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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 98.6, \sigma = 0.6, n = 36, s = \frac{0.6}{\sqrt{36}} = 0.1](https://tex.z-dn.net/?f=%5Cmu%20%3D%2098.6%2C%20%5Csigma%20%3D%200.6%2C%20n%20%3D%2036%2C%20s%20%3D%20%5Cfrac%7B0.6%7D%7B%5Csqrt%7B36%7D%7D%20%3D%200.1)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{98.4 - 98.6}{0.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B98.4%20-%2098.6%7D%7B0.1%7D)
![Z = -2](https://tex.z-dn.net/?f=Z%20%3D%20-2)
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.