Kim wants to know how many families in her small neighborhood of 70 homes would participate in a neighborhood sports event. she put all the addresses in a bag and drew a random sample of 35 addresses. she then asked those families if they would participate in the sports event. she found that 15% of the families would participate in the event. she claims that 15% of the neighborhood families would be expected to participate in the sports event. is this a valid inference? yes, this is a valid inference because she took a random sample of the neighborhood yes, this is a valid inference because the 35 families speak for the whole neighborhood no, this is not a valid inference because she asked only 35 families no, this is not a valid inference because she did not take a random sample of the neighborhood

My answer would be A. she did pick from the neighborhood so that means D is not an option. It is valid because she could host a small event so C is not an option. Just because you ask a couple dose not mean the many have the same answer so It isn't B.

A

Hopes this helps

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.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0