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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.

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Answer:

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

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