A set of data has a mean of 56.1. The data follows a normal distribution curve and has a standard deviation of 8.2. Find the probability that a randomly selected value is greater than 67.5

A.

0.0823

B.

1.39

C.

0.9177

D.

-1.08

Given that mean=56.1 and standard deviation=8.2, P (x>67.5) will be found as follows:

The z-score is given by:

z = (x-μ) / σ

thus the z-score will be given by:

z = (67.5-56.1) / 8.2

z=11.4/8.2

z=1.39

thus

P (z=1.39) = 0.9177

thus:

P (x>67.5) = 1-P (z>0.9177)

=1-0.9177

=0.0823

Answer: A. 0.0823