The proportion of households in a region that do some or all of their banking on the Internet is 0.41. In a random sample of 1000 households from this region, let x be the number that do some or all of their banking on the Internet a. Find the mean of x

Step-by-step explanation:Hello!

X: number of households in a region that do some or all of their banking on the internet.

The proportion of households is p=0.41

n= 1000

For this variable there are two possible outcomes, the household either "does some or all baking online" (success) or "does no make banking online" (failure).

The number of repetitions is fixed n= 1000 households.

Assuming each household is independent of the others and the probability of success p=0.41 is the same for all households, then:

X~Bi(n;p)

The mean for a variable with binomial distribution is:

E(X)= n*p= 1000*0.41= 410

The expected value of households that do some or all banking online is 410.

I hope it 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.