Answer to Question #216526 in Statistics and Probability for aasa

Question

Given that Z is a standard normal random variable, find Z for each

situation.

a) The area between -Z and Z is .9030.

b) The area between -Z and Z is .9528.

Answer 1

a)

$P(-z < Z < z) = 0.9030 \\ P(Z < z) - (1-P(Z < z)) = 0.9030 \\ 2P(Z < z) - 1 = 0.9030 \\ 2P(Z < z) = 1 + 0.9030 \\ 2P(Z < z) = 1.9030 \\ P(Z < z) = \frac{1.9030 }{ 2 } \\ P(Z < z) = 0.9515$

See the probability 0.9515 in the standard normal table the corresponding z value 1.66

$P(Z < 1.66) = 0.9515 \\ z = 1.66$

b)

$P(-z < Z < z) = 0.9528 \\ P(Z < z) - (1-P(Z < z)) = 0.9528 \\ 2P(Z < z) - 1 = 0.9528 \\ 2P(Z < z) = 1 + 0.9528 \\ 2P(Z < z) = 1.9528 \\ P(Z < z) = \frac{1.9528 }{ 2 } \\ P(Z < z) = 0.9764$

See the probability 0.9764 in the standard normal table the corresponding z value 1.98

$P(Z< 1.98) = 0.9764 \\ z= 1.98$