Question

According to a​ study, 47 47​% 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 99 99 of 208 208 respondents indicated that they live at home.​ (a) use the normal approximation to the binomial to approximate the probability that at least 99 99 respondents live at​ home? (b) do the results from part​ (a) contradict the​ study?

Answer 1

The probability is 0.4325.  This is close to 0.47, so no, the results do not really contradict the study.

To approximate a normal distribution, we use the fact that 

μ = np and

σ² = np(1-p).

This means that σ = √(np(1-p)).

Our formula for a z-score is 

z = (X-μ)/σ; using what we just stated, this makes it

z = (X-np)/(√(np(1-p))

For this problem, n=208, p=0.47, and X=99:

z = (99-(208)(0.47))/√((208)(0.47)(0.53))
z = (99-97.76)/√(51.8128) = 1.24/(√51.8128) = 0.17

Using a z-table (http://www.z-table.com) we see that the area under the curve to the left of, less than, this z-score is 0.5675.  We want the probability of at least 99 people; this is greater than or equal to, not less than.  To find what we need, we subtract from 1:
1 - 0.5675 = 0.4325.