Normal Distribution

Here is the problem..... A certain Dry cereal is packaged in 24-oz boxes. The machine that fills the boxes is set so that, on the average, a box contains 24.5 oz. The machine-filled boxes have contents weights that can be closely approximated by a normal curve. What percentage of the boxes will be underweight if the standard deviation is as follows?



So here is what I got. Mue is 24.5 oz

Std. Deviation is .5oz

24.5-24/.5 = .5/.5 = 1 which .341

24.5-25=-.5/.5 = -1 which is .341

From here my answer is really wrong. I know this should be simple but I can't get it. Please help!

Thanks!

Hello, hhardin!

No one is responding because the information is confusing and incomplete.

A certain cereal is packaged in 24-oz boxes.
The machine that fills the boxes is set so that, on the average, a box contains 24.5 oz.. ??
The machine-filled boxes have contents weights that can be closely approximated by a normal curve.
What percentage of the boxes will be underweight if the standard deviation is 0.5 oz?

Click to expand...

\(\displaystyle \text{They are packing }24\tfrac{1}{2}\text{ ounces of cereal (sometimes more) into a }24\text{-ounce box ?}\)


\(\displaystyle \text{And we need a definition of "underweight".}\)


\(\displaystyle \text{Or, if you prefer, we can reason as follows:}\)

\(\displaystyle \text{Half of the boxes weighs }less\text{ than the average}\)
. . \(\displaystyle \text{and the other half weighs }more\text{ than the average.}\)

\(\displaystyle \text{Therefore, 50\% of the boxes are underweight.}\)

Hello hhardin:

Hopefully the correct answer is .159 or 15.9 percent of the boxes will be less than 24 oz.

The work that you show provides the chance of being within one standard deviation of the mean....which is any where from 24 oz to 25 oz. or .341 + .341 or .682.

The z table show the cumulative area of the curve so one standard deviation from the mean would be the area from zero ounces to 25 oz or .841. Since the bell curve is symetrical the chance of the boxes being less than one standard deviation from the mean of 24.5 oz or under 24 oz is the same as the chances of the cereal weighing over one standard deviation over the mean or greater than 25 oz.

Since the table gives the cumulative area of one standard deviation as .841 and the total area under the curve is one than the chance of being either greater than one or less than one SD is 1-.841 or .159.

Or you get the same answer by .5 - .341 = .159.

Hope my answer is both helpful and correct.

Kind regards