Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 17 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17 weeks and that the population standard deviation is 2.2 weeks. Suppose you would like to select a random sample of 92 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 16.7.
P(X > 16.7) =  (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=92n=92 is randomly selected with a mean greater than 16.7.
P(M > 16.7) =  (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

This is a normal distribution question with

a)

P(x > 16.7)=?

The z-score at x = 16.7 is,

z = -0.1364

This implies that

P(x > 16.7) = P(z > -0.1364) = 1 - P(z < -0.1364) = 1 - 0.4457525368391769

b)

Sample size (n) = 92

Since we know that

P(x > 16.7)=?

The z-score at x = 16.7 is,

z = -1.3078

This implies that

P(x > 16.7) = P(z > -1.3078) = 1 - P(z < -1.3078) = 1 - 0.09547057711296442

PS: you have to refer z score table to find the final probabilities.

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