Question

Suppose certain coins have weights that are normally distributed with a mean of 5.395 g and a standard deviation of 0.058g.A vending machine is configured to accept those coins with weights between 5.325g and 5.465 g

If 290 different coins are inserted into the vending machine ,what is the expected number of rejected coins?!

The expected number of rejected coins is...(round to nearest integer)

Answer 1

= 5.395

= 0.058

To find P(X<5.325 OR X > 5.465):

Case 1: For X <5.325:

Z = (5.325 - 5.395)/0.058 = - 1.2069

Table of Area Under Standard Normal Curve gves area = 0.3869

So,

P(X<5.325) = 0.5 - 0.3869 = 0.1131

Case 2:P(X>5.465):

Z = (5.465 - 5.395)/0.058= 1.2069

Table gives area = 0.3869

So,

P(X>5.465) = 0.5 - 0.3869 = 0.1131

So,

P(X<5.325 OR X > 5.465) = 2 X 0.1131 = 0.2262

So,

Expected number of rejected coins = 0.2262 X 290 = 65.598 = 66 (Rounded to integer)

So,

Answer is:

66

answered by: anonymous