Question

Suppose certain coins have weights that are normally distributed with a mean of 5.854 g and a standard deviation of 0.071 g. A vending machine is configured to accept those coins with weights between 5.744 g and 5.964 g.

a. If 280 different coins are inserted into the vending machine, what is the expected number of rejected coins?

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 5.845 g

Standard deviation = 0.071 g

P(weight between 5.744 g and 5.964 g)

= P(5.744 < X < 5.964)

= P(X < 5.964) - P(X < (5.744)

= P(Z < (5.964 - 5.854)/0.071) - P(Z < (5.744 - 5.854)/0.071)

= P(Z < 1.55) - P(Z < -1.55)

= 0.9394 - 0.0606

= 0.8788

a) If 280 different coins are inserted into the vending machine, the expected number of rejected coins = (1-0.8788) x 280

= 33.94

= 34