Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Random Variables”.
1. Consider a dice with the property that that probability of a face with n dots showing up is proportional to n. The probability of face showing 4 dots is?
a) (frac{1}{7} )
b) (frac{5}{42} )
c) (frac{1}{21} )
d) (frac{4}{21} )
Answer: d
Clarification: P (n) is proportional to n where n=
1,2,3,…6 is random variable.
P(n) = kn
P(1)+P(2)….P (6) = 1
K(1+2+3+4+5+6) = 1
(K=frac{1}{21} )
Hence P(4) = 4K = (frac{4}{21}. )
2. Let X be a random variable with probability distribution function f (x)=0.2 for |x|<1
= 0.1 for 1 < |x| < 4
= 0 otherwise
The probability P (0.5 < x < 5) is _____
a) 0.3
b) 0.5
c) 0.4
d) 0.8
Answer: c
Clarification: P (0.5 < x < 5) = Integrating f (x) from
0.5 to 5 by splitting in 3 parts that is from 0.5 to 1
and from 1 to 4 and 4 to 5 we get
P (0.5 < x < 5) = 0.1 + 0.3 + 0
P (0.5 < x < 5) = 0.4.
3. Runs scored by batsman in 5 one day matches are 50, 70, 82, 93, and 20. The standard deviation is ______
a) 25.79
b) 25.49
c) 25.29
d) 25.69
Answer: a
Clarification: The mean of 5 innings is
(50+70+82+93+20)÷5 = 63
S.D = [1⁄n (x(n)-mean)2]0.5
S.D = 25.79.
4. Find median and mode of the messages received on 9 consecutive days 15, 11, 9, 5, 18, 4, 15, 13, 17.
a) 13, 6
b) 13, 18
c) 18, 15
d) 15, 16
Answer: b
Clarification: Arranging the terms in ascending order 4, 5, 9, 11, 13, 14, 15, 18, 18.
Median is (frac{(n+1)}{2} ) term as n = 9 (odd) = 13.
Mode = 18 which is repeated twice.
5. Mode is the value of x where f(x) is a maximum if X is continuous.
a) True
b) False
Answer: a
Clarification: For a continuous variable mode is defined as the value where f(x) is a maximum or it is defined as the quantity repeated maximum number of times.
6. E (XY)=E (X)E (Y) if x and y are independent.
a) True
b) False
Answer: a
Clarification: By the property of Expectation
E (XY) = E (X) E (Y).
That is the Expectation of a composite function XY is the product of the individual expectations of X and Y.
7. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is ______
a) (frac{1}{2} )
b) (frac{1}{3} )
c) (frac{1}{4} )
d) (frac{1}{6} )
Answer: a
Clarification: p=0.5 (Probability of tail)
q=1-0.5=0.5
n=4 and x is binomial variate.
P (X=x) = nCx px qn-x.
P (X=3) = 4C3 (0.5)3 = 1⁄2.
8. If E denotes the expectation the variance of a random variable X is denoted as?
a) (E(X))2
b) E(X2)-(E(X))2
c) E(X2)
d) 2E(X)
Answer: b
Clarification: By property of Expectation
V (X) = E (X2)-(E(X))2.
9. X is a variate between 0 and 3. The value of E(X2) is ______
a) 8
b) 7
c) 27
d) 9
Answer: d
Clarification: Integrating f(x) = x2 from 0 to 3 we get E(X2) = 32 = 9.
10. The random variables X and Y have variances 0.2 and 0.5 respectively. Let Z= 5X-2Y. The variance of Z is?
a) 3
b) 4
c) 5
d) 7
Answer: d
Clarification: Var(X) = 0.2, Var(Y) = 0.5
Z = 5X – 2Y
Var(Z) = Var(5X-2Y)
= Var(5X) + Var(2Y)
= 25Var(X) + 4Var(Y)
Var(Z) = 7.