Question

According to a? study, 71?% of all males between the ages of 18 and 24 live at home. ? (Unmarried college students living in a dorm are counted as living at? home.) Suppose that a survey is administered and 163 of the 223 respondents indicated that they live at home.?

(a) Use the normal approximation to the binomial to approximate the probability that at least 163 respondents live at?home?

A) P(X is greater than or equal to 163) = ?

(b) Do the results from part? (a) contradict the? study?

a) yes because the probability of P(X is greater than or equal to 163) is greater than 0.05.

b) No, because the probability of P(X is greater than or equal to 163) is less than 0.05

c) No because the probablility of P(X is greater than or equal to 163) is greater than 0.05

d) Yes, because the probability of P(X is greater than or equal to 163) is less than 005

Answer 1

a)
`P(X \geq 163) = 0.2451`

b) a) yes because the probability of P(X is greater than or equal to 163) is greater than 0.05.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Normal approximation to the binomial, with a sample of 223 respondents, so
`n = 223`.

The mean of males between 18 and 24 that live at home is 0.71, so
`p = 0.71`

We can approximate the normal to the binomial using the following values for the mean and standard deviation:

`\mu = np = 223*0.71 = 158.33`

`\sigma = √(np(1-p)) = √(223*0.71*(0.29)) = 6.78`

(a) Use the normal approximation to the binomial to approximate the probability that at least 163 respondents live at?home?

This is 1 subtracted by the pvalue of Z when
`X = 163`

`Z = (X - \mu)/(\sigma)`

`Z = (163 - 158.33)/(6.78)`

`Z = 0.69`

`Z = 0.69` has a pvalue of 0.7549.

So

`P(X \geq 163) = 1 - 0.7549 = 0.2451`

(b) Do the results from part? (a) contradict the? study?

Yes, because the probability of more than 163 respondents living at home is not unusual, that is, greater than 0.05. So the correct answer is:

a) yes because the probability of P(X is greater than or equal to 163) is greater than 0.05.