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 2.4 weeks. Suppose you would like
to select a random sample of 51 unemployed individuals for a
follow-up study.
Find the probability that a single randomly selected value is
greater than 17.7.
P(X > 17.7) =
(Enter your answers as numbers accurate to 4 decimal
places.)
Find the probability that a sample of size n=51 is randomly
selected with a mean greater than 17.7.
P(M > 17.7) =
(Enter your answers as numbers accurate to 4 decimal places.)
Solution :
P(x > 17.7) = 1 - P(x < 17.7)
= 1 - P[(x - ) / < (17.7 - 18) / 2.4)
= 1 - P(z < -0.125)
= 1 - 0.4503
= 0.5497
P(x > 17.7) = 0.5497
M = / n = 2.4 / 51 = 0.3361
P(M > 17.7) = 1 - P(M < 17.7)
= 1 - P[(M - M) / M < (17.7 - 18) / 0.3361)
= 1 - P(z < -0.893)
= 1 - 0.1859
= 0.8141
P(M > 17.7) = 0.8141
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