The mean hourly pay rate for financial managers in the East North Central region is \$32.62, and the standard deviation is \$2.32 (Bureau of Labor Statistics, September 2005). Assume that pay rates are normally distributed.
a. What is the probability a financial manager earns between \$30 and \$35 per hour?
b. How high must the hourly rate be to put a financial manager in the top 10% with respect to pay?
c. For a randomly selected financial manager, what is the probability the manager earned less than \$28 per hour?
Solution
Given that $\mu = 32.62$ and $\sigma = 2.32$. $X$ denote the hourly pay rate for financial managers in the East North Central region. $X\sim N(32.62, 2.32^2)$.
(a) The probability a financial manager earns between \$30 and \$35 per hour is
$$ \begin{aligned} P(30 < X < 35)&= P\bigg(\frac{30-32.62}{2.32} < \frac{X-\mu}{\sigma} < \frac{35-32.62}{2.32}\bigg)\\ &=P\bigg(-1.129 < Z < 1.026\bigg)\\ &= P(Z < 1.026) -P(Z < -1.129)\\ &=0.8476-0.1294\\ &= 0.7181 \end{aligned} $$
(b) Let $a$ be the hourly rate for a financial manager in the top 10% with respect to pay.
$$ \begin{aligned} &P(X > a) =0.1\\ &\Rightarrow P(X\leq a) =1-0.1\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}\leq \frac{a-32.62}{2.32}\big)=0.9\\ &\Rightarrow P(Z < \frac{a-32.62}{2.32}\big)=0.9\\ &\Rightarrow \frac{a-32.62}{2.32}= 1.282\\ &\Rightarrow a = 32.62 + 1.282* 2.32\\ &\Rightarrow a = 35.59424 \end{aligned} $$
(c) For a randomly selected financial manager, the probability that the manager earned less than \$28 per hour is
$$ \begin{aligned} P(X < 28) &= P\bigg(\frac{X-\mu}{\sigma} < \frac{28-32.62}{2.32}\bigg)\\ &= P(Z < -1.991)\\ &=0.0232 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
