Suppose certain coins have weights that are normally distributed with a mean of 5.159 g and a standard deviation of 0.079 g. A vending machine is configured to accept those coins with weights between 5.029 g and 5.289 g. If 270 different coins are inserted into the vending machine, what is the probability that the mean falls between the limits of 5.029 g and 5.289 g?
Solution : Given that mean ? = 5.159 and a standard deviation ? = 0.079
=> P(5.029 < x < 5.289) = P((x - ?)/? < Z < (x -
?)/?)
= P((5.029 - 5.159)/0.079 < Z < (5.289 - 5.159)/0.079)
= P(-1.6456 < Z < 1.6456)
= 0.901
=> Given n = 270
=> P(5.029 < x < 5.289) = P((x - ?)/(?/sqrt(n)) < Z
< (x - ?)/(?/sqrt(n)))
= P((5.029 - 5.159)/(0.079/sqrt(270)) < Z < (5.289 -
5.159)/(0.079/sqrt(270))
= P(-27.0395 < Z < 27.0395)
= 1
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