In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the mean and S.D. of the distribution.

I need some help on the above number. Thanx.

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- Feb 3rd 2009, 04:46 AMokellonormal distribution -help!
In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the mean and S.D. of the distribution.

I need some help on the above number. Thanx. - Feb 3rd 2009, 04:56 AMmr fantastic
$\displaystyle \Pr(X < 45) = 0.31$.

Find the value $\displaystyle a$ of $\displaystyle Z$ such that $\displaystyle \Pr(Z < a) = 0.31$.

From $\displaystyle Z = \frac{X - \mu}{\sigma}$ it follows that $\displaystyle a = \frac{45 - \mu}{\sigma}$.

Substitute your value of $\displaystyle a$ to get equation (1) with the unknowns $\displaystyle \mu$ and $\displaystyle \sigma$.

$\displaystyle \Pr(X > 64) = 0.08$.

Find the value $\displaystyle b$ of $\displaystyle Z$ such that $\displaystyle \Pr(Z > b) = 0.08$.

From $\displaystyle Z = \frac{X - \mu}{\sigma}$ it follows that $\displaystyle b = \frac{64 - \mu}{\sigma}$.

Substitute your value of $\displaystyle b$ to get equation (2) with the unknowns $\displaystyle \mu$ and $\displaystyle \sigma$.

Solve equations (1) and (2) simultaneously. - Feb 3rd 2009, 09:20 AMokellonormal distribution - help!
Hi Mr. Fantastic, thanx.

I got http://www.mathhelpforum.com/math-he...3c147d21-1.gif asand http://www.mathhelpforum.com/math-he...88de1d0a-1.gif as**-82.6**using a calculator.**70.6**

Kindly confirm. Thanks once again.

Regards,

Okello - Feb 3rd 2009, 11:42 AMmr fantastic
I doubt $\displaystyle \sigma$ is correct - for starters you have it negative!! Even ignoring the -ve as a typo it's size is way too big.

If you post your calculations I will review them for the mistake(s) (but be patient because I'm very busy and might take some time to post). - Feb 6th 2009, 04:00 AMokellonormal distribution - help!
Hi Mr. Fantastic,

I made http://www.mathhelpforum.com/math-he...48f3713a-1.gif = 0.31 ..............(i)

and http://www.mathhelpforum.com/math-he...dacbc5e0-1.gif =0.08 ...................(ii)

implying that

0.31http://www.mathhelpforum.com/math-he...3c147d21-1.gif + http://www.mathhelpforum.com/math-he...88de1d0a-1.gif = 45 -----iii

0.08http://www.mathhelpforum.com/math-he...3c147d21-1.gif + http://www.mathhelpforum.com/math-he...88de1d0a-1.gif = 64 ------iv

And used calc to find 2 unknowns. And I got http://www.mathhelpforum.com/math-he...3c147d21-1.gif asand http://www.mathhelpforum.com/math-he...88de1d0a-1.gif as**-82.6**using a calculator.**70.6**

Await for your response.

Thanks - Feb 6th 2009, 04:11 AMmr fantastic
If $\displaystyle \Pr(Z < a) = 0.31$ then you're expected to be able to get (using tables or technology) that a = -0.4959 since $\displaystyle \Pr(Z < -0.4959) = 0.31$

Similar with b.

This is the inverse normal problem - you must have met questions like this in your earlier work .... - Feb 15th 2009, 07:10 PMmatheagle
- Feb 16th 2009, 02:19 AMokellonormal distribution - help!
Hi Matheagle,

Have you calculated it?. What answer did you get?

Regards, - Feb 16th 2009, 02:28 AMmr fantastic
- Feb 16th 2009, 05:27 PMmatheagle
I'm going to put a similar problem on my exam next week.