# A form of standard deviation?

1. Jan 16, 2005

### Matt.D

*I have already posted this in another forum, but re-read the rules regarding homework questions.. Mods, I hope this is ok*

Hey guys, I've got this question from my Statistics Homework and wondered if someone could point me to a web site or supply some advice as to how to begin to solve the problem.

Bags of sweets are packed by a machine such that the masses (X) have a normal distribution with mean 250g and standard deviation 10g.
A bag is judged to be underweight and rejected if X<225g.
A bag is judged to be overweight and rejected if X>270g
What percentage of bags are rejected?

I've tried a few combinations, but without a formula I don't think I'm making any sense. Can an altered version of the formula for Standard Deviation be used?

Any help always appreciated : )

Matt

2. Jan 16, 2005

### jamesrc

You don't need to calculate the standard deviation in this problem; it's given to you. You should have (either in your textbook or look it up on the net) a plot and table of the normal distribution. As far as I know, it's pretty standard to see the area under the curve from zero to a given z-score tabulated. The z-score is defined as the distance away from the mean as a fraction of the standard deviation ($$z = \frac{x-\bar x}{\sigma}$$).

For your problem, you want to find the sum of the probability that a sample is higher than 270 and lower than 225. For a 270, the z-score is (270-250)/10 = 2 (that's how many standard deviations away from the mean it is). If you look up the area under the normal distribution for z = 2, you should get 0.47725 (unless I read off the wrong row or something). That means that ~48% of the data is between z = 0 and z = 2. But you want to know how much of the data is greater than z = 2, so your answer would be 50% - 0.47725 (because the total area under the curve from z = 0 to z = infinity is 50%, right?).

Now do the same thing for the lower rejection point and add the two probabilities together (and express your answer as a percentage).

I hope that gets you going with problems like this.

3. Jan 17, 2005

### Matt.D

"If you look up the area under the normal distribution for z = 2, you should get 0.47725 (unless I read off the wrong row or something)"

Hi James,

Thanks for your help so far but I've become a little unstuck trying to find 0.47725? I don't understand where that comes from.

Regards

Matt

4. Jan 17, 2005

### jamesrc

5. Jan 19, 2005

### Matt.D

Hi James,

Thanks for all your help. I now understand Normal Distribution that little bit better :)

Matt