SOLUTION: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 18 weeks. Assume that for the population of all unemployed i

Question 1181519: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 18 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 18 weeks and that the population standard deviation is 3.2 weeks. Suppose you would like to select a random sample of 73 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 17.6.
P(X > 17.6) =
(Enter your answers as numbers accurate to 4 decimal places.)
Find the probability that a sample of size n = 73 is randomly selected with a mean greater than 17.6.
P(M > 17.6) =
(Enter your answers as numbers accurate to 4 decimal places.)
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to calculate these probabilities:
**1. Probability for a single randomly selected value (X):**
* We're looking for P(X > 17.6).
* First, calculate the z-score: z = (X - μ) / σ = (17.6 - 18) / 3.2 = -0.125
* Now, look up the probability associated with this z-score in a standard normal distribution table or use a calculator. P(Z > -0.125) = 1 - P(Z < -0.125) ≈ 1-0.4503 = 0.5497
**2. Probability for a sample mean (M):**
* We're looking for P(M > 17.6).
* The standard error of the mean is: σ_M = σ / sqrt(n) = 3.2 / sqrt(73) ≈ 0.374
* Calculate the z-score for the sample mean: z = (M - μ) / σ_M = (17.6 - 18) / 0.374 ≈ -1.069
* Now, look up the probability associated with this z-score. P(Z > -1.069) = 1 - P(Z < -1.069) ≈ 1 - 0.1423= 0.8577
**Therefore:**
* P(X > 17.6) ≈ 0.5497
* P(M > 17.6) ≈ 0.8577

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