A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 38 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 38 weeks and that the population standard deviation is 2.8 weeks. Suppose you would like to select a random sample of 77 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 38.4 .
\[
P(X> 38.4)=
\]
(Enter your answers as numbers accurate to 4 decimal places.)
Find the probability that a sample of size $n=77$ is randomly selected with a mean greater than 38.4 .
\[
P(M> 38.4)=
\]
(Enter your answers as numbers accurate to 4 decimal places.)
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Answer
Final Answer: \[P(X>38.4) \approx \boxed{0.4432}\]
Step 1 ：The problem is asking for the probability that a single randomly selected value is greater than 38.4. This is a problem of finding the probability of a single value from a normal distribution.
Step 2 ：The formula for the z-score is: \[Z = \frac{X - \mu}{\sigma}\] where X is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 ：We can calculate the z-score for 38.4 using the given mean (\(\mu = 38\)) and standard deviation (\(\sigma = 2.8\)). This gives us a z-score of approximately 0.14285714285714235.
Step 4 ：We then use a z-table or a statistical function to find the probability that a randomly selected value is greater than this z-score. This gives us a probability of approximately 0.443201503183532.
Step 5 ：Final Answer: \[P(X>38.4) \approx \boxed{0.4432}\]
