If X is a normal variate with mean 30 and standard deviation 5, find the probability that

i) 26<x<40

ii) x>45

iii) |x-30|>5

Solution :

$\displaystyle \mu = 30$

$\displaystyle \sigma = 5$

am stuck on hw to solve the problem

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- Jan 2nd 2010, 04:39 AMwolfyparadiseIf X is a normal variate
*If X is a normal variate with mean 30 and standard deviation 5, find the probability that*

i) 26<x<40

ii) x>45

iii) |x-30|>5

Solution :

$\displaystyle \mu = 30$

$\displaystyle \sigma = 5$

am stuck on hw to solve the problem - Jan 2nd 2010, 09:51 AMANDS!
X comes from a population that is normally distributed. Therefore with a mean of 30 and a standard deviation of 5, you can easily convert your other statistics into Z-scores to calculate applicable probabilities. For the absolute value one, just create an interval like you normally would for absolute value: |x-30|<5 = 25<X<35.

- Jan 3rd 2010, 01:22 AMwolfyparadise
- Jan 3rd 2010, 03:12 AMmr fantastic
For part (iii),

*|x-30|>5*is equivalent to X > 35 or X < 25. And since the mean is 30, you can use symmetry to reduce the calculation: Pr(X > 35) + Pr(X < 25) = 2 Pr(X > 35) = 2 (1 - Pr(X < 35)) = 2 - 2 Pr(X < 35).

Now ...... surely you have been taught how to convert from a normal variable X to a standard normal variable Z. And you will have been taught how to calculate probability from a standard normal distribution (tables, technology, etc.). These are all routine questions - you will have examples in class notes and your textbook. Furthermore, the forum has dozens of worked examples. MHF is not a substitute for you going back to your classnotes and textbook and doing the necessary reading an revision.

If you need more help, please show that you have made some effort to nderstand the material. Show your work. Say specifically where you are stuck.