Question

g Healty 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 99∘F?

Answer 1

9.85% probability that he or she has a temperature above 99∘F.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 98.2, \sigma = 0.62`

If a healthy person is randomly selected, what is the probability that he or she has a temperature above 99∘F?

This is 1 subtracted by the pvalue of Z when X = 99. So

`Z = (X - \mu)/(\sigma)`

`Z = (99 - 98.2)/(0.62)`

`Z = 1.29`

`Z = 1.29` has a pvalue of 0.9015

1 - 0.9015 = 0.0985

9.85% probability that he or she has a temperature above 99∘F.