# Thread: Statistics: Normal Distribution

1. ## Statistics: Normal Distribution

Hi everyone... as you can tell I'm completely new to this forum. I'm studying A-level math in the UK and although I find it considerably more chalenging than my other subjects I do enjoy it. Hopefully I'll be able to give something back to this forum as I learn.

This is the question I'm stuck on:
Jam is packed in tins of nominal weight 1000g. The actual weight of Jam delivered to a tin by the filling machine is normally distributed about the mean weight set on the machine with a standard deviation of 12g. The average filling of Jam is 1000g.
It is a legal requirement that no more than 1% of tins contain less than the nominal weight.

b) Calculate the minimum setting of the filling machine which will meet this requirement.

The answer in the back of the textbook is 1028g
The book doesn't not provide a worked solution, and even though I think the working is pretty straightforward, I have no idea where to start. Any help would be greatly appreciated

2. Originally Posted by bb2120
Hi everyone... as you can tell I'm completely new to this forum. I'm studying A-level math in the UK and although I find it considerably more chalenging than my other subjects I do enjoy it. Hopefully I'll be able to give something back to this forum as I learn.

This is the question I'm stuck on:
Jam is packed in tins of nominal weight 1000g. The actual weight of Jam delivered to a tin by the filling machine is normally distributed about the mean weight set on the machine with a standard deviation of 12g. The average filling of Jam is 1000g.
It is a legal requirement that no more than 1% of tins contain less than the nominal weight.

b) Calculate the minimum setting of the filling machine which will meet this requirement.

The answer in the back of the textbook is 1028g
The book doesn't not provide a worked solution, and even though I think the working is pretty straightforward, I have no idea where to start. Any help would be greatly appreciated
This takes a bit of explaining - more than I have time for at the moment. If no-one else has bitten by the time I get a chance, I'll post my solution.

3. Look up the z score that corresponds to .99 in the body of the z-table.

Then use $\displaystyle z=\frac{x-1000}{12}$ and solve for x.

4. ## Thanks for the replies

I've spotted quite a few typos but can't edit my original post...

Galactus, thank you very much I didn't think it could be so simple...
But how did you come up with .99?

5. Originally Posted by bb2120
But how did you come up with .99?
0.99 = 99/100 = 99%

6. I see... taking into account the 1%
Thanks

7. Originally Posted by bb2120
Hi everyone... as you can tell I'm completely new to this forum. I'm studying A-level math in the UK and although I find it considerably more chalenging than my other subjects I do enjoy it. Hopefully I'll be able to give something back to this forum as I learn.

This is the question I'm stuck on:
Jam is packed in tins of nominal weight 1000g. The actual weight of Jam delivered to a tin by the filling machine is normally distributed about the mean weight set on the machine with a standard deviation of 12g. The average filling of Jam is 1000g.
It is a legal requirement that no more than 1% of tins contain less than the nominal weight.

b) Calculate the minimum setting of the filling machine which will meet this requirement.

The answer in the back of the textbook is 1028g
The book doesn't not provide a worked solution, and even though I think the working is pretty straightforward, I have no idea where to start. Any help would be greatly appreciated
I'll just toss in my 2 cents .....

We want to know the value of the mean so that Pr(X < 1000) = 0.01.

Let $\displaystyle \Pr(z < z_{1000}) = 0.01$. Then $\displaystyle z_{1000} = -2.32635$ using my TI-89.

This is the z-score corresponding to an x-score of 1000.

But you know that $\displaystyle Z = \frac{X - \mu}{\sigma}$.

So $\displaystyle z_{1000} = \frac{1000 - \mu}{12}$.

Therefore $\displaystyle -2.32635 = \frac{1000 - \mu}{12}$.

I get $\displaystyle \mu = 1027.9$, which checks out if you use that value to calculate Pr(X < 1000).