Question

question 9
the heights of nba basketball players have a normal distribution with a mean of 79 inches and a standard deviation of 3.89 inches. if a player reports his height had a z score of 3.5, would you believe this player? explain, showing math solution.
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12pt paragraph b i u a - 2 - t² - | :

Explanation:

Step1: Recall the z - score formula

The z - score formula is \(z=\frac{x - \mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. We know that \(\mu = 79\), \(\sigma=3.89\), and \(z = 3.5\).

Step2: Solve for \(x\)

Rearrange the formula \(z=\frac{x - \mu}{\sigma}\) to solve for \(x\). Multiply both sides by \(\sigma\): \(x-\mu=z\sigma\). Then add \(\mu\) to both sides: \(x=\mu + z\sigma\).
Substitute the given values: \(x=79+3.5\times3.89\).
First, calculate \(3.5\times3.89 = 13.615\).
Then, \(x=79 + 13.615=92.615\) inches.

Step3: Consider the context

In the context of human height (even for NBA players), a height of approximately \(92.615\) inches (\(7\) feet \(8.615\) inches) is extremely rare. While NBA players are tall, such a height is far beyond the typical range.

Answer:

I would not believe this player. Using the z - score formula \(x=\mu + z\sigma\) with \(\mu = 79\), \(\sigma = 3.89\), and \(z = 3.5\), we get \(x=79+3.5\times3.89=79 + 13.615 = 92.615\) inches. Such a height (over \(7\) feet \(8.6\) inches) is extremely unusual for an NBA player (or a human in general), so it is likely that the reported z - score is incorrect or there is an error in the claim.