Question

in 2019, the distribution of golfer lexi thompsons driving distance had a mean of 276 yards. assuming that her distribution of driving distance is approximately normal with an 80th percentile of 290 yards, calculate its standard deviation. standard deviation = yards (round to 2 decimal places.)

Explanation:

Step1: Find the z - score for the 80th percentile

Using a standard normal distribution table or calculator, the z - score $z$ corresponding to the 80th percentile is approximately $z = 0.84$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $x = 290$ yards (80th - percentile value), $\mu=276$ yards (mean), and $z = 0.84$. Rearranging the formula for $\sigma$ gives $\sigma=\frac{x - \mu}{z}$.

Step3: Substitute the values and calculate

Substitute $x = 290$, $\mu = 276$, and $z = 0.84$ into the formula: $\sigma=\frac{290 - 276}{0.84}=\frac{14}{0.84}\approx16.67$.

Answer:

$16.67$