Solve Normal Distrib. Homework: X mean 10, stdev 2, P(X<X<10)=0.2

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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.

 

[Mark44]

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

 

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

 

[Mark44]

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

 

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

 

[Mark44]

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.

 

[6021023]

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]

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.

 

[6021023]

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]

Because you want P(Z < zp) = 0.3. This probability represents the area under the bell curve from z = [itex]-\infty[/itex] 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?

 

[6021023]

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

 

[Mark44]

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.

 

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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]

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 = [itex]-\infty[/itex] 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 = [itex]\infty[/itex].

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

 

[6021023]

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]

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."

 

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