Question

The random variable X~(30,2^2)
Find p(X<33)
Find p(X>26)

Answer 1

i) P(X<33) = 0.9232

ii) P(X>26) = 0.001

Explanation:

Step(i):-

Given that the mean of the Population = 30

Given that the standard deviation of the Population = 4

Let 'X' be the Normal distribution

Step(ii):-

i)

Given that the random variable X = 33

`Z = (x-mean)/(S.D)`

`Z = (33-30)/(2) = 1.5` >0

P(X<33) = P( Z<1.5)

= 1- P(Z>1.5)

= 1 - ( 0.5 - A(1.5))

= 0.5 + 0.4232

P(X<33) = 0.9232

Step(iii) :-

Given that the random variable X = 26

`Z = (x-mean)/(S.D)`

`Z = (33-26)/(2) = 3.5` >0

P(X>26) = P( Z>3.5)

= 0.5 - A(3.5)

= 0.5 - 0.4990

= 0.001

P(X>26) = 0.001