Find the expected value, variance and standard deviation of the random variable whose p.m.f.’s are given below : x = x 0 1 2 3 4 5 P (X = x) 132 532 1032 1032 532 132 - Mathematics and Statistics

Advertisement
Advertisement
Advertisement
Sum

Find the expected value, variance, and standard deviation of the random variable whose p.m.f.’s are given below :

x = x 0 1 2 3 4 5
P (X = x) `1/32` `5/32` `10/32` `10/32` `5/32` `1/32`
Advertisement

Solution

We construct the following table to find the expected value, variance and standard deviation:

xi P (xi) xi·P (xi) xi 2·P (xi) = xi × xi·P (xi)
0 `1/32` 0 0
1 `5/32` `5/32` `5/32`
2 `10/32` `20/32` `40/32`
3 `10/32`   `30/32` `90/32`
4 `5/32` `20/32` `80/32`
5 `1/32` `5/32` `25/32`
Total `32/32` `80/32` `240/32`

From the table,

∑ xi . P (xi)  = `80/32 = 5/2` ,
∑ xi2 . P ( xi) = `240/32 = 15/2`

Expected value = E (X) = ∑ xi . P (xi

= `5/2` = 2.5

Variance = V(x) = ∑ xi2 . P (xi) - [ ∑ xi·P (xi) ]2

= `15/2 - (5/2)^2`

= `15/2 - 25/4`

= `30/4 - 25/4`

= `5/4`

= 1.25

Standard deviation = `sqrt(V(X)) = sqrt 1.25`
= 1.118

  Is there an error in this question or solution?
Chapter 7: Probability Distributions - Miscellaneous Exercise 2 [Page 244]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 10.4 | Page 244
Share
Notifications



      Forgot password?
Use app×