Percentange of bags rejected

[Matt.D]

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

 

[mathman]

Underweight is 2.5 s.d. too low, while overweight is 2.0 s.d. too high. Look up a table of values for the normal distribution (not the density function, which is the bell curve).

 

[tacman]

Hi Matt,

To solve this problem, you can use the standard normal distribution formula and z-scores. First, let's find the z-score for 225g and 270g using the given mean and standard deviation:
z-score for 225g = (225-250)/10 = -2.5
z-score for 270g = (270-250)/10 = 2

Next, we can use a z-score table (or a calculator) to find the percentage of values below -2.5 and above 2. The table shows that the area under the curve for z < -2.5 is approximately 0.0062 and the area for z > 2 is also 0.0062. This means that the percentage of bags rejected for being underweight or overweight is approximately 0.0062 + 0.0062 = 0.0124 or 1.24%.

Therefore, the percentage of bags rejected is approximately 1.24%.

I hope this helps! Let me know if you have any further questions. Good luck with your statistics homework!