A large auto manufacturing company in Easter Europe is experiencing increasing p
ID: 3150003 • Letter: A
Question
A large auto manufacturing company in Easter Europe is experiencing increasing problems with broken axles and wheels that freeze up during driving. A team of consultants from Germany conducts an analysis and suspect the problem lies with the wheel bearings manufactured and fitted to certain models. The measured inside diameter of the automobile wheel bearing sets manufactured at the factory is expected to be normally distributed with a mean of 1.30 inches and a standard deviation of 0 04 inches. What is the probability that a randomly selected wheel bearing will have an inside diameter of, Between 1.28 and 1.30 inches? Between 1.31 and 1.33 inches? Between what two values (in terms of inside diameter) will 60% of the wheel bearings fall? If random samples of 16 wheel bearings were selected, What proportion of the sample means would be between 1.31 and 1.33 inches? Compare your results in (b) and (d) Based on your findings, docs the company need to reevaluate and redesign the production process to build in tighter tolerance specifications for these parts?Explanation / Answer
4.
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 1.28
x2 = upper bound = 1.3
u = mean = 1.3
s = standard deviation = 0.04
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.5
z2 = upper z score = (x2 - u) / s = 0
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.308537539
P(z < z2) = 0.5
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.191462461 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 1.31
x2 = upper bound = 1.33
u = mean = 1.3
s = standard deviation = 0.04
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.25
z2 = upper z score = (x2 - u) / s = 0.75
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.598706326
P(z < z2) = 0.773372648
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.174666322 [ANSWER]
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c)
As the middle area is
Middle Area = P(x1<x<x2) = 0.6
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.2
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -0.841621234
By symmetry,
z2 = 0.841621234
As
u = mean = 1.3
s = standard deviation = 0.04
Then
x1 = u + z1*s = 1.266335151
x2 = u + z2*s = 1.333664849
Hence, between 1.266335151 and 1.333664849. [ANSWER]
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d)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 1.31
x2 = upper bound = 1.33
u = mean = 1.3
n = sample size = 16
s = standard deviation = 0.04
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = 1
z2 = upper z score = (x2 - u) * sqrt(n) / s = 3
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.841344746
P(z < z2) = 0.998650102
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.157305356 [ANSWER]
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e)
The answer in part b is slightly greater than in part d, because the interval 1.31 to 1.33 does not include the mean itself.
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