Hello, Im trying to find Pr(|X|>15) where X is approximately a normal distribution. But my answer doesnt match the correct answer. I have no idea whats wrong.
ok.. how did you solved it? once check 1-(your answer) matches with correct answer
I rearranged it as 2-2(Pr(|X|>15)) this may be a mistake but I dont see how this rearrangement is wrong. Afterwards I just found the z-score and looked it up on a table.
sorry I mean 2-2Pr(|X|<=15)
I think just 2P(X<15) should do it.
Do you have the mean and standard deviation of X?
@CliffSedge not x<15, it should be x>15
Ah, yes, I see the error in considering the ||
Though, I think because of symmetry, it doesn't matter.
@witcon .. see pr(|x|>15)= pr(x<-15) +pr(x>15) or 1- pr(-15<x<15) since pr(x<-15) = pr(x>15) 2*pr(x>15) will do, @CliffSedge since it is normal dist. mean = 0, SD =1
It's only N(0,1) if standard normal.
Its a normal dist. not normal standard, I have to convert using z-score
u=20 var=25
Also theloser's derivation gives the wrong answer
calculate this 2*pr(x<-15)
0.3174, the correct answer is 0.8413
Is there something else |X| could mean? Otherwise I dont see where Im going wrong.
mu= 20 and sigma = 5 Pr(x>15) is 0.8413 (see http://davidmlane.com/hyperstat/z_table.html) note: you might have to refresh the page before it takes input Pr(x<-15) is way off in the left tail, essentially 0
yes youre right Pr(X>15) is 0.8413 but thats not the same thing as Pr(|X|>15) is it?
you want the sum of Pr(x>15) + Pr(x<-15) but -15 is 7 std dev below the mean at 20
you change |x|>15 to 2 relations +x > 15 and -x > 15 --> x< -15
wait sorry Pr(X>15) is 1- Pr(X< 15) which is 1- 0.8413 or 0.1587
? 20 is the mean. the probability of being 20 or higher is 0.5 (half the samples are 20 or greater) the prob of 15 or greater is (obviously?) greater than 0.5
am I doing the z-score wrong, (15-20)/5 should be -1, meaning Z < -1 or Z > 1 or 1-(Z<1) which is 0.1587
Im making a mistake somewhere there
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