What percentage of bags are rejected due to weight?

AI Thread Summary
The discussion revolves around calculating the percentage of bags rejected due to weight based on a normal distribution with a mean of 250g and a standard deviation of 10g. Bags are rejected if they weigh less than 225g or more than 270g. To find the rejection percentage, the z-scores for these weights are calculated, and the area under the normal distribution curve is referenced. One participant explains how to determine the probabilities associated with these z-scores using a standard normal distribution table. The conversation concludes with the original poster gaining a better understanding of normal distribution concepts.
Matt.D
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*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 website 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
 
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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.
 
"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
 
Hi James,

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

Matt
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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