Question

A manufacturer of fishing line determines that the tensile strength of his product is normally distributed with a mean value of 10.0 lbs and standard deviation of 0.4 lbs. What percentage (%) of the manufactured product is expected to have a tensile strength of at least 9.5 lbs?

Answer 1

Let

X=tensile strength of the product

$X\sim N(\mu=10, \sigma=0.4)$

$\frac{X- \mu}{\sigma}=Z \sim N(\mu=0, \sigma=1)$

$P(X\geq 9.5)=1-P(X\leq 9.5) =1-P(\frac{X-\mu}{\sigma}\leq\frac{9.5-10}{.4} )$

  $=1-P(Z\leq -1.25)$

=1-0.10565

=0.89435

=89.435%