Answer: 1.94%
Step-by-step explanation:
Given: The weights of quarters are approximately normally distributed with [tex]\mu=5.67\ g[/tex] and [tex]\sigma=0.07\ g[/tex].
Machine will only accept coins weighing between 5.48 g and 5.82 g.
Let X be the weight of quarters.
The probability that legal quarters will be rejected = P(X<5.48)+P(X>5.82)
[tex]=P(\dfrac{X-\mu}{\sigma}<\dfrac{5.48-5.67}{0.07})+P(\dfrac{X-\mu}{\sigma}>\dfrac{5.82-5.67}{0.07})\\\\=P(Z<-2.714)+P(Z>2.142)\ \ \ \[Z=\dfrac{X-\mu}{\sigma}]\\\\=(1-P(Z<2.714))+(1-P(Z<2.142))\ \ \ \ [P(Z<-z)=1-P(Z<z)=P(Z>z)]\\\\=(1- 0.9967)+(1-0.9839)\ \ \ \ [\text{By p-value table}]\\\\= 0.0033+0.0161=0.0194=1.94\%[/tex]
Hence, the percentage of legal quarters will be rejected = 1.94%
