Area under a normal curve

  • Thread starter ChrisBlack
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  • #1
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Homework Statement


The mean is 17.8 and the standard deviation is 5.2. The shop does not want to give more than 1% of its customers free service, how long should the guarantee be? (cut the problem short, the rest was just story)


Homework Equations


I think the equation I would use is X = Z(standard deviation) + Mean


The Attempt at a Solution


I changed 1% into the Z score and got -2.33 and plugged that into the equation, -2.33(5.2) + 17.8 and ended up with 5.684. That is lower than the mean though which does not make sense to me, why would the shop guarantee they can complete the job in 6 minutes (rounding up) is the mean is 17.8. Where did I go wrong? Thanks in advance!
 

Answers and Replies

  • #2
Ray Vickson
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Are you sure you have the interpretation correct? The shop can say that no more than 1% of jobs will be finished in less than 5.68 minutes; that is, they can say "we guarantee we won't finish your job before 5.68 minutes". Wouldn't the shop want to say "we can guarantee your job won't take _longer than_ X minutes"? That would make X greater than 17.8, not less.

RGV
 
  • #3
34,687
6,393

Homework Statement


The mean is 17.8 and the standard deviation is 5.2. The shop does not want to give more than 1% of its customers free service, how long should the guarantee be? (cut the problem short, the rest was just story)


Homework Equations


I think the equation I would use is X = Z(standard deviation) + Mean


The Attempt at a Solution


I changed 1% into the Z score and got -2.33
The z score you want is +2.33.
and plugged that into the equation, -2.33(5.2) + 17.8 and ended up with 5.684. That is lower than the mean though which does not make sense to me, why would the shop guarantee they can complete the job in 6 minutes (rounding up) is the mean is 17.8. Where did I go wrong? Thanks in advance!
 
  • #4
10
0
Are you sure you have the interpretation correct? The shop can say that no more than 1% of jobs will be finished in less than 5.68 minutes; that is, they can say "we guarantee we won't finish your job before 5.68 minutes". Wouldn't the shop want to say "we can guarantee your job won't take _longer than_ X minutes"? That would make X greater than 17.8, not less.

RGV
The problem says that the shop does not want to give more than 1% of customers a free oil change, so I would think it would be over the mean, which makes sense if the Z score is positive, but i don't see why it's positive. Sorry, I should have written the whole problem
 
  • #5
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Right. So if the z-value for the time it takes the shop to do an oil change is larger than 2.33, that represents only 1% of the customers.

Now, you need to translate that z-value of 2.33 to an x-value, which represents the cutoff time for an oil change. If the shop takes longer than that cutoff value, the oil change is free.
 
  • #6
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The equation I posted at the top is the correct one to change the Z value to an X right? the answer i got it 29.91, or 30 minutes
 
  • #7
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6,393
Yes, your equation is fine, and barring any arithmetic errors (I didn't check), your answer looks fine.
 

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