WILL MARK BRAINLEIST HURRY
The box plots below show attendance at a local movie theater and high school basketball games: Two box plots shown. The top one is labeled Movies. Minimum at 60, Q1 at 65, median at 95, Q3 at 125, maximum at 150. The bottom box plot is labeled Basketball games. Minimum at 90, Q1 at 95, median at 125, Q3 at 145, maximum at 150. Which of the following best describes how to measure the spread of the data? The IQR is a better measure of spread for movies than it is for basketball games. The standard deviation is a better measure of spread for movies than it is for basketball games. The IQR is the best measurement of spread for games and movies. The standard deviation is the best measurement of spread for games and movies.

Hello, The answer it's D.

Step-by-step explanation:

Assume that the data for both movies and basketball games are normally distributed.

Therefore, the median and the mean are assumed equal.

The standard deviation, σ, is related to the interquartile range by

IQR = 1.35

From the data, we can say the following:

Movies:

Range = 150 - 60 = 90 (approx)

Q1 = 62 (approx), first quartile

Q3 = 120 (approx), third quartlie

Q2 (median) = 90 (approx)

IQR = Q3 - Q1 = 58

σ = IQR/1.35 = 58/1.35 = 43

Basketball:

Range = 150 - 90 = 60 approx

Q1 = 95 (approx)

Q3 = 145 (approx)

Q2 = 125 (approx)

IQR = 145 - 95 = 50

σ = 50/1.35 = 37

Test the given answers.

A. The IQRs are approximately equal, so they are not good measures of spread. This is not a good answer.

B. The std. deviation is a better measure of spread for basketball. This is not a good answer.

C. IQR is not a better measure of spread for basketball games. This is not a good answer.

D. The standard deviation is a good measure of spread for both movies and basketball. This is the best answer.

D

Have a Great Day!

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared