Question

Recent survey data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. Their parents must have a basement! A random sample of 125 young adults in this age group was selected. What is the probability that between 13 and 17 of these young adults lived with their parents? Hint: use 14.2% to determine the standard error and the p-bar would be the 13/125 and the 17/125.

Answer 1

38.76% probability that between 13 and 17 of these young adults lived with their parents

Explanation:

I am going to use the normal approxiation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`p = 0.142, n = 125`

So

`\mu = E(X) = np = 125*0.142 = 17.75`

`\sigma = √(V(X)) = √(np(1-p)) = √(125*0.142*0.858) = 3.9025`

What is the probability that between 13 and 17 of these young adults lived with their parents?

Using continuity correction, this is
`P(13 - 0.5 \leq X \leq 17 + 0.5) = P(12.5 \leq 17.5)`, which is the pvalue of Z when X = 17.5 subtracted by the pvalue of Z when X = 12.5. So

X = 17.5

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

`Z = (17.5 - 17.75)/(3.9025)`

`Z = -0.06`

`Z = -0.06` has a pvalue of 0.4761

X = 12.5

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

`Z = (12.5 - 17.75)/(3.9025)`

`Z = -1.35`

`Z = -1.35` has a pvalue of 0.0885

0.4761 - 0.0885 = 0.3876

38.76% probability that between 13 and 17 of these young adults lived with their parents