# normal distribution question

• May 22nd 2011, 09:15 AM
wintersoltice
normal distribution question
I'm looking at an example. There's one part that I don't understand.

The random variable Y has a normal distribution with mean 10 and standard deviation 2. Find the value of the constant a such that P(10-a<Y<10+a)=0.95.

Y~N(10,4)
Given P(10-a<Y<10+a)=0.95,
Therefore, P(Y<10-a)=0.025
10-a=6.080072028
a=3.92

What I don't get it is how does P(10-a<Y<10+a)=0.95
gives me P(Y<10-a)=0.025 ???

Any help is much appreciated. Thanks.
• May 22nd 2011, 09:33 AM
TKHunny
A little arithmetic.

P(10-a < Y < 10+a) = 0.95

P(10-a < Y AND Y < 10+a) = 0.95

1 - P(10-a < Y AND Y < 10+a) = 1 - 0.95

P(10-a > Y OR Y > 10+a) = 0.05

P(10-a > Y) + P(Y > 10+a) = 0.05

2*P(10-a > Y) = 0.05

P(Y<10-a)=0.025
• May 22nd 2011, 09:37 AM
SpringFan25
see the attached picture.

http://www1.picturepush.com/photo/a/...mg/5704989.jpg

The area under the distribution is 100%, and you are told that 95% is in the range (10-a,10+a). That leaves 5% outside this range.

The normal distribution is symmetric about the mean so there must be 2.5% on either end.

Hence P(Y<10-a) = 2.5% = 0.0025