Parking at a large university has become a very big problem. university administrators are interested in determining the average parking time (e. g. the time it takes a student to find a parking spot) of its students. an administrator inconspicuously followed 280 students and carefully recorded their parking times. which of the following graphs should not be used to display information concerning the students parking times? a. pie chartb. stem-and-leaf displayc. histogramd. box plot

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.