Question

suppose certain coins have weights that are normally distributed with a mean of 5.134g and a standard deviation of 0.073g. A vending machine is configured to accept those coins with weights between 5.024g and 5.244g.
if 280 different coins are inserted into the vending machine, what is the expected number of rejected coins?

Answer 1

@CliffSedge do you think you could help me on this one??

Answer 2

Just a min. Going to go grab a snack.
I'll take a look at it and let you know.

Answer 3

ok thanks man

Answer 4

Alright, so doesn't look too bad.
Do you have an idea about how to go about solving this?

Answer 5

Let's write out the info in mathematical shorthand to keep it organized:
The distribution is N(5.134, 0.073)
Tolerance is 5.024 < X < 5.244
Number of trials, n = 280
Expected value, E(not X) = n*P(X < 5.024 OR X > 5.244)

Answer 6

well at first, i was thinking this normalcdf(5.024,5.244,5.134,0.073/sqrt280) but that didnt work lol

Answer 7

okay i get see what you did.

Answer 8

how does that help solve this problem by reversing the question.

Answer 9

It's looking for the expected number of rejected coins, so if you find the proportion of accepted ones, then you can subtract that from 1 to get the proportion you want.

Answer 10

In other words, E = n*P(X < 5.024 OR X > 5.244) = n(1-P(5.024 < X < 5.244)).

You can use normalcdf on the tolerance range, then subtract that p-val. from 1.

Answer 11

i dont get it :(

Answer 12

Hmm, ok well, let's just get our info step-by-step.
What is the p-value of N(5.024, 5.244, 5.134, 0.073)?