Question

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds,What percent of players weigh between 180 and 220 pounds

Answer 1

Answer:

57.62% of players weigh between 180 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:

$\mu = 200, \sigma = 25$

What percent of players weigh between 180 and 220 pounds

We have to find the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 180.

X = 220

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

$Z = \frac{220 - 200}{25}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881

X = 180

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

$Z = \frac{180 - 200}{25}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% of players weigh between 180 and 220 pounds