SOLUTION: Suppose certain coins have weights that are normally distributed with a mean of 5.078 g and a standard deviation of 0.072 g. A vending machine is configured to accept those coins w

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Question 1152431: Suppose certain coins have weights that are normally distributed with a mean of 5.078 g and a standard deviation of 0.072 g. A vending machine is configured to accept those coins with weights between 4.948 g and 5.208 g.
a. If 270 different coins are inserted into the vending​ machine, what is the expected number of rejected​ coins?
The expected number of rejected coins is
Answer by VFBundy(438) About Me  (Show Source):
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Probability coin is less than 4.948 grams:

z-score = %284.948+-+5.078%29%2F0.072 = %28-0.130%29%2F0.072 = -1.81

Look up -1.81 on a z-table. You get an answer of 0.0351. This is the probability the a coin is less than 4.948 grams.

Probability coin is more than 5.208 grams:

z-score = %285.208+-+5.078%29%2F0.072 = 0.130%2F0.072 = 1.81

Look up 1.81 on a z-table. You get an answer of 0.9649. This is the probability a coin is LESS than 5.208 grams. We want to find the probability a coin is MORE than 5.208 grams. So, if the probability a coin is less than 5.208 grams is 0.9649, the probability it is more than 5.208 grams is 0.0351.

So, the probability the coin does NOT fall in the accepted range is 0.0351 + 0.0351...or 0.0702.

Because there are 270 coins, and the probability a coin will be rejected is 0.0702, the number of expected rejected coins is:

270 * 0.0702 = 18.954...or, rounded off, 19.