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 97 miles per hour (mph) and the standard deviation of the serve speeds was 10 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) 87 mph to 107 mph

B) 117 mph to 137 mph

C) 67 mph to 127 mph

D) 77 mph to 117 mph

D) 77 mph to 117 mph

Step-by-step explanation:

Given that:

Mean serve speed(m) = 97 miles per hour

Standard deviation of serve speed = 10 mph

Shape of distribution is not known ; interval that will contain the speeds of at least three-fourths of the player's serves

Using Chebyshev's rule:

(1 - 1/k²) = 3 / 4

1 / k² = 3/4 - 1

- 1/k² = - 1 / 4

1/k² = 1/4

Cross multiply:

k² = 4

Square root of both sides :

k = 2

Hence, number of standard deviation = 2

(-k*sd) + m ; (k*sd)+m

(-2*10) + 97 ; (2*10)+97

-20 + 97; 20 + 97

77 ; 117

77mph ; 117mph

is there a picture

step-by-step explanation:

answer

a) the midrange of the data set shown is 85.5

step by step explanation

let's find the midrange.

the midrange is nothing but add the greatest value and the least value, then divide it by 2.

here the greatest value = 99

the least value = 72

midrange = (99 + 72)/2

= 171/2

midrange = 85.5

you.