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bagirrra123 [75]
3 years ago
13

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds,What perc

ent of players weigh between 180 and 220 pounds
Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
4 0

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

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