Test Index

Statistics Part 2

Section: Mathematics

© examsnet.com

Question : 16 of 50

Marks: +1, -0

A random variable

X

0,1,2,3

with probabilities

\; \frac{2a+1}{30},\; \frac{8a-1}{30},\; \frac{4a+1}{30}, b

respectively, where

a, b \in R

\mu

\sigma

respectively be the mean and standard deviation of

X

\sigma^2+\mu^2=2

\; \frac{a}{b}

[21 Jan 2026 Shift 2]

12
60
30
3

Solution:

Concept:

Use properties of probability distribution: sum of probabilities equals 1, and

E(X^2) = \sigma^2 + \mu^2 = 2

.

Explanation:

Let

p(0)=\frac{2a+1}{30}

,

p(1)=\frac{8a-1}{30}

,

p(2)=\frac{4a+1}{30}

,

p(3)=b

.

Since

\sum p = 1

, we have

\frac{2a+1}{30}+\frac{8a-1}{30}+\frac{4a+1}{30}+b=1

.

Combine numerators:

\frac{14a+1}{30}+b=1 \Rightarrow 14a+1+30b=30 \Rightarrow 14a+30b-29=0

...(1)

Given

\sigma^2+\mu^2=2

and

\sigma^2=E(X^2)-\mu^2

, so

E(X^2)=2

.

E(X^2)=\sum x^2 p(x) = 0^2\cdot p(0)+1^2\cdot p(1)+2^2\cdot p(2)+3^2\cdot p(3)

.

= \frac{8a-1}{30}+4\times\frac{4a+1}{30}+9b = \frac{8a-1+16a+4}{30}+9b = \frac{24a+3}{30}+9b

.

Set equal to 2:

\frac{24a+3}{30}+9b=2

. Multiply by 30:

24a+3+270b=60 \Rightarrow 24a+270b=57

.

Divide by 3:

8a+90b=19 \Rightarrow 8a+90b-19=0

...(2)

Solve (1) and (2): Multiply (1) by 3:

42a+90b-87=0

. Subtract (2):

(42a+90b-87)-(8a+90b-19)=0 \Rightarrow 34a-68=0 \Rightarrow a=2

.

Substitute

a=2

in (1):

14(2)+30b-29=0 \Rightarrow 28+30b-29=0 \Rightarrow 30b=1 \Rightarrow b=\frac{1}{30}

.

Thus

\frac{a}{b}=2 \div \frac{1}{30}=60

.

Answer:

\boxed{60}

(Option B)

© examsnet.com

Go to Question:

Prev Question

Next Question