A normal population has a mean $\mu = 30$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than 31 ?
A) 0.4443
B) 0.7881
C) 0.5557
D) 0.7580

Question

A normal population has a mean $\mu = 30$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than  31 ? 
A)  0.4443 
B)  0.7881 
C)  0.5557 
D)  0.7580

Quizplus · Accepted Answer

The probability that a randomly chosen value will be greater than 31 is 0.4443. This is because the z-score for a value of 31 is (31 - 30) / 7 = 0.1429, and the area under the normal curve to the right of a z-score of 0.1429 is 0.4443.

A Normal Population Has a Mean μ=30\mu = 30μ=30 And Standard Deviation

A Normal Population Has a Mean $\mu = 30$ And Standard Deviation