Question

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

Answer 1

Answer:

The percentage of legal quarters that will be rejected is 1.954 %.

Explanation:

Here we are required to find

P(X<5.48 or X>5.82)

We have μ = 5.67 and  σ = 0.07 where the range is between 5.48 and 5.82 we each z-score given by

$z =\frac{x-\mu}{\sigma}$

Therefore for 5.48 we have $z_{5.48} = \frac{5.48 -5.67}{0.07}$ = -2.714

and for 5.82 we have   $z_{5.82} = \frac{5.82 -5.67}{0.07}$ = 2.143

Therefore we have the area of interest on the normal distribution chart given by

P(Z<-2.714) + P(Z >  2.143) =  P(Z<-2.714) + (1 - P(Z <  2.143))

+  0.00336 + 1 - 0.98382 = 0.01954

= 0.01954 × 100 = 1.954 %

Therefore 1.954 percentage of legal quarters will be rejected.