Question

1. According to Consumer Response Annual Report, millennials spent an average of $103 on monthly dining in 2016. Let the amount spent on a monthly dining be normally distributed with unknown standard deviation. Assume that the probability of a randomly-selected millennial that spends more than $133 is 10%, and the probability that a randomly-selected millennial that spends less than $97 pounds is 40%. What is the probability that a randomly-selected millennial will spend between $73 and $109 on a monthly dining?

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answer.

Given mean = 103

Z = (x-mean)/s.d

And it is given that

P(x<97) = 0.4

From z table, p(z<-0.25) = 0.4

Z = -0.25

-0.25 = (97-103)/s.d

S.d = 24

Now we have mean = 103

S.d = 24

And we need to find

P(73<x<109) = p(x<109) - p(x<73)

P(x<109)

Z = (109-103)/24= 0.25

From z table,p(z<0.25) = 0.5987

P(x<73)

Z = (73-103)/24 = -1.25

From z table, P(z<-1.25) = 0.1056

Required answer is 0.5987 - 0.1056 = 0.4931