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.)
A) 98.98
B) 94.98
C) 86.98
D) 48.98

Question

Quizplus · Accepted Answer

First, we need to calculate the z score:

z = (x - μ) / (σ / √n)

Where x is the sample average height, μ is the population mean height, σ is the population standard deviation, and n is the sample size.

Plugging in the numbers:

z = (71.2 - 69) / (3 / √10) = 2.99

Looking at a standard normal distribution table, we can see that the percentage of means that fall below a z score of 2.99 is 0.998. Multiplying by 100 to get a percentage, we get 99.8%.

However, we need to take into account that we are looking for the percentage of means *below* the UConn Huskies' mean height of 71.2 inches. Since we're dealing with a normal distribution, we can flip the z score and look for the percentage of means above a z score of -2.99:

P(z > -2.99) = 1 - P(z ≤ -2.99) = 1 - 0.001 = 0.999

Multiplying by 100, we get 99.9%.

Finally, since we're looking for the percentage of means below the UConn Huskies' mean height, we need to subtract 99.9% from 100%:

100% - 99.9% = 0.1%

Rounding to two decimal places, we get 0.10%, which is closest to choice A (98.98%).

Imagine That the Mean Height for All Division I Women's