Combining Random Variables

Jake and Anna work in the same store, but in different departments. Jake expects to sell an average of \(5\) shirts per day and Anna expects to sell an average of \(3\). What is the total expected average number of shirts sold in the store per day?

Answer:

Let \(X\) be the random variable representing how many shirts Jake sells, and \(Y\) be the random variable representing Anna's sales. You would hope these are independent random variables! Call \(T\) the random variable of the total sales in the store, so \(T = X + Y\).

From the problem statement,

\[ \mu_X = 5 \text{ and } \mu_Y = 3.\]

or in other words a total of \(8\) shirts.

Jake and Anna work in the same store, but in different departments. Jake expects to sell an average of \(5\) shirts per day and Anna expects to sell an average of \(3\). How many more shirts can Jake expect to sell per day?

Solution:

Just like before, let \(X\) be the random variable representing how many shirts Jake sells, and \(Y\) be the random variable representing Anna's sales, where you are reasonably assuming they are independent. Call \(T\) the random variable of the difference between Jake and Anna's sales in the store. Then since \(T = X - Y\),

\[ \begin{align} \mu_T &= \mu_X - \mu_Y \\ &= 5 - 3 \\ &= 2. \end{align}\]

So Jake can expect to sell \(2\) more shirts than Anna.

Suppose you had looked at the difference between Anna and Jake's sales instead? Then you would have found a mean of \(-2\)! That can happen, and you need to look at the actual combined distribution to figure out what it implies in real life. If you find a negative number when looking at the difference in the sales, it just implies that in general Anna sells fewer shirts than Jake does.

Jake and Anna work in the same store, but in different departments. Jake expects to sell an average of \(5\) shirts per day and Anna expects to sell an average of \(3\). However, Jake has a standard deviation in his sales of \(1\) shirt, while Anna has a standard deviation of \(4\) shirts. Is the standard deviation of their combined shirt totals the same as the sum of the standard deviation of their individual totals?

Solution:

Setting up some variables:

• \(X\) is the random variable of the number of shirts Jake sells;
• \(Y\) is the random variable of the number of shirts Anna sells; and
• \(T\) is the random variable of the number of shirts they sell combined.

As you have already seen, \(\mu_T = 8\). What about the variance and standard deviation? From the statement of the problem, their individual standard deviations are

\[ \sigma_X = 1 \mbox{ and } \sigma_Y = 4.\]

Then for the variance,

\[ \begin{align} \sigma^2_T &= \sigma^2_X + \sigma^2_Y \\ &= 1^2 + 4^2 \\ &= 17, \end{align} \]

but

\[ \sigma_T = \sqrt{17} \approx 4.1\]

which is not the same as

\[ \sigma_X + \sigma_Y = 1 + 4 = 5.\]

In fact,

\[ \sigma_T < \sigma_X + \sigma_Y.\]

So while the average number of shirts they sell per day stays the same if they work together, the standard deviation of the number of shirts they sell together is smaller than if they stay separate.

Suppose you have a business where you are making and delivering pizzas, where both making and delivering the pizzas are normal distributions, with

• making the pizza has an average time of \(18\) minutes with a standard deviation of \(1.5\) minutes; and
• delivering the pizzas has an average time of \(25\) minutes with a standard deviation of \(8\) minutes.

(a) What is the probability that making and delivering a pizza takes more than an hour?

(b) What percentage of the pizzas take longer to make than to deliver?

Solution:

(a) In this part of the question you are looking for the total time, in other words, the sum of two normally distributed independent random variables. First, let's define the random variables:

• \(X\) is the random variable for the time it takes to make a pizza;
• \(Y\) is the random variable for the time it takes to deliver a pizza; and
• \(T\) is the random variable to the total time to make and deliver a pizza.

You are told that both of the random variables are normal, and you would expect that making the pizza and delivering the pizza are independent of each other. So \(T\) is also normally distributed, with \(T = X + Y\).

The average time to make and deliver a pizza would be

\[ \begin{align} \mu_T &= \mu_X + \mu_Y \\ &= 18 + 25 \\ &= 43 \, min. \end{align}\]

Since the times are independent,

\[ \begin{align} \sigma^2_T &= \sigma^2_X + \sigma^2_Y \\ &= 1.5^2 + 8^2 \\ &= 66.25,\end{align} \]

so

\[ \sigma_T = \sqrt{66.25} \approx 8.1 \, min.\]

In other words, \(T\) is a normal distribution with mean \(43\) and standard deviation \(8.1\).

