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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%2034.6%2C%20%5Csigma%20%3D%2010.2)
Sample of 36
This means that ![n = 36, s = \frac{10.2}{\sqrt{36}} = 1.7](https://tex.z-dn.net/?f=n%20%3D%2036%2C%20s%20%3D%20%5Cfrac%7B10.2%7D%7B%5Csqrt%7B36%7D%7D%20%3D%201.7)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{37 - 34.6}{1.7}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B37%20-%2034.6%7D%7B1.7%7D)
![Z = 1.41](https://tex.z-dn.net/?f=Z%20%3D%201.41)
has a pvalue of 0.9207
X = 30
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{30 - 34.6}{1.7}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30%20-%2034.6%7D%7B1.7%7D)
![Z = -2.71](https://tex.z-dn.net/?f=Z%20%3D%20-2.71)
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.