Question

the body temperatures of adults are normally distributed with a mean of 98.6°f and a standard deviation of 0.60°f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4°f. a. 0.9360 b. 0.9772 c. 0.0228 d. 0.8188

Explanation:

Step1: Calculate the standard error

The standard error formula is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 0.60$ and $n = 36$, then $\sigma_{\bar{x}}=\frac{0.60}{\sqrt{36}}=\frac{0.60}{6}=0.1$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=98.4$, $\mu = 98.6$, and $\sigma_{\bar{x}}=0.1$. So $z=\frac{98.4 - 98.6}{0.1}=\frac{- 0.2}{0.1}=-2$.

Step3: Find the probability

We want $P(\bar{X}>98.4)$. Since $P(\bar{X}>98.4)=P(Z>-2)$. And $P(Z > - 2)=1 - P(Z\leqslant-2)$. From the standard normal table, $P(Z\leqslant-2)=0.0228$. So $P(Z>-2)=1 - 0.0228 = 0.9772$.

Answer:

B. 0.9772