Students in a class were surveyed about the number of children in their families. The results of the survey are shown in the table. A 2-column table has 5 rows. The first column is labeled Number of children in family with entries 1, 2, 3, 4, 5 or more. The second column is labeled Number of surveys with entries 9, 18, 22, 8, 3.

Two surveys are chosen at random from the group of surveys. After the first survey is chosen, it is returned to the stack and can be chosen a second time. What is the probability that the first survey chosen indicates four children in the family and the second survey indicates one child in the family?

StartFraction 1 Over 50 EndFraction
StartFraction 2 Over 15 EndFraction
StartFraction 3 Over 20 EndFraction
StartFraction 17 Over 60 EndFraction

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.

im pretty sure its

(8,15)

step-by-step explanation:

in if you ask questions, the points are taken from your total and given to the one who answered it..

so if i have 30 points and i ask a question for 5 points, then when someone asks a question i will lose 5 points and the answerer earns 5

happy to