Question

Healthy people have body temperatures that are normally distributed with a mean of 98.20∘F and a standard deviation of 0.62∘F

(a) If a healthy person is randomly selected, what is the probability that he or she has a temperature above 98.7∘F? answer:

(b) A hospital wants to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 1.5 % of healthy people to exceed it? answer:

NormalDistribution

Answer 1

Solution:

We are given

µ = 98.20

σ = 0.62

Part a

We have to find P(X>98.7)

P(X>98.7) = 1 - P(X<98.7)

Z = (X - µ)/σ

Z = (98.7 - 98.2)/0.62

Z = 0.806452

P(Z<0.806452) = P(X<98.7) = 0.790009

(by using z-table)

P(X>98.7) = 1 - P(X<98.7)

P(X>98.7) = 1 - 0.790009

P(X>98.7) = 0.209991

Required probability = 0.209991

Part b

The z value for the top 1.5% area by using z-table is given as below:

Z = 2.17009

X = µ + Z*σ

X = 98.20 + 2.17009*0.62

X = 99.54546

Answer: 99.5 ∘F