You want to know the probability that making and delivering a pizza takes more than an hour. The graph below shows the normal distribution for the total time, and the shaded region represents the time over \(60\) minutes.

Fig. 4. Normal Distribution showing time longer than an hour

Then the \(z\)-score associated with \(60\) minutes is

\[ z = \frac{60-43}{8.1} = 2.099\]

which, using a standard normal table, gives you the probability of taking more than \(60\) minutes is

\[ P(T>60) = P(z>2.099) = 0.0179.\]

In other words, there is only a \(1.79\%\) chance that a pizza will take longer than an hour to make and deliver!

(b) Next you want to know the percentage of the pizzas take longer to make than to deliver. This time you want to know about the difference between \(X\) and \(Y\), so you need a new random variable, call it \(D\), to represent this. In other words \(D = X - Y\). It is still true that both \(X\) and \(Y\) are independent random variables that follow a normal distribution.

The average time difference between making and delivering a pizza would be

\[ \begin{align} \mu_D &= \mu_X - \mu_Y \\ &= 18 - 25 \\ &= -8 \, min. \end{align}\]

Since the times are independent,

\[ \begin{align} \sigma^2_D &= \sigma^2_X + \sigma^2_Y \\ &= 1.5^2 + 8^2 \\ &= 66.25,\end{align} \]

so

\[ \sigma_D = \sqrt{66.25} \approx 8.1 \, min.\]

In other words, \(D\) is a normal distribution with mean \(-8\) and standard deviation \(8.1\). If a pizza takes longer to make than to deliver, what you want to find is \(P(D>0)\). In the graph below, the shaded region represents when the pizza takes longer to make than to deliver.

Fig. 5. Normal distribution showing time greater than 0

Then the \(z\)-score associated with \(0\) minutes is

\[ z = \frac{0-(-8)}{8.1} = 0.988\]

which, using a standard normal table, gives you the probability of taking more than \(60\) minutes is

\[ P(D>0) = P(z>0.988) = 0.1611.\]

In other words, about \(16\%\) of the time, the pizza will take longer to make than to deliver.

Suppose you have two inspectors working for you. If either of them inspects an item, it takes an average of \(5.8\) minutes to do the inspection, with a standard deviation of \(8\) minutes. However, if both of them work together to inspect the same item, it takes an average of \(11.6\) minutes with a standard deviation of \(17\) minutes. Is it better for you to have the inspectors working separately or together?

Solution:

First, let's give the variables some names:

• \(X\) is the variable for inspector A;
• \(Y\) is the variable for inspector B; and
• \(T\) is the variable for their combined times.

Then \(T = X + Y\), so

\[ \begin{align} \mu_T &= \mu_X + \mu_Y \\ &= 5.8 + 5.8 \\ &= 11.6 \, min. \end{align}\]

That means it doesn't matter if they work together or separately, in either case, their average time is going to be \(11.6\) minutes.

In order for you to look at their combined variances, you need to know that they are independent variables. So for the rest of this example, you will need to assume that two people can inspect an item at the same time without interfering with each other, making them independent variables. Then the variance is

\[\begin{align} \sigma^2_T &= \sigma^2_X + \sigma^2_Y \\ &= 8^2 + 8^2 \\ &= 128, \end{align} \]

and the standard deviation is

\[ \begin{align} \sigma_T &= \sqrt{ \sigma^2_T} \\ & = \sqrt{134.8} \\ &\approx 11.3 \, min. \end{align} \]

So when the two inspectors work separately, they have a much smaller variation in their inspection time.

What does that mean in terms of you having them work together or separately? Given that their mean inspection time is the same either way, it pays you to choose the option that gives you the least variation in inspection times. That means you want the two inspectors working separately since when they work together their standard deviation is \(17\) minutes rather than \(11.3\) minutes when they work separately.

A local shop sells toy cars. The probability of selling between \(0\) and \(5\) toy cars is given in the table below.

Table 1. Probability of selling.

Remember that you can't just add to get the standard deviation! Instead, you must find the variance for the week, then take the square root. The variance for the weekly toy car sales is additive, so

\[ \begin{align} \text{variance for weekly car sales} &= 5(2.95) \\ &= 14.75 \end{align}\]

which gives you

\[ \begin{align} \text{standard deviation for weekly car sales} &= \sqrt{14.75} \\ & \approx 3.84 . \end{align}\]