Question

Suppose X is a normal random variable with mean μ = 100 and standard deviation σ = 7. Find b such that

P(100 ≤ X ≤ b) = 0.3.

HINT [See Example 3.] (Round your answer to one decimal place.)
b =

Answer 1

Using standard normal conversion,

P(X < x) = P(Z < (x - ) / )

P(100 < X < b) = 0.3

P(X < b) - P(X < 100) = 0.3

P(Z < ( b - 100) / 7 ) - P(Z < (100 - 100) / 7 ) = 0.3

P(Z < ( b - 100) / 7 ) - P(Z < 0) = 0.3

P(Z < ( b - 100) / 7 ) - 0.5 = 0.3

P(Z < ( b - 100) / 7 ) = 0.8

From Z table, z-score for the probability of 0.8 is 0.8416

( b - 100) / 7 = 0.8416

b = 105.9