Question

Suppose certain coins have weights that are normally distributed with a mean of 5.571 g and a standard deviation of 0.058 g. A vending machine is configured to accept those coins with weights between 5.461 and 5.681 g.

If 260 different coins are inserted into the vending machine, what is the expected number of rejected coins?

Answer 1

Mean = 5.571

S.D. = 0.058

P( 5.461 < X < 5.681)

Z score value at 5.461

Z = (X - μ) / σ
Z = (5.461 - 5.571) / 0.058
Z = -1.89655

Z score value at 5.681

Z= (X - μ) / σ
Z = (5.681 - 5.571) / 0.058
Z = 1.89655

From the Z score value

P( -1.89655 < x < 1.89655)

= P( X < 1.89655) - P( X < -1.89655)

= 0.9711 - 0.0289

P( 5.461 < X < 5.681)=0.9421

No of rejected coins = (1-0.9421)*260=15.054 rejected coins