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# Normal Distribution Watch

1. Hey, I have a question. Say if X is distributed normally X~N(6600, 5500) where 6600 is the average time in minutes and 5500 is the variance.
Then when the question says find P(X>180) where the 180 is in hours. So i decided to divide the mean and variance by 60 to get the time in hours but the answer comes out wrong. Can't I do this? Thanks!
EDIT: When i convert 180 into minutes, it works just fine but I dont understand why...
2. Convert 180 hours to minutes from the beginning.
3. (Original post by Glutamic Acid)
Convert 180 hours to minutes from the beginning.
Why doesn't it work the other way? I just don't understand it.
4. (Original post by JBKProductions)
Why doesn't it work the other way? I just don't understand it.
Ah, ok. Let X be the time in minutes, and Z the time in hours.

So Z = X/60.

Thus E[Z] = E[X/60] = E[X]/60, which is ok.
But Var[Z] = Var[X/60] = Var[X]/3600, which is presumably what went wrong.

{I'm using Var[aX] = a^2Var[X].}
5. The variance is squared; you can't just divide it by 60.
6. Oh I see. Thanks!
7. The distances travelled to work D km, by the employees at a large company are normally distributed with D ˜ N (30, 8 squared)

(a) Find the upper quartile, Q3, of D.

(b) Write down the lower quartile, Q1, of D.

(c) An outlier is defined as any value of D such that D < 8.4 or D > 51.6
An employee is selected at random.
Find the probability that the distance travelled to work by this employee is an outlier.
________________________________ ________________________________ ______________

For (a) you do P (D < Q3) and make it equal to something that’s what someone told me! I was wondering why the sign was < and not > to find the P (D < Q3)?

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Updated: December 18, 2010
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