Normal Distribution question...

This is from a college problem sheet. I'm having quite a bit of difficulty with it. So any help would be greatly appreciated.

Let the random variable X have the N(7,37) distribution. Determine constants s, t so that the probability that |X-s|>t is 0.01. The value to return is t.

Now, to solve this i've tried this:
Set s = mean = 7.
Divide by sqrt(37) to normalise the distribution to N(0,1)

This gives: P[ (|X-7|)/(sqrt(37)) > t/(sqrt(37)) ] = 0.01.

From this i should be able to work out the answer, but i can't seem to get it. As far as i can tell, this equation is telling me that 1% of the area is shaded, meaning that i should be looking up my Z-tables to find out the area beneath 0.995 and then take away the area beneath 0.990, then adding the area beneath 0.005 (this is because this is the modulus... so i need the top 1/2% and the bottom 1/2%, if ya know what i mean)

Or something like that :eek:


TS Contributor
I'm working on this one....

An equivalent statement would be:

P(|Z| > 2.576 ) = 0.01

and since sigma is sqrt(37) = 6.083, then we have:

P(|X-s| > (2.576*6.083)) = 0.01
P(|X-s| > 15.67) = 0.01

So maybe working backwords from this to find values for t and s that work...