Question

Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.65 degrees and a standard deviation of 0.8 degrees. If 50 healthy people are selected at random, what is the probability that the average temperature for these people is 98.45 degrees or higher?

Answer 1

$\mu = 98.65 \\ \sigma = 0.8 \\ n = 50$

The mean of the sampling distribution of the sample mean is equal to the population mean:

$\mu_{\bar{x}} = \mu$

The standard deviation of the sampling distribution of the sample mean:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{50}} = 0.113$

The z-score is the value decreased by the mean, divided by the standard deviation.

$z = \frac{\bar{x} - \mu_{bar{x}}}{\sigma_{\bar{x}}} = \frac{98.45 -98.65}{0.113} = -1.769 P(\bar{x}> 98.45) = 1 -P(\bar{x}<98.45) \\ = 1 -P(z< -1.769) \\ = 1 -0.0384 \\ = 0.9616$