In: Statistics and Probability
A leading magazine (like Barron's) reported at one time that the
average number of weeks an individual is unemployed is 28 weeks.
Assume that for the population of all unemployed individuals the
population mean length of unemployment is 28 weeks and that the
population standard deviation is 2.5 weeks. Suppose you would like
to select a random sample of 72 unemployed individuals for a
follow-up study.
Find the probability that a single randomly selected value is
greater than 28.5.
P(X > 28.5) = (Enter your answers
as numbers accurate to 4 decimal places.)
Find the probability that a sample of size n=72n=72 is randomly
selected with a mean greater than 28.5.
P(M > 28.5) = (Enter your answers
as numbers accurate to 4 decimal places.)
Solution :
Given that ,
mean = = 28
standard deviation = = 2.5
P ( x > 28.5 ) = 1 - P( x < 28.5 )
= 1- P[(x - ) / < ( 28.5 - 28 ) / 2.5 ]
= 1 - P ( z < 0.2 )
Using z table,
= 1 - 0.5793
= 0.4207
Probability = 0.4207
n = 72
m = 28
m = / n = 2.5 / 72 = 0.2946
P( M > 28.5 ) = 1 - P ( < 28.5 )
= 1 - P[(M - m) / m < ( 28.5 - 28 ) / 0.2946 ]
= 1 - P( z < 1.70)
Using z table,
= 1 - 0.9554
= 0.0446
Probability = 0.0446