Question

At the U.S. Open Tennis Championship a statistician keeps track of every serve that a player hits during the tournament. The statistician reported that the mean serve speed of a particular player was 98 miles per hour (mph) and the standard deviation of the serve speeds was 13 mph. If nothing is known about the shape of the distribution, give an interval that will contain the speeds of at least three-fourths of the player's serves.

a. 124 mph to 150 mph.
b. 85 mph to 111 mph.
c. 59 mph to 137 mph.
d. 72 mph to 124 mph.

Answer 1

d. 72 mph to 124 mph.

Explanation:

We solve for this question using the Chebyshev's inequality(x-μ)/σ

For k > 1, 1 - 1/k²

P(x-μ) ≤ kσ≥ 1 - 1/k²

For k = 2

Mean(μ) = 98mph

Standard deviation(σ) = 13mph

P(x-μ) ≤ 2 × 13 mph≥ 1 -1/2²

P(x-μ) ≤ 2 × 13 mph≥ 1 -1/4

P(x-μ) ≤ 26 mph≥ 3/4

According to Chebyshev's theorem,at least 3/4 of the data is 1 standard deviation from the mean

= μ ± σ :μ - σ, μ + σ

The interval is given as: μ - σ, μ + σ

98mph - 26 mph = 72 mph

98 mph + 26 mph = 124 mph

Therefore, the interval = 72mph to 124mph

Option D is correct