Question

hapter 8: Sampling Distribution 1. The distribution of i is normal: n 2 30. 2. Be able to find the mean of sample means: Hx =H 3. Be able to find the standard deviation of sample means: Ox = %3D 4. Be able to distinguish and find the probability for an individual value x and a group x. 5. Be able to distinguish and find an individual value x or a group average £ from a given probability. 6. Examples a. Furnace repair bills are normally distributed with a mean of 273 dollars and a standard deviation of 25 dollars. If 100 of these repair bills are randomly selectéd, find the probability that they have a mean cost between 273 dollars and 275 dollars. b. The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 3.5. Suppose 49 golfers played the course today. Find the probability that the average score of the 49 golfers exceeded 76. Popala c. Assume that the heights of men are normally distributed with a mean of 69.8 inches and a standard deviation of 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height greater than 70.8 inches. lion 20 of iMme Stat

Answer 1

Answer:

a).

standard deviation of sample means = sd/sqrt(n) = 25/sqrt(100) =2.5

z value for 273, z =(273-273)/2.5 =0

z value for 275, z =(275-273)/2.5 =0.8

P( 273< mean x < 275) = P( 0<z<0.8)

=P( z < 0.8) – P( z < 0)

=0.7881 -0.5000

=0.2881

Excel function used to probability

P( z <0.8): =NORM.S.DIST(0.8,TRUE)

P( z <0): =NORM.S.DIST(0,TRUE)

b).

standard deviation of sample means = sd/sqrt(n) = 3.5/sqrt(49) =0.5

z value for 76, z =(76-75)/0.5 =2

P( mean x > 76) = P( z > 2)

= 0.0228

Excel function: P( z >2): =1-NORM.S.DIST(2,TRUE)

c).

standard deviation of sample means = sd/sqrt(n) = 2.8/sqrt(64) =0.35

z value for 70.8, z =(70.8-69.8)/0.35 =2.86

P( mean x > 70.8) = P( z > 2.86)

=0.0021

Excel function: P( z >2.86): =1-NORM.S.DIST(2.86,TRUE)