Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.0606.
A) −1.55
B) 1.55
C) 126
D) 374

Question

Quizplus · Accepted Answer

We need to convert the probability of P(X ≤ x) = 0.0606 to a z-score using the standard normal distribution table.

First, we calculate the z-score:

z = (x - µ) / σ

0.0606 = P(X ≤ x) => 1 - P(X > x) = 0.0606
P(Z ≤ z) = 0.0606
From the standard normal distribution table, the closest value to 0.0606 is 0.0618, which corresponds to a z-score of -1.56.

-1.56 = (x - 250) / 80

x = 126.72

Therefore, the value x such that P(X ≤ x) = 0.0606 is x = 126.72, which is closest to answer C: 126.

Let X Be Normally Distributed with Mean µ = 250