# backwards Normal distribution problem

• April 5th 2008, 12:40 PM
BklynKid
backwards Normal distribution problem
I feel like I should be able to do this but I feel absolutely lost.

Assume that the weight, X, of male college students is normally distributed. Assume further that P(X <= 160) = 1/2 and P(X <= 140) = 1/4.

a.) What find the value of 'mu' and 'sigma'
b.) Calculate P(X >= 190)
c.) Calculate what percentage of students who weigh over 190 pounds will weigh more than 200 pounds.

I think b and c should be simple enough once I have mu and sigma but how do I get those? Any help would be appreciated. (Bow)
• April 5th 2008, 01:55 PM
mr fantastic
Quote:

Originally Posted by BklynKid
I feel like I should be able to do this but I feel absolutely lost.

Assume that the weight, X, of male college students is normally distributed. Assume further that P(X <= 160) = 1/2 and P(X <= 140) = 1/4.

a.) What find the value of 'mu' and 'sigma'
b.) Calculate P(X >= 190)
c.) Calculate what percentage of students who weigh over 190 pounds will weigh more than 200 pounds.

I think b and c should be simple enough once I have mu and sigma but how do I get those? Any help would be appreciated. (Bow)

P(X <= 160) = 1/2 tells you straight away that $\mu = 160$.

To get $\sigma$:

$Z = \frac{X-\mu}{\sigma} \Rightarrow z_{140} = \frac{140 - 160}{\sigma} = \frac{-20}{\sigma}$

where $\Pr(Z < z_{140}) = \frac{1}{4} \Rightarrow z_{140} = -0.6745$ (from inverse tables, correct to four decimal places).

Therefore $-0.6745 = \frac{-20}{\sigma} \Rightarrow \sigma = ......$.

The slightly more difficult question would have been if they'd given you P(X <= 200) = 0.9113 and P(X <= 140) = 1/4.

Then you'd need to solve the following simultaneous equations:

$z_{200} = \frac{200 - \mu}{\sigma}$ .... (1)

$z_{140} = \frac{140 - \mu}{\sigma}$ .... (2)

for $\mu$ and $\sigma$ ......
• April 5th 2008, 02:46 PM
BklynKid
Wow that's amazing, thanks! Yea that makes sense! :) I was able to do part (b) just fine using that.

A few questions though, what is the reason behind $\mu = 160$ ? Just because the probability is 1/2 that makes it an automatic?
And how would I then approach part (c) of this problem when P(X >= 190) = 0.154

Again, thank you.
• April 5th 2008, 02:58 PM
mr fantastic
Quote:

Originally Posted by BklynKid
Wow that's amazing, thanks! Yea that makes sense! :) I was able to do part (b) just fine using that.

A few questions though, what is the reason behind $\mu = 160$ ? Just because the probability is 1/2 that makes it an automatic?
And how would I then approach part (c) of this problem when P(X >= 190) = 0.154

Again, thank you.

The normal distribution is a symmetric around its mean. It's always true for a normal distribution that $\Pr(X \geq \mu) = \Pr(X \leq \mu) = \frac{1}{2}$.

c.) Conditional probability: You want $\Pr(X > 200 | X > 190) = \frac{\Pr(X > 200 \, \text{and} \, X > 190)}{\Pr(X > 190)} = \frac{\Pr(X > 200)}{\Pr(X > 190)} = .....$
• April 5th 2008, 03:19 PM
BklynKid
Yea that makes sense also. Thanks a ton! :D :)

(I'm gonna donate right now)