Given:
[tex]\begin{gathered} mean(\mu)=8.21 \\ standard-deviation(\sigma)=1.90 \end{gathered}[/tex]To Determine: The probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase
Solution
P(x<5)
The z score formula is given as
[tex]z=\frac{x-\mu}{\sigma}[/tex]Substitute the given into the formula
[tex]z=\frac{5-8.21}{1.90}[/tex][tex]\begin{gathered} z=-\frac{3.21}{1.9} \\ z=-1.68947 \end{gathered}[/tex]The probabilty would be
[tex]\begin{gathered} P(x<-1.68947)=0.04556 \\ \approx0.045 \\ \approx4.5\% \end{gathered}[/tex]Hence, the probability is 0.045, which is unusual as it is less than 5%