Question

the distribution of height for national basketball association (nba) players is approximately normal with a mean of 78.4 inches. if 5.7% of players have heights greater than 84 inches, calculate the standard deviation of the distribution. (round to 3 decimal places.)

Explanation:

Step1: Identify the given values

We know that the height distribution is normal with mean $\mu = 78.4$ inches. Let the standard deviation be $\sigma$. We are given that $P(X > 84)=0.057$. For a normal distribution, we can use the z - score formula $z=\frac{x - \mu}{\sigma}$. First, we need to find the z - score corresponding to the right - tailed probability of $0.057$. The left - tailed probability corresponding to $x = 84$ is $P(X\leq84)=1 - 0.057 = 0.943$.

Step2: Find the z - score

We look up the z - score in the standard normal distribution table (or use a calculator) that corresponds to a cumulative probability of $0.943$. Using a standard normal table or a calculator (e.g., the inverse of the standard normal CDF), we find that the z - score $z$ such that $P(Z\leq z)=0.943$ is approximately $z = 1.58$ (we can verify this: $P(Z\leq1.58)\approx0.9429\approx0.943$).

Step3: Use the z - score formula to solve for $\sigma$

We know that $z=\frac{x-\mu}{\sigma}$, where $x = 84$, $\mu = 78.4$, and $z = 1.58$. Rearranging the formula for $\sigma$, we get $\sigma=\frac{x - \mu}{z}$.
Substitute the values: $x = 84$, $\mu = 78.4$, and $z = 1.58$ into the formula.
$\sigma=\frac{84 - 78.4}{1.58}=\frac{5.6}{1.58}\approx3.544$

Answer:

$3.544$