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**

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