Question

Sean and Evan are college roommates who have part-time jobs as servers in restaurants.

The distribution of Sean’s weekly income is approximately normal with mean $\$225$ and standard deviation $\$25$.

The distribution of Evan’s weekly income is approximately normal with mean $\$240$ and standard deviation $\$15$.

Assuming their weekly incomes are independent of each other, which of the following is closest to the probability that Sean will have a greater income than Evan in a randomly selected week?

• A. $0.67$
• B. $0.7000$
• C. $0.227$
• D. $0.303$
• E. $0.354$

Answer 1

The correct option is D

$0.303$

Explanation for the correct answer:

Step-1: Problems of normal distributions can be solved using the z-score formula:

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure $X$ is given by:

The $Z$-score measures how many standard deviations the measure is from the mean.

After finding the $Z$-score, we look at the $z$-score table and find the $p$-value associated with this $z$-score.

This $p$-value is the probability that the value of the measure is smaller than $X$, that is, the percentile of $X$. Subtracting $1$ by the $p$-value, we get the probability that the value of the measure is greater than $X$.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We want the probability of Sean having a greater income than Evan, which is the probability of the subtraction of Sean's income by Evan's income is greater than $0$.

Step 2: Distribution of the difference between Sean's and Evan's income:

Sean has mean $225$, Evan $240$. So

$$\begin{gathered} \mu =225-240 \\ =-15 \end{gathered}$$

Sean's standard deviation is of $25$, Evan's of $15$. So

$$\begin{gathered} \sigma =\sqrt{{25}^2+{15}^2} \\ =\sqrt{625+225} \\ =\sqrt{850} \\ =29.15 \end{gathered}$$

Step 3: Probability that Sean will have a greater income than Evan in a randomly selected week:

Probability of the subtraction being greater than $0$, which is $1$ subtracted by the $p$ value of $Z$ when $X=0$.

So

$$\begin{gathered} Z=\frac{X-\mu }{\sigma } \\ \Rightarrow Z=\frac{0-(-15)}{29.15} \\ \Rightarrow Z=\frac{15}{29.15} \\ \Rightarrow Z=0.51 \end{gathered}$$

$Z=0.51$ has a $p$-value of $0.695$

$$\begin{gathered} 1-0.695=0.305 \\ \approx 0.303 \end{gathered}$$

Therefore, the correct answer is option $\left(D\right)$.