Question

Assume that the population of human body temperatures has a mean of 98.6 degrees F and a standard deviation of 0.62 degrees F. If a sample of size n = 106 is randomly selected, find the probability of getting a mean temperature of 98.2 degrees F or lower.

Answer 1

Answer:

0% probability of getting a mean temperature of 98.2 degrees F or lower.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

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.

In this problem, we have that:

$\mu = 98.6, \sigma = 0.62, n = 106, s = \frac{0.62}{\sqrt{106}} = 0.06$

Find the probability of getting a mean temperature of 98.2 degrees F or lower.

This is the pvalue of Z when X = 98.2. So

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

$Z = \frac{98.2 - 98.6}{0.06}$

$Z = -6.67$

$Z = -6.67$ has a pvalue of 0.

So there is a 0% probability of getting a mean temperature of 98.2 degrees F or lower.