Question

According to the Labor Department, the average duration of unemployment for adults ages 20 to 24 was 34.6 weeks during a recent month. Assume that the standard deviation for this population is 10.2 weeks. A random sample of 36 adults in this age group was selected. What is the probability that the average duration of unemployment was between 30 and 37 weeks

Answer 1

Answer:

0.9173 = 91.73% probability that the average duration of unemployment was between 30 and 37 weeks.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 34.6, standard deviation of 10.2

This means that $\mu = 34.6, \sigma = 10.2$

Sample of 36

This means that $n = 36, s = \frac{10.2}{\sqrt{36}} = 1.7$

What is the probability that the average duration of unemployment was between 30 and 37 weeks?

This is the pvalue of Z when X = 37 subtracted by the pvalue of Z when X = 30.

X = 37

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

By the Central Limit Theorem

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

$Z = \frac{37 - 34.6}{1.7}$

$Z = 1.41$

$Z = 1.41$ has a pvalue of 0.9207

X = 30

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

$Z = \frac{30 - 34.6}{1.7}$

$Z = -2.71$

$Z = -2.71$ has a pvalue of 0.0034

0.9207 - 0.0034 = 0.9173

0.9173 = 91.73% probability that the average duration of unemployment was between 30 and 37 weeks.