Question

Let "Z" be a random variable from the standard normal distribution. Find the value for ? that satisfies each of the following probabilities.
(Round all answers to two decimal places)

A) P(Z < ?) = 0.6829.
    ? =  

B) P(Z > ?) = 0.3087.
    ? =  

C) P(-? < Z < ?) = 0.7402.
    ? = ±

Answer 1

Using Excel we find, z-scores for the following cases:

A) P(Z < z_{​​​​​​0}​​​​​) = 0.6829

Excel Command: = NORMSINV(0.6829)= 0.4758 (approx 0.48)

=> P(Z < 0.48) = 0.6829

=> z_{​​​​​​0} = 0.48

B) P(Z > z_{​​​​​​0}​​​) = 0.3087

=> 1 - P(Z < z_{​​​​​​0}​​​) = 0.3087

=> P(Z < z_{​​​​​​0}​​​) = 0.6913

Excel Command: = NORMSINV(0.6913)=0.4995 (approx 0.50)

=> P(Z > 0.50) = 0.3087

=> z_{​​​​​​0} = 0.50

C) P(-z_{​​​​​​0} < Z < z_{​​​​​​0}​​​​​) = 0.7402

=> 2 * P(Z < z_{​​​​​​0}​​​) - 1 = 0.7402

=> 2 * P(Z < z_{​​​​​​0}​​​​​) = 0.2598

=> P(Z < z_{​​​​​​0}) = 0.1299

Excel Command: = NORMSINV(0.1299)=-1.1269 (approx -1.13)

=> P(-1.13 < Z < 1.13) = 0.7402

=> z_{​​​​​​0} = ± 1.13