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 67.5A. 0.0823B. 1.39C. 0.9177D. –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