How Likely Is It That Mechanics Take Over 8.7 Hours to Rebuild a Transmission?

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Homework Help Overview

The discussion revolves around the statistical analysis of the time mechanics take to rebuild a transmission, specifically focusing on a sample of 40 mechanics and the probability that their mean rebuild time exceeds 8.7 hours. The context involves concepts of mean, standard deviation, and the calculation of z-scores in relation to sample size.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the sample mean and the standard deviation, questioning how to properly calculate the z-score and the standard error of the mean. There are inquiries about the correct application of formulas and the implications of sample size on these calculations.

Discussion Status

The discussion is active, with participants providing guidance on the importance of the standard error in calculations. Some participants express confusion about the calculations and seek clarification on the correct approach to determining the z-score. There is acknowledgment of differing interpretations and methods being explored.

Contextual Notes

Participants note the importance of using the correct standard deviation for the sample size and raise concerns about the accuracy of their calculations. There is mention of homework guidelines regarding posting in multiple sections, which adds to the context of the discussion.

xsgx
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Mod note: This post was moved from another forum section, so doesn't use the homework template.
A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.7 hours.

Is this the same as P(X-bar > 8.7) and P(Z > .16666666666..) because then the answer would be about 43% and the answers given are:

0.1469
0.1346
0.1946
0.1285

Can someone explain what I am doing wrong? Thanks in advance for your replies.
 
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xsgx said:
Mod note: This post was moved from another forum section, so doesn't use the homework template.
A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.7 hours.

Is this the same as P(X-bar > 8.7) and P(Z > .16666666666..) because then the answer would be about 43% and the answers given are:

0.1469
0.1346
0.1946
0.1285

Can someone explain what I am doing wrong? Thanks in advance for your replies.
Please show your calculation for z. I'm pretty sure that's where the problem lies.
 
Mark44 said:
Please show your calculation for z. I'm pretty sure that's where the problem lies.

Mean= 8.4
x=8.7
SD=1.8

1.8%3D%200.16666666666666666666666666666667.gif
 
Last edited by a moderator:
You have a sample of 40 mechanics. How does that need to fit into your calculations?
 
Mark44 said:
You have a sample of 40 mechanics. How does that need to fit into your calculations?
That is where I am stuck and unsure of what to do. Do I raise the probability of getting one mechanic whose time exceeds 8.7 to the power of 40?

So that would be
gif.latex?.433816%5E4%5E0.gif
?
 
No, that's not it. Your sample size (40) affects the standard deviation you use to calculate z. Your book should have this formula. As I recall, it's called the standard error of the mean, and represents the s.d. of all samples of size n.

BTW, I noticed that you also posted this question in the Precalc section. That's a no-no to post the same question in two or more forum sections.
 
Mark44 said:
No, that's not it. Your sample size (40) affects the standard deviation you use to calculate z. Your book should have this formula. As I recall, it's called the standard error of the mean, and represents the s.d. of all samples of size n.

BTW, I noticed that you also posted this question in the Precalc section. That's a no-no to post the same question in two or more forum sections.
I just found the equation in my book they dedicated such a little piece of text to it that I missed it when reading the first time. I calculated the standard error to of the mean to be
40%3D%200.28460498941515413987990041899894.gif
Where do I go from this step?
 
Mark44 said:
I think you read it wrong. See http://en.wikipedia.org/wiki/Standard_error

Both my book and the Wikipedia link gives the same formula for calculating the error of the mean: Error of the mean= variance/sample size. When I do the calculations I get
ex%3F3.24%2F40%253D%25200.28460498941515413987990041899894&hash=b7bb1ba2c30b7b6dfd3eb8b7ebaa6ef7.png
. I don't see how I can apply that number to solving this problem. Is this the new standard deviation that I use to calculate the z-score?
 
  • #10
^ Actually that's exactly what I needed to do haha. I don't quite understand why it works that way though. Can anyone explain it?
 
  • #11
xsgx said:
Both my book and the Wikipedia link gives the same formula for calculating the error of the mean: Error of the mean= variance/sample size. When I do the calculations I get
ex%3F3.24%2F40%253D%25200.28460498941515413987990041899894&hash=b7bb1ba2c30b7b6dfd3eb8b7ebaa6ef7.png
.
In the Wiki link, it gives this: ##S.E. = \frac{s}{\sqrt{n}}##
Your calculation apparently doesn't include the square root.
xsgx said:
I don't see how I can apply that number to solving this problem. Is this the new standard deviation that I use to calculate the z-score?
You use it when you're looking at how samples of a given size are distributed.
From the Wiki page (emphasis added):
The standard error (SE) is the standard deviation of the sampling distribution of a statistic.
 
  • #12
xsgx said:
Both my book and the Wikipedia link gives the same formula for calculating the error of the mean: Error of the mean= variance/sample size. When I do the calculations I get
ex%3F3.24%2F40%253D%25200.28460498941515413987990041899894&hash=b7bb1ba2c30b7b6dfd3eb8b7ebaa6ef7.png
. I don't see how I can apply that number to solving this problem. Is this the new standard deviation that I use to calculate the z-score?

Your elementary arithmetic is wrong: 3.24/40 = 0.081 exactly, nowhere near 0.28460... . Anyway, it is unscientific nonsense to keep 20-30 decimal digits of accuracy in such problems, especially as the 3.24 might not have even full two-decimal accuracy (being based on limited observations). Sometimes you should keep several additional decimals of accuracy during a calculation, to guard against roundoff errors, but keeping the kind of (20+ digits) accuracy you done---especially in a number that is so wrong---is something you should avoid.
 
  • #13
Sorry I forgot to say that I took the square root of the answer 0.081. Anyways I have solved this problem. This thread can be closed.