Question

Suppose X is a normal random variable with mean μ = 5 and P(X > 9) = 0.2005.

(i) Find approximately (using the Z-table) what is Var(X).

(ii) Find the value c such that P(X > c) = 0.1.

Answer 1

Solution :

mean = = 5

standard deviation = = ?

x = 9

Using standard normal table,

(i)

P(Z > z) = 0.2005

1 - P(Z < z) = 0.2005

P(Z < z) = 1 - 0.2005 = 0.7995

P(Z < 0.84) = 0.7995

z = 0.84

Using z-score formula,

x = z *  +

= (x - ) / z = (9 - 5) / 0.84 = 4 / 0.84 = 4.7619

variance = Var(X) =  ² = 22.68

(ii)

P(Z > z) = 0.1

1 - P(Z < z) = 0.1

P(Z < z) = 1 - 0.1 = 0.9

P(Z < 1.28) = 0.9

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 4.7619 + 5 = 11.1

c = 11.1