anonymous
  • anonymous
Mean lifetime of a wristwatch is 25 months, with a standard deviation of 5 months. If the distribution is normal, for how many months should a guarantee be made if the manufacturer does not want to exchange more than 10% of watches? Just want to make sure I'm doing the process right. The answer is 23.75 months, correct? That's from (-0.25*5) + 25
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Let me Go over it. and ill Give you a Yes or no. One sec @DurableToaster
anonymous
  • anonymous
You need to look up in Normal distribution tables the amount that leaves 90 % outside. (I think that is -1.28 from http://www.maths.leeds.ac.uk/~sta6ajb/ma... But that table is where the mean = 0 and variance =1 in our table, critical value is mean - 1.28 sqrt(variance) = 25 - 1.28 times 5 =25 - 6.4 = 18.6 I always found the easiest way to understand was to sketch the bell curve / standard normal distribution with the values of the mean (0) and the standard deviation pencilled in - virtually all the curve lies between 0 +/- 4 then immediately underneath, write teh corresponding values for the given curve - the mean 25 will be written beneath 0; the value 25 +5 (mean plus one standard deviateion) is immediately below the 1; 25 + 2 times 5 is immediately below the 2 of the standard normal. It should help you make sense of it. For this curve almost 100% lies between 25 + 4 times standard deviation(5) or 45 and at the lower end 25 - 4 times 5 or 5months.
anonymous
  • anonymous
Make since?

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anonymous
  • anonymous
I... I suppose you were taught differently than I. I was only given a table for half of the normal distribution area, not taught terms like critical value and variance.
anonymous
  • anonymous
Oh. Sorry if i made it confusing! D:
anonymous
  • anonymous
No, no. Using google I found a table that goes up to 99.99%. I use the same methods as you, just without a more lengthy table
anonymous
  • anonymous
Thanks for the help
anonymous
  • anonymous
np!

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