Question

The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds. If a college football player is randomly​ selected, find the probability that he weighs between 170 and 220 pounds. Round to four decimal places.

Answer 1

Answer:

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 200, \sigma = 50$

Find the probability that he weighs between 170 and 220 pounds.

This is the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 170.

X = 220

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

$Z = \frac{220 - 200}{50}$

$Z = 0.4$

$Z = 0.4$ has a pvalue of 0.6554

X = 170

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

$Z = \frac{170 - 200}{50}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

0.6554 - 0.2743 = 0.3811

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.