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

Answer:

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

Step-by-step 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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$p = 0.142, n = 125$

So

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.142*0.858} = 3.9025$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17.5 - 17.75}{3.9025}$

$Z = -0.06$

$Z = -0.06$ has a pvalue of 0.4761

X = 12.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12.5 - 17.75}{3.9025}$

$Z = -1.35$

$Z = -1.35$ 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