Question

Suppose that the weights of the basketball players of a tournament are approximately normally distributed with a mean of 190lb and a standard deviation of 15lb.Which of the following is a good approximation of the probability that a randomly selected player weighs more than 220lb?a. 1%b. 2.5%c. 5%d. 16%

Answer 1

Answer:

b. 2.5%

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 190, \sigma = 15$

Which of the following is a good approximation of the probability that a randomly selected player weighs more than 220lb?

This is 1 subtracted by the pvalue of Z when X = 220. So

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

$Z = \frac{220 - 190}{15}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

Close to 2.5%

So the correct answer is:

b. 2.5%