Question

According to a​ study, 45​% 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 118 of 223 respondents indicated that they live at home.​

(a) Use the normal approximation to the binomial to approximate the probability that at least 118

respondents live at home.​

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

Answer 1

Solution:

Given: 45​% of all males between the ages of 18 and 24 live at home. ​ Thus p = 0.45

n   = Sample size = 223

x = Number of males between the ages of 18 and 24 live at home = 118

Part a) We have to use the normal approximation to the binomial to approximate the probability that at least 118 respondents live at home.​

That is:

Since sample size = n = 223 is large sample size and

n*p = 223 * 0.45 = 100.35 > 5

n*(1-p) = 223 * ( 1 - 0.45) = 223 * 0.55 = 122.65 > 5

Thus we can use the normal approximation to the binomial to approximate the probability that at least 118 respondents live at home.​

Thus find mean and standard deviation:

Mean = n*p = 223 * 0.45 = 100.35

Standard Deviation :

Also we need to use continuity correction, that is we add or subtract 0.5 from 118

Here we need to subtract 0.5 so that 118 will be included in the range of x values.

Thus we get:

Now find z score:

Thus we get:

Look in z table for z =2.3 and 0.01 and find corresponding area.

P( Z < 2.31) = 0.9896

Thus

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

Yes, because the probability of P(X is greater than or equal to 118) is less than 0.05