# If X is a normal variate

• Jan 2nd 2010, 05:39 AM
If 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 :
$\mu = 30$
$\sigma = 5$

am stuck on hw to solve the problem
• Jan 2nd 2010, 10:51 AM
ANDS!
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, 02:22 AM
Quote:

Originally Posted by ANDS!
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.

i am having trouble with starting the question .....I dont know what the first step should be ......
do u mean i integrate the from 26 to 40 for x ...........but i dont even know the value of f(x)
• Jan 3rd 2010, 04:12 AM
mr fantastic
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