1. ## Stats

Q. If all the 1 pound cans of beans filled by a food processor have a mean weight of 16.00 ounces with a standard deviation of 0.02 ounces, at least what percentage of the cans must contain between 15.80 and 16.20 ounces of beans, assuming the weights are normally distributed?

A.
(16.20-15.80) x SQRT 16 / 0.02

= (0.4)(4)/0.02
= 1.6/0.02
= 80%

Is this correct or have i used the wrong equation?

2. I'm confused. With such a low standard deviation, practically all cans should contain between 15.8 and 16.2 ounces.

3. yeah... thats what the question states on my past paper though. Am i using the correct formula though?

4. I don't know if I'm heading in the right direction and if there isn't an easier, perhaps approximate, way to do this, but as far as I see it, you would have to look at the (cumulative) distribution function of the normal distribution.

$\Phi_{\mu,\sigma^2}(x) = \frac{1}{\sigma\sqrt{2\pi}}
\int_{-\infty}^x \exp \Bigl( -\frac{(u - \mu)^2}{2\sigma^2} \ \Bigr)\, du$

Then compute $\Phi_{16,.02^2}(16.2) - \Phi_{16,.02^2}(15.8)$, which is approximately 1, i.e. 100%.

5. Originally Posted by tone999
Q. If all the 1 pound cans of beans filled by a food processor have a mean weight of 16.00 ounces with a standard deviation of 0.02 ounces, at least what percentage of the cans must contain between 15.80 and 16.20 ounces of beans, assuming the weights are normally distributed?

A.
(16.20-15.80) x SQRT 16 / 0.02

= (0.4)(4)/0.02
= 1.6/0.02
= 80%

Is this correct or have i used the wrong equation?
You're meant to calculate Pr(15.80 < X < 16.20). Convert each value of X to a z-value and use your tables. As has already been implied, the interval is 10 standard deviations either side of the mean so for all practical purposes the answer is 100%. This should be obvious (and the z-values would leave no room for doubt had you calculated them).