Question

the distribution of golfer lexi thompsons driving distance has 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 corresponding to the 80th percentile

Using the standard normal distribution table or a calculator, the z - score \(z\) for the 80th percentile (\(P(X\leq x)=0.8\)) is approximately \(z = 0.84\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x = 290\) (the value at the 80th percentile), \(\mu=276\) (the mean), and \(\sigma\) is the standard deviation.
Substitute the values into the formula: \(0.84=\frac{290 - 276}{\sigma}\).

Step3: Solve for \(\sigma\)

First, simplify the numerator: \(290−276 = 14\).
Then, from \(0.84=\frac{14}{\sigma}\), we can solve for \(\sigma\) by cross - multiplying: \(\sigma=\frac{14}{0.84}\).
Calculate \(\frac{14}{0.84}\approx16.67\).

Answer:

\(16.67\)