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A discrete random variable X has the probability functions as:

X

0

1

2

3

4

5

6

7

8

f(x)

K

2k

3k

5k

5k

4k

3k

2k

k

The value of E(X) is:

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SSC CGL Tier-II ( JSO ) 2019 Official Paper ( Held On : 17 Nov 2020 )
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  1. 97/26
  2. 107/26
  3. 93/26
  4. 103/26

Answer (Detailed Solution Below)

Option 4 : 103/26
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Detailed Solution

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Explanation:

x

F(x)

xf(x)

0

k

0

1

2k

2k

2

3k

6k

3

5k

15k

4

5k

20k

5

4k

20k

6

3k

18k

7

2k

14k

8

k

8k

Total

26k

103k

 

We know that for Fpr random variable X sum of Probaboility = 1

⇒ ∑Pi = 1

⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1

⇒ 26k = 1

⇒ k = 1/26

∴ xf(x) = 103k = 103 × 1/26

∴ E(x) is 103/26

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