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.

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 and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.6554

X = 170

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.

282 6/7 cm^2

step-by-step explanation:

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