Explanation:
standard deviation = σ = 1.8 hours
average = mean = μ = 42.6 hours
Probability that a randomly selected automobile worker works less than 40 hours per week:
P(X < 40)
We apply the z score formula:
[tex]\begin{gathered} z=\frac{X-\mu}{\sigma} \\ \text{let X = 40} \end{gathered}[/tex][tex]\begin{gathered} z\text{ = }\frac{40-42.6}{1.8} \\ z\text{ = }\frac{-2.6}{1.8} \\ z\text{ = }-1.4444 \\ \\ P(X<\text{ 40})\text{ = P}(z\text{ < }-1.4444) \end{gathered}[/tex][tex]\begin{gathered} p\text{ value of z }=\text{ -1.4444} \\ p\text{ value is 0}.074369 \end{gathered}[/tex]