Question

The body temperatures of adults are normally distributed with a mean of 98.6 degrees Fahrenheit and a standard deviation of 1.2 degrees Fahrenheit. If 31 adults are randomly selected, find the probability that their mean body temperature is greater than 99.3 degrees Fahrenheit.

Answer 1

Solution :

Given that ,

mean = = 98.6

standard deviation = = 1.2

= / n = 1.2 / 31 = 0.2155

P( > 99.3) = 1 - P( < 99.3)

= 1 - P[( - ) / < (99.3 - 98.6) / 0.2155]

= 1 - P(z < 3.25)

= 1 - 0.9994

= 0.0006

Probability = 0.0006