Probability and Statistics Multiple Choice Questions & Answers (MCQs) 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) )
Answer: a
Clarification: 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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} )
Answer: c
Clarification: (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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: (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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: The mean and variance of Poisson’s distribution are the same which is equal to λ.