Answer :
Solution
Mean = 0.98mm
Standard Deviation = 0.35mm
Let X be a random variable
We want to find
p(X<0.5mm)
[tex]\begin{gathered} p(X<0.5)=p(z<\frac{0.5-0.98}{0.35}) \\ p(X<0.5)=p(z<-1.37) \\ p(X<0.5)=0.085343 \\ p(X<0.5)=0.09\text{ (to two decimal places)} \end{gathered}[/tex]The answer is
[tex]p(X<0.5)=0.09\text{ (to two decimal places)}[/tex]Probability of 0.09 and would warrant a refund