \(\sqrt{7} \)
View Answer
Explanation: \(P(x) = \frac{(e^{-λ} λ^x)}{x!} \)
Therefore, \(P(3) = \frac{(e^{-λ} λ^3)}{3!} \)
and \(P(1) = \frac{(e^{-λ} λ^1)}{1!} \)
P(1) = P(2)
\(λ=\frac{λ^3}{6} \)
Therefore, \(λ=\sqrt{6}. \)
5. What is the mean and variance for standard normal distribution?
a) Mean is 0 and variance is 1
b) Mean is 1 and variance is 0
c) Mean is 0 and variance is ∞
d) Mean is ∞ and variance is 0
View Answer
Explanation: The mean and variance for the standard normal distribution is 0 and 1 respectively.
6. Find λ in Poisson’s distribution if the probabilities of getting a head in biased coin toss as \(\frac{3}{4} \) and 6 coins are tossed.
a) 3.5
b) 4.5
c) 5.5
d) 6.6
View Answer
Explanation: p = 3⁄4
λ = np = (6) 3⁄4 = 4.5.
7. If P(6) = λP(1) in Poisson’s distribution, what is the mean?(Approximate value)
a) 4
b) 6
c) 5
d) 7
View Answer
Explanation: \(\frac{e^{-λ} λ^6}{6!}= λ \frac{e^{-λ} λ^1}{1!} \)
λ4 = 6! = 720
Therefore λ = 5.18 = 5.
8. Find f(2) in normal distribution if mean is 0 and variance is 1.
a) 0.1468
b) 0.1568
c) 0.1668
d) 0.1768
View Answer
Explanation: Given mean = 0
Variance = 1
\(f(2) = \frac{1}{(\sqrt{2π})} e^{\frac{-1}{2} \frac{2}{1}}= 0.1468. \)
9. Find the mean of tossing 8 coins.
a) 2
b) 4
c) 8
d) 1
View Answer
Explanation: p = 1⁄2
n = 8
q = 1⁄2
Therefore, mean = np = 8 * 1⁄2 = 4.
10. Mean and variance of Poisson’s distribution is the same.
a) True
b) False
View Answer
Explanation: The mean and variance of Poisson’s distribution are the same which is equal to λ.
Sanfoundry Global Education & Learning Series – Probability and Statistics.
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