Question

Suppose X ~ N(6, 3^2).

a. Compute P(X > 9.9)

b. Determine the 95th percentile of X, that is, the constant c such that P(X < c)= 0.95

c. Find the mean and variance of

Y = 2X -1

Answer 1

Given that X follows normal distribution with mean = 6 and variance = 3^2 = 9

So standard deviation = 3

a) We have to find P( X> 9.9)

Using Z table we get P( Z <= 1.1) = 0.8665

So

b) Now we have to find 95Th percentile

i.e we have to find c such that P( X < c) = 0.95

using Z table we get ,

P( Z < 1.64) = 0.9495 and P( Z < 1.65) = 0.9505 ...................... Using Z table

So taking average of 1.64 and 1.65 we get P( Z < 1.645 ) = 0.95

So z = 1.645

c = 6 + 1.645 *3 = 10.935

So 95 Th percentile is 10.935

c)

Y = 2X - 1

mean of y is E(Y) = 2*E(X) - 1 = 2*6-1 = 12-1 = 11

So mean of y is 11

variance of y is V(X) = 2^2 * V(X) = 2^2 *3^2 = 4*9 =36

Variance of y is 36