Question

2. After 1964, quarters were manufactured so that their weights have a mean of 5.67 g and a standard deviation of 0.06 g. Some vending machines are designed so that you can adjust the weights of quarters that are accepted. If many counterfeit coins are found, you can narrow the range of acceptable weights with the effect that most counterfeit coins are rejected along with some legitimate quarters.

a. If you adjust your vending machines to accept weights between 5.60 g and 5.74 g, what percent of legal quarters are rejected? Draw a graph and shade the area/region.

b. If you adjust vending machines to accept all legal quarters except those with weights in the top 2.5% and the bottom 2.5%, what are the limits of the weight that are accepted?

Answer 1

Solution:

Given:

Mean =

Standard Deviation =

Part a) If you adjust your vending machines to accept weights between 5.60 g and 5.74 g, what percent of legal quarters are rejected? Draw a graph and shade the area/region.

That is find:

P( 5.60 < X < 5.74) =.............?

Find z scores:

thus we get:

P( 5.60 < X < 5.74) = P( -1.17 < Z < 1.17)

P( 5.60 < X < 5.74) = P( Z < 1.17) - P( Z < -1.17)

Look in z table for z = 1.1 and 0.07 as well as for z = -1.1 and 0.07 and find corresponding area.

P( Z < -1.17) =0.1210

P( Z < 1.17) =0.8790

Thus

P( 5.60 < X < 5.74) = P( Z < 1.17) - P( Z < -1.17)

P( 5.60 < X < 5.74) = 0.8790 - 0.1210

P( 5.60 < X < 5.74) = 0.7580

Part b) If you adjust vending machines to accept all legal quarters except those with weights in the top 2.5% and the bottom 2.5%, what are the limits of the weight that are accepted?

That is find x values such that:

P( X < x) =2.5% = 0.025 and P( X > x ) = 2.5% = 0.025

thus find z values such that:

P( Z< z) = 0.0250 and P(Z > z ) = 0.0250

Thus look in z table for Area = 0.0250 or its closest area and find z value.

Area 0.0250 corresponds to -1.9 and 0.06

thus z = -1.96

Similarly z value for upper tail 0.0250 area would be positive , that is: 1.96

Now use these z values to find x values by using following formula:

and

Thus the limits of the weight that are accepted are : ( 5.55 g and 5.79 g)