Question 1037574: Suppose certain coins have weights that are normally distributed with a mean of 5.368 g and a standard deviation of 0.057 g. A vending machine is configured to accept those coins with weights between 5.258 g and 5.478 g.
A. If 270 different coins are inserted in the vending machine, what is the expected number of rejected coins?
B. If 270 different coins are inserted in the vending machine, what is the probability that the mean falls between the limits of 5.258 g and 5.478.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose certain coins have weights that are normally distributed with a mean of 5.368 g and a standard deviation of 0.057 g. A vending machine is configured to accept those coins with weights between 5.258 g and 5.478 g.
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z(5.258) = (5.258-5.368)/0.057 = -1.93
z(5.478) = (5.478-5.368)/0.057 = +1.93
P(5.258 < x < 5.478) = P(-1.93< z < 1.93) = 0.9464
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A. If 270 different coins are inserted in the vending machine, what is the expected number of rejected coins?
Ans: 0.9464*270 = 256 when rounded up
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B. If 270 different coins are inserted in the vending machine, what is the probability that the mean falls between the limits of 5.258 g and 5.478.
Comment:: Same procedure but the standard deviation is 0.057/sqrt(270) = 0.0035
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Cheers,
Stan H.
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