This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses 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) \(\frac{4}{21} \)

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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) \(\frac{1}{6} \)

View Answer

Explanation: p=0.5 (Probability of tail)

q=1-0.5=0.5

n=4 and x is binomial variate.

P (X=x) =

^{n}C

_{x}p

^{x}q

^{n-x}.

P (X=3) =

^{4}C

_{3}(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(X^{2})-(E(X))^{2}

c) E(X^{2})

d) 2E(X)

View Answer

Explanation: By property of Expectation

V (X) = E (X

^{2})-(E(X))

^{2}.

9. X is a variate between 0 and 3. The value of E(X^{2}) is ______

a) 8

b) 7

c) 27

d) 9

View Answer

Explanation: Integrating f(x) = x

^{2}from 0 to 3 we get E(X

^{2}) = 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

View Answer

Explanation: 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.

**Sanfoundry Global Education & Learning Series – Probability and Statistics.**

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