normal distribution -help!

**okello** · February 3rd 2009, 04:46 AM

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.

**mr fantastic** · February 3rd 2009, 04:56 AM

Originally Posted by okello

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.

$\Pr(X < 45) = 0.31$ .

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

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

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

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

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

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

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

Solve equations (1) and (2) simultaneously.

**okello** · February 3rd 2009, 09:20 AM

Hi Mr. Fantastic, thanx.

I got

as -82.6 and

as 70.6 using a calculator.

Kindly confirm. Thanks once again.

Regards,
Okello

**mr fantastic** · February 3rd 2009, 11:42 AM

Originally Posted by okello

Hi Mr. Fantastic, thanx.

I got

as -82.6 and

as 70.6 using a calculator.

Kindly confirm. Thanks once again.

Regards,
Okello

I doubt $\sigma$ 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).

**okello** · February 6th 2009, 04:00 AM

Hi Mr. Fantastic,
I made

= 0.31 ..............(i)

and

=0.08 ...................(ii)

implying that

0.31

+

= 45 -----iii
0.08

+

= 64 ------iv

And used calc to find 2 unknowns. And I got

as -82.6 and

as 70.6 using a calculator.

Await for your response.
Thanks

**mr fantastic** · February 6th 2009, 04:11 AM

Originally Posted by okello

Hi Mr. Fantastic,
I made

= 0.31 ..............(i)

and

=0.08 ...................(ii)

implying that

0.31

+

= 45 -----iii
0.08

+

= 64 ------iv

And used calc to find 2 unknowns. And I got

as -82.6 and

as 70.6 using a calculator.

Await for your response.
Thanks

Originally Posted by mr fantastic

$\Pr(X < 45) = 0.31$ .

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

Mr F adds: This does NOT mean that a = 0.31.

[snip]

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

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

Mr F adds: This does NOT mean that b = 008.

[snip]

If $\Pr(Z < a) = 0.31$ then you're expected to be able to get (using tables or technology) that a = -0.4959 since $\Pr(Z < -0.4959) = 0.31$

Similar with b.

This is the inverse normal problem - you must have met questions like this in your earlier work ....

**matheagle** · February 15th 2009, 07:10 PM

Originally Posted by okello

Hi Mr. Fantastic, thanx.

I got

as -82.6 and

as 70.6 using a calculator.

Kindly confirm. Thanks once again.

Regards,
Okello

A negative standard deviation?
This is why I don't like seeing calculators in my class, especially integral calculators.

**okello** · February 16th 2009, 02:19 AM

Hi Matheagle,
Have you calculated it?. What answer did you get?

Regards,

**mr fantastic** · February 16th 2009, 02:28 AM

Originally Posted by okello

Hi Matheagle,
Have you calculated it?. What answer did you get?

Regards,

Wrong questions.

Here are the correct questions:

Hi okello,
Have you calculated b? What answer did you get?

**matheagle** · February 16th 2009, 05:27 PM

I'm going to put a similar problem on my exam next week.

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