Question

QUESTION 9 Imagine that the mean height for all Division I women's basketball programs is 69 inches with a standard deviation of 3 inches. The 2010–2011 women's basketball team at the University of Connecticut, with 10 players listed on the roster, had an average height of 71.2 inches. Using the z statistic, what percent of means would fall below that for these UConn Huskies? (Round z score to two decimal places.)

Answer 1

Answer:

98.98% of means would fall below that for these UConn Huskies

Step-by-step explanation:

To solve this qustion, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 69, \sigma = 3, n = 10, s = \frac{3}{\sqrt{10}} = 0.9487$

The 2010–2011 women's basketball team at the University of Connecticut, with 10 players listed on the roster, had an average height of 71.2 inches. Using the z statistic, what percent of means would fall below that for these UConn Huskies?

This is the pvalue of Z when X = 71.2. So

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

By the Central limit theorem

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

$Z = \frac{71.2 - 69}{0.9487}$

$Z = 2.32$

$Z = 2.32$ has a pvalue of 0.9898

98.98% of means would fall below that for these UConn Huskies