Question

At a tennis tournament a statistician keeps track of every serve. The statistician reported that the mean serve speed of a particular player was 101 miles per hour (mph) and the standard deviation of the serve speed was 14 mph. If nothing is known about the shape the distribution, give an interval that will contain the speeds approximately three-fourths of the player’s serves.

Answer 1

Chebyshev's Theorem:

For any set of data and for any number k, greater than 1, the proportion of the data that lies within k standard deviation of the mean is at least 1 - ( 1 / k² )

For k = 2 at least 1 - (1/4) = 3/4 = 0.75 = 75% of the data falls within 2 standard deviation of the mean.

We have mean = = 101 and standard deviation = 14

Hence an interval that will contain the speeds approximately three-fourths of the player’s serves is -

( - 2* , + 2* )

- 2* = 101 - (2 * 14) = 101 - 28 = 73

+ 2* = 101 + ( 2 * 14 ) = 101 + 28 = 129

Hence an interval that will contain the speeds approximately three-fourths of the player’s serves is ( 73 , 129 )

Hope this will help you. Thank you :)