Adding the Expected Sales of Two Stores
A business that operates two stores may want to estimate its average weekly sales across both locations. Each store’s sales can vary from week to week, but if we know the mean sales for each store, we can find the mean of their combined sales with a direct rule.
Let \(X\) be Store 1’s sales in a week and \(Y\) be Store 2’s sales in that same week. The total weekly sales are the random variable \(X+Y\). As in The Mean of a Discrete Random Variable, a random variable’s mean is its probability-weighted average. For a sum, the means add:
This rule is called the linearity of expected value. Importantly, it does not require \(X\) and \(Y\) to be independent. The stores might experience shared influences, such as a local event or a change in weather. Those influences can affect how their sales vary together, but they do not change the rule for adding their means.
The units also add. If each mean is measured in dollars per week, their sum is expected dollars per week across both stores. The result describes an average over many comparable weeks, as discussed in Interpreting Expected Value as a Long-Run Average. It does not promise that the stores will earn exactly that total in a particular week.
Using the Sum Rule
When each store’s probability distribution is provided, calculate its mean by multiplying each possible sales amount by its probability and adding the products. Then add the two means. Define the variables carefully so the time period and units match: both variables should represent sales over the same kind of week and in the same currency.
Let \(X\) and \(Y\) represent the two stores’ sales over the same week, in the same units.
Use the value-probability pairs for each store to calculate \(\mu_X\) and \(\mu_Y\).
Calculate \(\mu_{X+Y}=\mu_X+\mu_Y\) and report the total in context with units.
Describe the expected total as a long-run average, not a guaranteed sales amount for one week.
If a joint distribution showing pairs of sales is available, you can also find the mean total directly: add the two stores’ sales within each pair, then calculate the probability-weighted mean of those totals. The result agrees with adding the separate means. The rule is especially useful when the joint distribution is not given or is difficult to construct.
Worked Example: Combining Two Stores’ Weekly Sales
Worked Example: Combining Two Stores’ Weekly Sales
A fictional retailer models weekly sales at Store A and Store B with the distributions below. Find the expected total weekly sales across both stores. No information is given about whether the stores’ sales are related.
| Store A sales, dollars | Probability | Store B sales, dollars | Probability |
|---|---|---|---|
| 800 | 0.20 | 600 | 0.25 |
| 1,000 | 0.50 | 900 | 0.50 |
| 1,200 | 0.30 | 1,200 | 0.25 |
State. Let \(X\) be Store A’s weekly sales, in dollars, and \(Y\) be Store B’s weekly sales, in dollars. The total weekly sales are \(X+Y\). We want to find \(\mu_{X+Y}\).
Plan. Each store’s probabilities are between 0 and 1 and sum to 1: Store A’s probabilities total \(0.20+0.50+0.30=1\), and Store B’s total \(0.25+0.50+0.25=1\). Find each mean from its distribution, then add them. The sum rule does not require an independence assumption.
Do. Store A’s mean weekly sales are:
As an arithmetic check, the distribution is centered at $1,000, with possible sales $200 below or above that amount. The weighted adjustment from $1,000 is \((-200)(0.20)+(200)(0.30)=20\), giving \(1000+20=1020\).
Store B’s mean weekly sales are:
For a check, the $600 and $1,200 outcomes are equally likely and are equally far below and above $900. Their weighted adjustments cancel: \((-300)(0.25)+(300)(0.25)=0\). The middle outcome is $900, so the mean is $900.
Now add the means:
The addition can be checked by grouping the amounts: \(1020+900=(1000+900)+20=1900+20=1920\).
Conclude. According to the model, the expected combined weekly sales for Stores A and B are $1,920. Over many comparable weeks, the average of their combined sales would be about $1,920 per week. This is an expected value, not a guarantee about sales in any one week. We do not need to know whether the stores’ sales are independent in order to find this mean.
Worked Example: Finding the Mean from Shared Weekly Conditions
Worked Example: Finding the Mean from Shared Weekly Conditions
A fictional pair of stores experiences three kinds of weeks. The stores’ sales may be related because both locations respond to the same conditions. The table gives each pair of weekly sales and the probability of that kind of week. Find the mean total sales in two ways.
| Week type | Probability | Store A sales | Store B sales | Total sales |
|---|---|---|---|---|
| Quiet | 0.20 | $700 | $500 | $1,200 |
| Typical | 0.30 | $900 | $800 | $1,700 |
| Busy | 0.50 | $1,300 | $1,100 | $2,400 |
State. Let \(X\) and \(Y\) be the stores’ sales in a week, in dollars, and let \(T=X+Y\) be total weekly sales. The listed probabilities are all between 0 and 1 and sum to \(0.20+0.30+0.50=1\).
Plan. First calculate \(\mu_X\) and \(\mu_Y\), then add them. As a separate check, calculate the probability-weighted mean of the total sales shown in the table. Because the joint week types are provided, both calculations are possible.
Do. Store A’s mean sales are:
Checking by measuring each outcome’s difference from $1,000 gives \((-300)(0.20)+(-100)(0.30)+(300)(0.50)=-60-30+150=60\). Adding this adjustment to $1,000 also gives $1,060.
