distribution probability

WillB · #1 $Old$ June 29th, 2009, 09:10 PM

Hello everyone!

I would like to know how to do this question.
Let X have the distribution N(80,225). Find
a. $P(75 \le X \le 85)$
b. P(X<50)
c. P(X>95)
d. P(|X-80|>18)

Thanks

mr fantastic · #2 $Old$ June 29th, 2009, 09:31 PM

Quote:

Originally Posted by WillB $View Post$

Hello everyone!

I would like to know how to do this question.
Let X have the distribution N(80,225). Find
a. $P(75 \le X \le 85)$
b. P(X<50)
c. P(X>95)
d. P(|X-80|>18)

Thanks

Are you using tables?

Your textbook or class notes should have salient examples to follow. You will need to be much more specific in stating what you can and can't do.

Note:

If using tables, you first need to convert X-values to Z-values.

a. = Pr(X < 85) - Pr(X < 75) = ....

b. $\Pr(X < 50) = \Pr \left(Z < \frac{50 - 80}{15}\right) = \Pr(Z < -2) = \Pr(Z > 2) = 1 - \Pr(Z < 2)$ .

c. = 1 - Pr(X < 95) = ....

d. = Pr(X > 98) + Pr(X < 62) = 2 Pr(X > 98) (since 80 is the mean) = 2 (1 - Pr(X < 98)) = ....

WillB · #3 $Old$ June 30th, 2009, 02:38 PM

I'm sorry. I am using tables to find the values. This is my first post and I am not really sure how this works. I will show you what I did so far.

a.

$P(\frac{75-80}{15}\le\frac{x-80}{\sqrt{225}}\le\frac{85-80}{15})=\Phi(0.3)-\Phi(-0.3)=0.6293-0.3707=0.2586$

b. P(X<50)

$\frac{50-80}{15}=-2$

$\Phi 2=0.9772$ (I'm not sure which table to use for >)

1-0.9772=0.0228

c. P(X>95)

$\frac{95-80}{15}=0.33333$

$1-\Phi 0.33333=1-0.6293=0.3707$

d. P(|X-80|>18)

I'm not exactly sure how you did it, but going from your work:
2(1-P(X<98)

$\frac{98-80}{15}=1.2$

$\Phi 1.2=0.8849$

2(1-0.8849)=0.2302

mr fantastic · #4 $Old$ June 30th, 2009, 04:17 PM

Quote:

Originally Posted by WillB $View Post$

I'm sorry. I am using tables to find the values. This is my first post and I am not really sure how this works. I will show you what I did so far.

a.

$P(\frac{75-80}{15}\le\frac{x-80}{\sqrt{225}}\le\frac{85-80}{15})=\Phi(0.3)-\Phi(-0.3)=0.6293-0.3707=0.2586$

Mr F says: ${\color{red} \frac{1}{3} \neq 0.3}$ . The accuracy of your answer has suffered as a consequence. Since your tables will be 'four figure' maths tables you should at least use ${\color{red} \frac{1}{3} = 0.3333}$ (correct to four decimal places).

b. P(X<50)

$\frac{50-80}{15}=-2$

$\Phi 2=0.9772$ (I'm not sure which table to use for >)

1-0.9772=0.0228

Mr F says: Looks OK.

c. P(X>95)

$\frac{95-80}{15}=0.33333$ Mr F says: Last time I checked, ${\color{red}\frac{95-80}{15}= 1}$ .

$1-\Phi 0.33333=1-0.6293=0.3707$

d. P(|X-80|>18)

I'm not exactly sure how you did it, but going from your work: Mr F says: I used the symmetry of the normal distribution arund its mean.

2(1-P(X<98)

$\frac{98-80}{15}=1.2$

$\Phi 1.2=0.8849$

2(1-0.8849)=0.2302 Mr F says: Looks OK. Note that the calculation is the same as Pr(Z > 1.2) + Pr(Z < -1.2). Perhaps the symmetry is more obvious now ....

..

WillB · #5 $Old$ June 30th, 2009, 04:57 PM

You have been a great help! The tables in the book only show two decimal points, so for part a I used 0.33 as my value. I could calculate it using the formula, but what does the 'w' mean?

I don't know how I messed up c (answer should be 0.1587) and YES, now I understand where you got d.

mr fantastic · #6 $Old$ June 30th, 2009, 05:07 PM

Quote:

Originally Posted by WillB $View Post$

You have been a great help! The tables in the book only show two decimal points, so for part a I used 0.33 as my value. I could calculate it using the formula, but what does the 'w' mean?

[snip]

What w? In your tables? I'm afraid I don't know without seeing the context.

WillB · #7 $Old$ June 30th, 2009, 05:23 PM

Yes, they give this equation at the top of the normal distibution table
$P(Z \le z)=\int^{z}_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-w^{2}/2}dw$ I see that $W=\frac{X-\mu}{\sigma}$ , so w should be the same thing.

mr fantastic · #8 $Old$ June 30th, 2009, 05:36 PM

Quote:

Originally Posted by WillB $View Post$

Yes, they give this equation at the top of the normal distibution table
$P(Z \le z)=\int^{z}_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-w^{2}/2}dw$ I see that $W=\frac{X-\mu}{\sigma}$ , so w should be the same thing.

This intergral is the cdf of the normal distribution. w is a dummy variable of integration. The integral can't be done exactly (unless z = 0 or +oo), it has to be done numerically using technology (in which case you may as well just use the technology to directly get the probability in the first place).

matheagle · #9 $Old$ July 1st, 2009, 01:44 AM

Here, try this table
Free Cumulative Area Under the Standard Normal Curve Calculator