Question

A state park has normally distributed daytime temperatures in the month of June. The mean temperature is 75F and the standard deviation of 7F. What is the probability the temperature is above 82F?

Answer 1

Answer:

0.1587 or 15.87%

Step-by-step explanation:

• Standardize and find your Z value as: Z = (82-75)/7 = 1.00
• Find in the Z score table the probability P( Z< 1.00), which is = 0.8413 = 84.13%
• Compute the complementary as (1-P), which is = 1 - 0.8413 = 0.1587 = 15.87%