Question

A. The amount of snowfall in a certain mountain range is normally distributed with a mean of 91 inches and a standard deviation of 15 inches. What is the probability that the mean annual snowfall during 64 randomly chosen years will exceed 93.8 inches? Is this a rare event?

B. Assume that the population of human body temperatures has a mean of 98.6° F, and standard deviation is 0.62° F. If 106 people are randomly selected for evaluation, what is the probability that they will have a mean temperature of 98.4° or lower? Show work.

Answer 1

Solution :

A ) Given that,

mean = = 91

standard deviation = = 15

n = 64

=  91

₌ / n =  15 64 = 1.875

P ( >93.8 )

= 1 - P (< 93.8 )

= 1 - P ( - /) < ( 93.8 - 91 / 1.875)

= 1 - P ( z < 2.8 / 1.875 )

= 1 - P ( z < 1.49)

Using z table

= 1 - 0.9319

= 0.0681

Probability = 0.0681

B ) Given that,

mean = = 98.6

standard deviation = = 0.62

n = 106

=  98.6

₌ / n =  0.62 106 = 1.875

P( < 98.4 )

P ( - /) < ( 98.4 - 98.6 /0.0602)

P ( z < - 0.2 / 0.0602 )

P ( z < -3.23 )

= 0.0006

Probability = 0.0006