Normal Distribution and Population Mean

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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)
= 0.3618
 d. What is the likelihood of selecting a sample with a mean of more than $100,000? z = (100000-110000)/5656.85
z = -1.767768
P(z > -1.767768)
= 0.9615
 e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000 = 0.9615 - 0.3618
= 0.5997
Chapter 9 #32
A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled 3 pounds. Of course he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds with a standard deviation of .03 pounds. a. What is the estimated population mean? b. Determine a 95% confidence interval for the population mean mean - z*sd/sqrt(N) to mean + z*sd/sqrt(N)

The estimated mean is given in the problem as 3.01 pounds.
 The interval goes from:
z (from a table) is 1.96
N = 36
sd = 0.03
mean = 3.01
 
3.01 - 1.96*0.03/sqrt(36) to 3.01 + 1.96*0.03/sqrt(36)
 
3.0002 to 3.0198
Chapter 9 #34
A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer. n = 50, m = 26, s = 6.2

 
Standard error of the mean, SE = s/sqrt n = 6.2/sqrt 50 = 0.8768  
The z- score for 95% confidence level is z = 1.96
 
Margin of error, E = z * SE = 1.96 * 0.8768 = 1.7186
 
Lower Limit of the confidence interval = m - E = 24.2814
 
Upper Limit of the confidence interval = m + E = 26 + 1.7186 = 27.7186  
The confidence interval is [24.28 weeks, 27.72 weeks]
 
Since 28 weeks falls outside the above interval, it is not reasonable to assume  
that the population mean is 28 weeks.
 
[But it is a close decision since, if we round off to a whole number of weeks,  
then we have 28 weekd as the upper...
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