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 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:
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
has a pvalue of 0.7881
X = 180
has a pvalue of 0.2119
0.7881 - 0.2119 = 0.5762
57.62% of players weigh between 180 and 220 pounds
