A leading magazine (like Barron\'s) reported at one time that the average number

Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 24.4 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 24.4 weeks and that the population standard deviation is 8.7 weeks. Suppose you would like to select a random sample of 78 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is between 21.5 and 25.7.
P(21.5 < X < 25.7) =

Find the probability that a sample of size n=78n=78 is randomly selected with a mean between 21.5 and 25.7.
P(21.5 < ¯xx¯ < 25.7) =

Enter your answers as numbers accurate to 4 decimal places.

Explanation / Answer

Mean ( u ) =24.4
Standard Deviation ( sd )=8.7
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 21.5) = (21.5-24.4)/8.7
= -2.9/8.7 = -0.3333
= P ( Z <-0.3333) From Standard Normal Table
= 0.36944
P(X < 25.7) = (25.7-24.4)/8.7
= 1.3/8.7 = 0.1494
= P ( Z <0.1494) From Standard Normal Table
= 0.55939
P(21.5 < X < 25.7) = 0.55939-0.36944 = 0.1899                  

b.
when Number ( n ) = 78
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 21.5) = (21.5-24.4)/8.7/ Sqrt ( 78 )
= -2.9/0.9851
= -2.9439
= P ( Z <-2.9439) From Standard Normal Table
= 0.00162
P(X < 25.7) = (25.7-24.4)/8.7/ Sqrt ( 78 )
= 1.3/0.9851 = 1.3197
= P ( Z <1.3197) From Standard Normal Table
= 0.90653
P(21.5 < X < 25.7) = 0.90653-0.00162 = 0.9049  

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