Solving a normal distribution problem

[6021023]

Homework Statement

Assume X is normally distributed with a mean of 10 and a standard deviation of 2. Determine the value for x that solves P(-x < X - 10 < x) = 0.95

Homework Equations

P(X < x) = P(Z < z) = P(Z < (x - mean)/stdev)

The Attempt at a Solution

P(10 -x < X < 10 + x) = 0.95
P(X < 10 + x) - P(X < 10 - x) = 0.95
P(Z < (10+x-10)/2) - P(Z < (10-x-10)/2) = 0.95
P(Z < x/2) - P(Z < -x/2) = 0.95

That's as far as I get, and then I don't know what to do next, or if what I did was correct.

 

[Focus]

I hope you have some normal tables for this. You are making some mistakes in that P(X < 10 + x) - P(X < 10 - x) = 0.95 isn't true. First of all, what is the distribution of X-10?
Can you use symmetry in any way to simplify P(-x < X - 10 < x)? After that can you make your distribution N(0,1)?

If you have any problems on these then give me a shout :)

 

[6021023]

Is P(-x < X - 10 < x) the same as P(10 -x < X < 10 + x)?

 

[Mark44]

Yes, sure.

 

[6021023]

Hi Mark. Is P(10 -x < X < 10 + x), the same as P(X < 10 + x) - P(X < 10 - x)?

 

[Mark44]

Yes.

 

[6021023]

So if P(10 -x < X < 10 + x) is the same as P(X < 10 + x) - P(X < 10 - x), then if P(-x < X - 10 < x) = 0.95 then does that mean that P(X < 10 + x) - P(X < 10 - x) = 0.95?

 

[Mark44]

That works for me. I got into this late, but I can't see how Focus's comment in post 2 could be correct.

 

[6021023]

So I guess that what I have down so far is correct, right? It appears that there is symmetry where I left off at P(Z < x/2) - P(Z < -x/2) = 0.95. Could I then do the following?

2[P(Z < x/2) - P(Z < 0)] = 0.95

 

[Mark44]

Absolutely. Did you make this realization after making a sketch of the bell curve?

 

[6021023]

Yeah, -x/2 and x/2 are like -a and a. I drew those points on a bell curve and saw the symmetry.

So would the following be correct?

2[P(Z < x/2) - P(Z < 0)] = 0.95
P(Z < x/2) - P(Z < 0) = 0.475
P(Z < x/2) - 0.5 = 0.475
P(Z < x/2) = 0.975
x/2 = 1.96
x = 3.92

 

[Mark44]

Looks good. There are some rules of thumb you can use for N(0, 1) distributions:
Probability within 1 sigma of the mean (i.e., -1 < z < 1), $\approx$ .68
Probability within 2 sigmas of the mean, $\approx$ .95
Probability within 3 sigmas of the mean, $\approx$ .99+

Your x values are a little under 2 s.d. away, so the probability/area should be around .95, which is right on the money for your problem.

 

[6021023]

What's an N(0, 1) distribution? Does that just mean normal distribution? What does the 0 and the 1 mean?

 

[Mark44]

Sorry. That is just notation for the standard normal distribution: mean = 0, and standard deviation = 1. Your original distribution was N(10, 2).

 

[6021023]

So can I use that checking probability trick for N(0,1) distributions for the N(10,2) distribution? Here the z value is 1.96, which is just under 1 s.d. away.

 

[Mark44]

The idea is that you can transform a random variable X that is N(mean, sd) to a z distribution (i.e., N(0, 1)) in this way:

z = (x - mean)/s.d.

You can go the other way, too:
x = z*s.d. + mean

Any probability written in terms of X can be rewritten in terms of z, which is preferable, since there are tables of z values.

 

[Focus]

You are making this much too complicated. You can't justify the step for P(X < 10 + x) - P(X < 10 - x) = 0.95. X-10 is normal with mean zero and standard deviation 2. You can use the symmetry about zero to say that's the same as 2P(X-10<x). I'll let you finish it off :)

 

[Mark44]

You are making this much too complicated. You can't justify the step for P(X < 10 + x) - P(X < 10 - x) = 0.95.

Sure he can. The expression above represents the probability that X lies between 10 - x and 10 + x.

X-10 is normal with mean zero and standard deviation 2.

We're way past that. You must not have read the other posts in this thread.

You can use the symmetry about zero to say that's the same as 2P(X-10<x). I'll let you finish it off :)

If you think about it at all, 2P(X - 10 < x) is larger than 1 if x < 10, so this is not a helpful start.

 

[Focus]

Sorry you are right I'm way off the ball, I meant to say 2P(X - 10 < x)-1.