1. ## The normal distribution

Okay, so this may seem a bit simple for here, and I wasn't sure, but this is from my University past paper. The question reads:

A chocolate company produce boxes of chocolates with mean $1kg$. It is decided that all boxes weighing less than $990g$ or more than $1020g$ will be re-packed. If the weight of the boxes is Normally distributed with standard deviation $20g$, what percentage of boxes will be repacked?

So I got from this that...

$X\sim N(1,0.02^{2})$ and I was trying to find $1-\left(P(0.99\leq X\leq 1.02)\right)$ (because the stuff in the $P$ would be the boxes that were the right weight, so 1 minus that would be boxes to be repacked). However, after standardising and using the tables I'm given and what not, with this I get a percentage that, to me, seems a bit high something like 43%.

Can anyone help? Have I made a misunderstanding somewhere?

2. Originally Posted by chella182
Okay, so this may seem a bit simple for here, and I wasn't sure, but this is from my University past paper. The question reads:

A chocolate company produce boxes of chocolates with mean $1kg$. It is decided that all boxes weighing less than $990g$ or more than $1020g$ will be re-packed. If the weight of the boxes is Normally distributed with standard deviation $20g$, what percentage of boxes will be repacked?

So I got from this that...

$X\sim N(1,0.02^{2})$ and I was trying to find $1-\left(P(0.99\leq X\leq 1.02)\right)$ (because the stuff in the $P$ would be the boxes that were the right weight, so 1 minus that would be boxes to be repacked). However, after standardising and using the tables I'm given and what not, with this I get a percentage that, to me, seems a bit high something like 43%.

Can anyone help? Have I made a misunderstanding somewhere?
$\Pr (0.99 \leq X \leq 1.02) = 0.5328$ (I used technology, using tables will probably give a slightly different answer). So your answer looks OK.

In fact, this large a probability is hardly surprising given that the interval (0.99, 1.02) is equivalent to $\left( \mu - \frac{\sigma}{2}, \, \mu + \sigma \right)$ ....

3. Originally Posted by mr fantastic
$\Pr (0.99 \leq X \leq 1.02) = 0.5328$ (I used technology, using tables will probably give a slightly different answer). So your answer looks OK.

In fact, this large a probability is hardly surprising given that the interval (0.99, 1.02) is equivalent to $\left( \mu - \frac{\sigma}{2}, \, \mu + \sigma \right)$ ....
Cheers. I guess I just thought that (realistically rather than statistically) repacking 43% of the boxes seems a lot of extra work is all. Thanks again.