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?
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 <u>at least</u> 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.
Let's say for example this pizza had 6 slices to start with. Two-thirds of this is (2/3)*12 = 4 slices. So you have 4 slices left over from yesterday. Then let's say you hand one slice to each of your four friends. Each friend would have 1/6 of the original pizza.
