A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 33.9 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 33.9 weeks and that the population standard deviation is 6.7 weeks. Suppose you would like to select a random sample of 119 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is between 35.3 and 35.4. P(35.3 < X < 35.4) = Find the probability that a sample of size n = 119 is randomly selected with a mean between 35.3 and 35.4. P(35.3 < M < 35.4) =

P(35.3 < M < 35.4) = 0.0040.

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 normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

Find the probability that a single randomly selected value is between 35.3 and 35.4

This is the pvalue of Z when X = 35.4 subtracted by the pvalue of Z when X = 35.3. So

X = 35.4

By the Central Limit Theorem

has a pvalue of 0.9927

X = 35.3

has a pvalue of 0.9887

0.9927 - 0.9887 = 0.0040

So the answer is:

P(35.3 < M < 35.4) = 0.0040.

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