Question

Weights of American adults are normally distributed with a mean of 180 pounds and a standard deviation of 8 pounds. What is the probability that a randomly selected individual will be between 185 and 190 pounds? Solution: pnorm(190-185,180,8) = 2.24756e-106, which is essentially 0. What was done wrong in this solution?

Answer 1

Answer:

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 180, \sigma = 8$

What is the probability that a randomly selected individual will be between 185 and 190 pounds?

This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So

X = 190

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

$Z = \frac{190 - 180}{8}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

X = 185

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

$Z = \frac{185 - 180}{8}$

$Z = 0.63$

$Z = 0.63$ has a pvalue of 0.7357

0.8944 - 0.7357 = 0.1587

15.87% probability that a randomly selected individual will be between 185 and 190 pounds