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A random variable X has the probability ...

A random variable X has the probability distribution given below
Its variance is

A

`(16)/3`

B

`4/3`

C

`5/3`

D

`(10)/3`

Text Solution

Verified by Experts

The correct Answer is:
B
Given distribution is
`therefore` Variance `=Sigmax_i^2p-(Sigmax_ip)^2`
`=(1k+8k+27k+32k+25k)-(k+4k+9k+8k+5k)^2`
`=(93k)-(27k)^2=(93xx1/9)-(27xx1/9)^2" "[thereforeSigmap=1,"so "k=1/9]`
`=(93)/9-9=(93-81)/9=(12)/9=4/3`
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