Normal Distribution and Population Mean
Chapter 7 #42
The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.
a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect? z(29) = (29-32)/2 = -3/2 z(34) = (34-32)/2 = 1 z(32) = 0
P(32 < x < 34) = P(0< z < 1) = 0.34
b. What percent of the garages take between 29 hours and 34 hours to erect?
P(29 < x < 34) = P(-1.5 < z < 1) = 0.7745
c. What percent of the garages take 28.7 hours or less to erect? z(28.7) = (28.7-32)/2 = -3.3
P(0 < x < 28.7) = P (-10 < z < -3.3) = 0.00048348...
d. Of the garages, 5 percent take how many hours or more to erect? find the z-value corresponding to an area of 95% to the left and only
5% to the right under the curve.
Use your z-chart or InvNor(0.95) = 1.645 on your calculator.
Now find the x-value corresponding to that z-value.
1.645 = (x-32)/2 x-32 = 2*1.645 x= 35.29 hours
5% of the houses require 35.29 or more hours to erect.
Chapter 8 #21
What is a sampling error?
Sampling error is the expected chance difference, variation, or deviation between a random sample and the population.
Chapter 8 #34
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
SE = sd/sqrt(N)
= 40000/sqrt(50)
= 5656.85
b. What is the expected shape of the distribution of the sample mean?
It will be normally distributed. c. What is the likelihood of selecting a sample with a mean of at least $112,000? z = (112000-110000)/5656.85 z = 0.353553
P(z > 0.353553)
The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.
a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect? z(29) = (29-32)/2 = -3/2 z(34) = (34-32)/2 = 1 z(32) = 0
P(32 < x < 34) = P(0< z < 1) = 0.34
b. What percent of the garages take between 29 hours and 34 hours to erect?
P(29 < x < 34) = P(-1.5 < z < 1) = 0.7745
c. What percent of the garages take 28.7 hours or less to erect? z(28.7) = (28.7-32)/2 = -3.3
P(0 < x < 28.7) = P (-10 < z < -3.3) = 0.00048348...
d. Of the garages, 5 percent take how many hours or more to erect? find the z-value corresponding to an area of 95% to the left and only
5% to the right under the curve.
Use your z-chart or InvNor(0.95) = 1.645 on your calculator.
Now find the x-value corresponding to that z-value.
1.645 = (x-32)/2 x-32 = 2*1.645 x= 35.29 hours
5% of the houses require 35.29 or more hours to erect.
Chapter 8 #21
What is a sampling error?
Sampling error is the expected chance difference, variation, or deviation between a random sample and the population.
Chapter 8 #34
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
SE = sd/sqrt(N)
= 40000/sqrt(50)
= 5656.85
b. What is the expected shape of the distribution of the sample mean?
It will be normally distributed. c. What is the likelihood of selecting a sample with a mean of at least $112,000? z = (112000-110000)/5656.85 z = 0.353553
P(z > 0.353553)