Answer:
34%
Step-by-step explanation:
The body temperatures of adults have a mean of 98.6° F and a standard deviation of 0.60° F.
We want to find the probability that the mean body temperature of 36 randomly selected adults is greater than 98.5° F.
The mean of the sampling distribution of the sample means is the same as the population mean,
[tex] \mu = 98.6 \degree[/tex]
The standard error of the mean becomes the standard deviation of the sampling distribution of the means.
[tex]SE= \frac{ \sigma}{ \sqrt{n} } [/tex]
[tex]SE= \frac{0.6}{ \sqrt{36} } = 0.1[/tex]
By the Central Limit Theorem, the distribution of mean of means is approximately normal
The z-score of 98.5° F
[tex] z= \frac{98.5 - 98.6}{0.1} = - 1[/tex]
By the empirical rule 68% of the distribution is within one standard deviation (-1 to 1).
Therefore from (-1 to 0), it will be approximately 34%
