dear community members, I got this question in my handbook for exam preparation. I was trying to calculate the output but I'm stuck with figuring out the correct value.
The question is:
According to the Census, self-employed individuals in Romania worked an average 44.6 hours per week, with a variance 210.25 hours. The average of hours worked among wage workers was 40 hours per week. Assuming this variable is approximately normally distributed, find the proportion who worked longer than wage workers on average (consider only two decimal points in computation).
So will it be just 5% standard deviation, and, thus, 5% of workers?
They tell you that $p$, the proportion of self-employed is approximately normal. Hence, \begin{align*} P(p>40) = 1-P(p<40) &= 1-P\left(Z<\frac{40-44.6}{\sqrt{210.25}}\right) \\ &= 1-\Phi(-0.3172414)\\ &= 0.6244698, \end{align*} where $Z$ is the standard normal distribution.