# Normal distribution

## 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) = 0.2

## Homework Equations

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

## The Attempt at a Solution

P(X < 10) - P(X < x) = 0.2
P(Z < (10-10)/2) - P(Z < (x-10)/2) = 0.2
P(Z < 0) - P(Z < (x-10)/2) = 0.2
0.5 - P(Z < (x-10)/2) = 0.2
P(Z < (x-10)/2) = 0.3
(x-10)/2 = 0.617911
x - 10 = 1.235822
x = 11.235822

The answer doesn't make sense. x is supposed to be smaller than 10.

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## Answers and Replies

Mark44
Mentor
P(x < X < 10) = 0.2 <==> P(z < Z < 0) = 0.2, where Z is the usual standard normal distribution.

From a table of areas under the standard normal distribution, I find a z value of about -.525.

z = (x - 10)/2 ==> 2z = x - 10 ==> x = 2z + 10

Substituting the value of z = -.525 that I found earlier, I get an x value of 8.95.

So, P(8.95 < X < 10) = 0.2, approximately

So when I tried to solve the problem, what did I do wrong?

Mark44
Mentor

## The Attempt at a Solution

P(X < 10) - P(X < x) = 0.2
P(Z < (10-10)/2) - P(Z < (x-10)/2) = 0.2
P(Z < 0) - P(Z < (x-10)/2) = 0.2
0.5 - P(Z < (x-10)/2) = 0.2
P(Z < (x-10)/2) = 0.3
Your error is in the next line. Your z-value (which is what you're getting from the table) should be negative. Apparently you picked the wrong value. If you recall, my value was -.525.
(x-10)/2 = 0.617911
x - 10 = 1.235822
x = 11.235822

In working these kinds of problems, I find that it is much easier to switch right away to a probability involving z (or t, or whatever), do my calculation and look up the value, and then change back to the original random variable X.

Mark

I checked my book again, and it shows that the z value for 0.3 is 0.511967. Is that incorrect?

Mark44
Mentor
I think you might be using your table incorrectly. The table I used has probability values (areas under the bell curve) only to 4 decimal places, but that's just a detail.

For z = 0.3, my table shows a probability of 0.6179.
For z = 1.0, it shows 0.8413.

Yes, I was looking at the table incorrectly. For z = 0.3, the probability is 0.617911. This is similar to what you got and it's what I originally found.

So for z = 0.3, if the probability is 0.617911, then shouldn't the following steps be correct?

P(Z < (x-10)/2) = 0.3
(x-10)/2 = 0.617911

Mark44
Mentor
No. Since the probability P(Z < zp) = 0.3 (zp is the particular z value you're looking for), you have to be looking for a z-value in the left half of the bell curve. IOW, for negative values of z. Keep in mind that for z = 0, half of the area is to the left, and half to the right.

If you're not working with a sketch of the bell curve, with the area you want shaded in, you should be.

Why does it have to be in the left half of the curve? 0.3 is still a positive number. What part in my answer should I change, and how should I change it?

Mark44
Mentor
Because you want P(Z < zp) = 0.3. This probability represents the area under the bell curve from z = $-\infty$ to some z value to the left of zero. If the inequality went the other way, as in P(Z > zp) = 0.3, you would be looking for a positive z-value.

If you had to solve P(Z < zp) = .5, what would you get for what I'm calling zp?

For P(Z < zp) = .5
zp = 0.691462

Mark44
Mentor
No. zp = 0
I don't think you get the connection between probability and area under the bell curve. For example, P(-1 < Z < 0) represents the area under the curve between z = -1 and z = 0. The area/probability is about .34.

I thought that the probability is the area under the curve.

Although now I can see how you're using the table. So I guess for P(Z < zp) = .3, then zp = -0.53 or -0.52?

Mark44
Mentor
I thought that the probability is the area under the curve.

Although now I can see how you're using the table. So I guess for P(Z < zp) = .3, then zp = -0.53 or -0.52?

Yes, probability is the area under the curve, but if you have a probability like P(Z< a), the probability is the area under the curve between z = $-\infty$ and z = a. If it's a probability like P(a < Z < b), it's the area under the curve between z = a and z = b. Finally, for a probability like P(Z > b), it's the area under the curve from z = a to z = $\infty$.

As for the values, if you recall, I first said that it was about -.525.

What I was asking is how do you know if the answer is -0.53 or -0.52? The z value doesn't correspond to one of those numbers but rather a number in between. Which value do I use, or does it not matter?

Mark44
Mentor
Since the probability I was looking for was about midway between those numbers, I interpolated to get -.525, which is a better choice than either -.52 or -.53.

If you want to find out more about this, do a search on "linear interpolation."

I think I've pretty much got this problem figured out now. Thanks!