• December 8th 2008, 03:54 PM
lava1980
Can someone please explain this question to me:

In grading eggs into small, medium and large, the Nancy Farms packs the eggs that weigh more than 3.6 oz into packages marked "large," and the eggs that weigh less than 2.4 oz into packages marked "small". The remainder are packaged as "medium". If a day's packaging contained 10.2% large eggs and 4.18% small eggs, determine the mean and the standard deviation for the eggs' weights. Assume the distribution is normal.

Which values do I find the z scores for? .102 and .418? or 3.6 and 2.4? Then which equation do I put the values in to solve?

Help!
• December 8th 2008, 04:11 PM
mr fantastic
Quote:

Originally Posted by lava1980
Can someone please explain this question to me:

In grading eggs into small, medium and large, the Nancy Farms packs the eggs that weigh more than 3.6 oz into packages marked "large," and the eggs that weigh less than 2.4 oz into packages marked "small". The remainder are packaged as "medium". If a day's packaging contained 10.2% large eggs and 4.18% small eggs, determine the mean and the standard deviation for the eggs' weights. Assume the distribution is normal.

Which values do I find the z scores for? .102 and .418? or 3.6 and 2.4? Then which equation do I put the values in to solve?

Help!

Let X denote the random variable weight of an egg.

X ~ Normal $(\mu, \sigma^2)$.

Given data:

1. Pr(X > 3.6) = 0.102.

2. Pr(X < 2.4) = 0.0418.

1. $\Pr( Z > z*) = 0.102 \Rightarrow z^* = 1.2702$ (correct to four decimal places).

Therefore $z^* = \frac{3.6 - \mu}{\sigma} \Rightarrow 1.2702 = \frac{3.6 - \mu}{\sigma}$ .... (1)

2. $\Pr( Z < z*) = 0.0418 \Rightarrow z^* = -1.7302$ (correct to four decimal places).

Therefore $z^* = \frac{2.4 - \mu}{\sigma} \Rightarrow -1.7302 = \frac{2.4 - \mu}{\sigma}$ .... (2)

Solve equations (1) and (2) simultaneously for $\mu$ and $\sigma$.
• December 9th 2008, 02:34 PM
lava1980
but how do you solve the equations with two variables?
Thank you so much for your help.

I am with you so far, but I don't know how to solve the equation with two variables missing (the mean and the standard deviation).

Where do I go from here?
• December 9th 2008, 07:17 PM
mr fantastic
Quote:

Originally Posted by lava1980
Thank you so much for your help.

I am with you so far, but I don't know how to solve the equation with two variables missing (the mean and the standard deviation).

Where do I go from here?

$1.2702 = \frac{3.6 - \mu}{\sigma}$ .... (1)

$-1.7302 = \frac{2.4 - \mu}{\sigma}$ .... (2)

These are simultaneously equations. You solve them for the unknowns $\mu$ and $\sigma$.

It's no different to solving

$1.2702 = \frac{3.6 - x}{y}$ .... (1)

$-1.7302 = \frac{2.4 - x}{y}$ .... (2)

for x and y ....