Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Mathematical Expectation”.
1. The expectation of a random variable X(continuous or discrete) is given by _________
a) ∑xf(x), ∫xf(x)
b) ∑x2 f(x), ∫x2 f(x)
c) ∑f(x), ∫f(x)
d) ∑xf(x2), ∫xf(x2)
Answer: a
Clarification: The expectation of a random variable X is given by the summation (integral) of x times the function in its interval. If it is a continuous random variable, then summation is used and if it is discrete random variable, then integral is used.
2. Mean of a random variable X is given by _________
a) E(X)
b) E(X2)
c) E(X2) – (E(X))2
d) (E(X))2
Answer: a
Clarification: Mean is defined as the sum of the function in its domain multiplied with the random variable’s value. Hence mean is given by E(X) where X is a random variable.
3. Variance of a random variable X is given by _________
a) E(X)
b) E(X2)
c) E(X2) – (E(X))2
d) (E(X))2
Answer: c
Clarification: Variance of a random variable is nothing but the expectation of the square of the random variable subtracted by the expectation of X (mean of X) to the power 2. Therefore the variance is given by E(X2) – (E(X))2.
4. Mean of a constant ‘a’ is ___________
a) 0
b) a
c) a/2
d) 1
Answer: b
Clarification: Let f(x) be the pdf of the random variable X.
Now, E(a) = ∫af(x)
= a∫f(x)
= a(1) = a.
5. Variance of a constant ‘a’ is _________
a) 0
b) a
c) a/2
d) 1
Answer: a
Clarification: V(a) = E(a2) – (E(X))2
= a2 – a2
= 0.
6. Find the mean and variance of X?
| x | 0 | 1 | 2 | 3 | 4 |
| f(x) | 1/9 | 2/9 | 3/9 | 2/9 | 1/9 |
a) 2, 4/3
b) 3, 4/3
c) 2, 2/3
d) 3, 2/3
Answer: a
Clarification: Mean = (E(X) = ∑f(x) = 0(frac{1}{9}) + 1(frac{2}{9}) + 2(frac{3}{9}) + 3(frac{2}{9}) + 4(1/9) )
= 2
Variance ( = E(X^2)-(E(X))^2 = (0 + frac{2}{9} + frac{12}{9} + frac{28}{9} + frac{26}{9}) – 4 )
( = frac{4}{3} ).
7. Find the expectation of a random variable X?
| x | 0 | 1 | 2 | 3 |
| f(x) | 1/6 | 2/6 | 2/6 | 1/6 |
a) 0.5
b) 1.5
c) 2.5
d) 3.5
Answer: b
Clarification: (E(X) = 0(frac{1}{6}) + 1(frac{2}{6}) + 2(frac{2}{6}) + 3(frac{1}{6}) = 1.5. )
8. Find the expectation of a random variable X if f(x) = ke-x for x>0 and 0 otherwise.
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: (int_0^∞ ke^{-x} dx = 1 )
kГ(1) = 1
k = 1
Now, (E(X) = int_0^∞ xe^{-x} dx = Г(2) = 1.)
9. Find the mean of a random variable X if f(x) = x – 5⁄2 for 0
b) 3.75
c) 2.5
d) 2.75
Answer: b
Clarification: (E(X) = int_0^1 (x-5/2)dx+∫_1^2(2x)dx+0 )
(= (frac{x^3}{3} – frac{5x^2}{4}) ) {from 0 to 1} ( + (frac{2x^3}{3}) ) {from 1 to 2}
(= frac{1}{3} – frac{5}{4} + frac{16}{3} – frac{2}{3} )
= 3.75.
10. Find the mean of a continuous random variable X if f(x) = 2e-x for x>0 and -ex for x<0.
a) 0
b) 1
c) 2
d) 3
Answer: d
Clarification: (E(X) = int_0^∞ 2xe^{-x} dx + int_{-∞}^0 xe^x dx )
= 2 Г(2) + Г(2) = 3.
11. What is moment generating function?
a) Mx(t) = E(etx)
b) Mx(t) = E(e-tx)
c) Mx(t) = E(e2tx)
d) Mx(t) = E(et)
Answer: a
Clarification: Moment generating function is nothing but the expectation of etX. So, the function is multiplied with etX before performing the integration or summation.
12. Find the Moment Generating Function of f(x) = x for 0
b) ((frac{e^{-t}-1}{t})^2 )
c) ((frac{e^{2t}-1}{t})^2 )
d) ((frac{e^{2t}-1}{t^2}) )
Answer: a
Clarification: Mx(t) = E(etx) = (int_0^1 xe^{tx} dx+int_1^2 (2-x) e^{tx} dx + 0 = (frac{e^t-1}{t})^2. )
13. E(X) = npq is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: b
Clarification: In binomial distribution, probability of success is given by p and that of failure is given by q and the event is done n times. The mean of this distribution is given by npq.
14. E(X) = λ is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: c
Clarification: In Poisson’s distribution, there is a positive constant λ which is the mean of the distribution and variance of the distribution.
15. E(X) = μ and V(X) = σ2 is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: d
Clarification: In Normal distribution, the mean and variance is given by μ and σ2 respectively. In case of standard normal distribution the mean is 0 and the variance is 1.