Question

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees° F. If 36 adults are randomly​ selected, find the probability that their mean body temperature is greater than 98.4degrees° F.

Answer 1

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be 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:

$Z = \frac{X - \mu}{\sigma}$

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 98.6, \sigma = 0.6, n = 36, s = \frac{0.6}{\sqrt{36}} = 0.1$

If 36 adults are randomly​ selected, find the probability that their mean body temperature is greater than 98.4degrees° F.

This is 1 subtracted by the pvalue of Z when X = 98.4. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{98.4 - 98.6}{0.1}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that their mean body temperature is greater than 98.4degrees° F.