Store B’s mean sales are:
For a check, relative to $900 the weighted adjustments are \((-400)(0.20)+(-100)(0.30)+(200)(0.50)=-80-30+100=-10\). Thus the mean is \(900-10=890\).
Adding the two means gives:
For the direct check, the expected total is:
The direct calculation also checks by using $2,000 as a reference: the weighted adjustments are \((-800)(0.20)+(-300)(0.30)+(400)(0.50)=-160-90+200=-50\), so the mean total is \(2000-50=1950\).
Conclude. Both methods give expected combined weekly sales of $1,950. The stores’ sales are described together by shared week types, but that relationship does not prevent us from adding their means. The result is the long-run average total under this model.
Worked Example: Same Store Means, Different Pairings
Worked Example: Same Store Means, Different Pairings
Consider two fictional stores. Each has weekly sales of either $500 or $1,500, with probability 0.50 for each amount. Find the mean total sales under two possible ways the stores’ sales could be paired: they always have the same sales, or whenever one has high sales the other has low sales.
State. Let \(X\) and \(Y\) be the weekly sales, in dollars, for the two stores. Each store has mean sales \((500)(0.50)+(1500)(0.50)=250+750=1000\) dollars per week. As a check, the outcomes are equally spaced below and above $1,000, so their weighted adjustments cancel. The sum rule gives \(1000+1000=2000\) dollars per week for the expected total in either pairing.
Plan. To see why the pairings do not change the mean total, list the possible paired sales and calculate the probability-weighted mean of their totals. Each pairing model must assign probabilities that are between 0 and 1 and add to 1.
Do. If the stores always have the same sales, the paired outcomes are \((500,500)\) and \((1500,1500)\), each with probability 0.50. Their totals are $1,000 and $3,000. The direct mean is:
As a check, the totals are equally far below and above $2,000, so their weighted adjustments cancel and the mean remains $2,000.
If one store always has low sales when the other has high sales, the paired outcomes are \((500,1500)\) and \((1500,500)\), each with probability 0.50. The total is $2,000 in both cases, so:
Here the check is immediate: every possible total is $2,000, so its probability-weighted mean must be $2,000.
Conclude. In both pairing models, each store’s mean is $1,000 and the mean total is $2,000, as the sum rule predicts. However, the possible totals differ: one model gives totals of $1,000 or $3,000, while the other always gives $2,000. The individual means are enough to find the mean total, but they do not by themselves describe every possible total or its probability.
What the Sum of Means Does—and Does Not—Tell You
The expected total is useful for planning, budgeting, and describing long-run average sales. It is not enough to reconstruct the distribution of total sales. For that, information about which sales values can occur together may matter. The third example showed that stores with the same individual distributions can produce quite different patterns of weekly totals.
The next tutorial considers the standard deviation of the sum of independent random variables. Keep that question separate from the one answered here: adding means follows the sum rule without an independence requirement, while a spread calculation for a sum needs additional information and assumptions.
Common Mistakes and AP Exam Tips
- Assuming independence is needed to add means. It is not. State \(\mu_{X+Y}=\mu_X+\mu_Y\); do not claim the stores must be independent for this mean calculation.
- Adding possible sales amounts instead of means. A particular week’s total is the sum of the two stores’ sales in that week. The expected total is found by adding the stores’ probability-weighted means.
- Treating the mean as a guaranteed weekly total. An expected value is a long-run average under the model. A single week can have a different total.
- Mixing time periods or units. Define both variables for the same period and use compatible units. Weekly sales in dollars can be added to weekly sales in dollars; a weekly amount should not be casually combined with a monthly amount.
- Claiming the sum of means gives the entire distribution of the total. The rule determines the mean only. It does not identify all possible totals or their probabilities when the way store sales occur together is unknown.
- Giving a number without context. A complete response identifies the total as expected sales across both stores, states the time period, and includes units.
A full-credit explanation can define \(X\) and \(Y\), calculate their means correctly, use \(\mu_{X+Y}=\mu_X+\mu_Y\), and interpret the result as the long-run average combined sales for one week. If independence is raised, make clear that it is not required for this expected-value rule.
Key Takeaway
To find the expected total weekly sales from two stores, find each store’s expected weekly sales and add them. The stores’ sales may be related; independence is not needed for this rule. The result is a mean, not a prediction that the combined sales will equal that amount in every week.
Check Your Understanding
Use the sum rule for expected value to answer these questions. Show calculations and include units where appropriate.
- Store North has weekly sales of $700 and $1,100 with probabilities 0.40 and 0.60. Store South has weekly sales of $500 and $900 with probabilities 0.25 and 0.75. Find each store’s mean and the expected total weekly sales.
- In question 1, no information is given about how the stores’ sales are related. Can you still find the mean total? Explain why.
- Two stores each have mean weekly sales of $2,400. What is the expected combined weekly sales? Does your answer mean the stores will earn that total next week?
- Suppose two stores’ weekly sales are paired in a table of possible outcomes and probabilities. Describe how to find the expected total directly from that table, and explain how the answer relates to adding the separate means.
- Store A’s mean sales are $1,800 per week, and Store B’s mean sales are $2,100 per week. What additional information would be needed to calculate the mean total sales? What does the sum rule say about that need?