Question

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?

Answer #1

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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