Let Z be the standard normal variable Find the values of a t

Let Z be the standard normal variable. Find the values of a that satisfy the given probabilities. (Give your answers to four decimal places.)

(a) P(Z > a) = 0.1810

(b) P(-a < Z < a) = 0.3981

Solution

a)

P( z > a ) = 0.7731

1 - 0.7731 = 0.2269

so since the probability is greater than 0.50 we suppose that is negative

a = -0.75

b)

1 - 0.4098 = 0.5902 / 2 = 0.2951

a = 0.54