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. β1.08
Accepted Solution
A:
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-ΞΌ)/Ο 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