SOLUTION: Suppose certain coins have weights that are normally distributed with a mean of 5.338 g and a standard deviation of 0.068 g. A vending machine is configured to accept those coins w

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Question 675624: Suppose certain coins have weights that are normally distributed with a mean of 5.338 g and a standard deviation of 0.068 g. A vending machine is configured to accept those coins with weights between 5.208 g and 5.468 g.
a. If 250 different coins are inserted into the vending machine, what is the expected number of rejected coins?
The expected number of rejected coins is____(Round to the nearest integer)
______________________________________________
If you could tell me how to do it rather than just give the answer that be helpful considering that kind of the point. Since i been following the example and i keep getting the wrong answer(online homework, infinite # of tries at the cost of 0.5 reduced per retry).
Its kinda frustrating, since i been following the damn example and even other methods that i found googling. Also if you can explain it by using the TI-83 calculator that would be even better.
What i been doing is (x-median)/Standard devuation, so:
(5.208-5.338)/0.068 and (5.468 -5.338)/0.068
to get the Z score. After that i find the area to the right and left.
Since i don't remember the formula or i don't know it. I google to look for it and i think its this one.
http://homepage.smc.edu/mcgraw_colleen/math_52/Finding%20%20p-values.pdf
So i do this after.
Calculator TI-83
normalcdf(-10,result of (5.208-5.338)/0.068,) and normalcdf(result of(5.468 -5.338)/0.068,10)
After that i sum their answers and finally i multiply it by 250. I haved tried but it says its wrong(only 3 tries till the right answer)
Ty for your help.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

μ = 5.338  ,  σ = .068 

vending machine is configured to accept those coins with weights between 5.208 g and 5.468 g.

SAMPLE: 250 different coins

P(5.208 < x < 5.468) = P(z+%3C=+%285.468-5.338%29%2F%28.068%2Fsqrt%28250%29%29  -   P(z+%3C=+%285.208-5.338%29%2F%28.068%2Fsqrt%28250%29%29

find this difference and multiply times 250 (rounding UP to the nearest integer )

TI normalcdf(z)  gives the portion of the area under the standard normal curve to the LEFT of the z-value entered.

Actually can do in one step for between two z-values: normalcdf(smaller z, larger z)

Important to Understand z -values as they relate to the Standard Normal curve:

Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  

For ex: normalcdf(2) - normalcdf(-2) would give the portion of the area under the curve between those two z-values 

Note: z = 0 , 50% of the area under the curve is to the left and 50%  to the right