Combining the Spread of Independent Times
A delivery’s total time may include several parts, such as time spent preparing an order and time spent traveling to its destination. In Mean of the Sum of Two Random Variables, we learned that the expected values of two times add. The spread of the total has a different rule: when the two random variables are independent, their variances add, and the standard deviation of the total is the square root of that sum.
Let \(X\) be the order-preparation time and \(Y\) be the travel time. Their total is \(X+Y\). The standard deviations \(\sigma_X\) and \(\sigma_Y\) describe the spread of the individual times in minutes. To find the standard deviation of the total, first square each standard deviation, add those variances, and then take the square root.
This formula has a Pythagorean form: square the two component standard deviations, add the squares, and take the square root. The analogy is about the calculation, not about the possible delivery outcomes. The standard deviations are not being treated as travel directions or as individual times.
The independence condition is essential. If one part of the delivery tends to be long whenever another part is long, the components may vary together. In that situation, simply adding their variances is not generally valid. As discussed in Interpreting Independence in a Context Sentence, independence must be supported by the chance model or context; it should not be assumed merely because two quantities have different names.
Standard deviation keeps the original units. If both component times are measured in minutes, then their variances are in squared minutes, but the square root returns the standard deviation of the total in minutes. The formula works the same way for other units, provided both random variables use compatible units and refer to parts of the same total.
A Reliable Calculation Plan
Before applying the formula, define the total and check whether the variables are independent. If the individual standard deviations are given, square them and combine the variances. If they are not given but probability distributions are available, use the methods from Variance of a Discrete Random Variable and The Standard Deviation of a Discrete Random Variable to find each component’s standard deviation first.
State what \(X\) and \(Y\) measure, and identify \(X+Y\) as the combined quantity.
Use the stated model or context to establish that the component random variables are independent. Do not use the variance-addition rule without this condition.
Calculate \(\sigma_X^2+\sigma_Y^2\). If needed, find each variance from its distribution before adding.
Calculate \(\sigma_{X+Y}\), include the original units, and describe the spread of the total in context.
The means and standard deviations answer different questions. The means add whether or not \(X\) and \(Y\) are independent, as in the previous tutorial. This standard-deviation rule requires independence. In either case, define the variables over the same time period and use the same units so their sum makes sense.
Worked Example: Standard Deviation of a Delivery Time
Worked Example: Standard Deviation of a Delivery Time
A fictional delivery model treats order-preparation time and travel time as independent. Let \(X\) be preparation time, with mean 18 minutes and standard deviation 4 minutes. Let \(Y\) be travel time, with mean 12 minutes and standard deviation 3 minutes. Find the standard deviation of the total delivery time \(X+Y\).
State. \(X\) is preparation time in minutes, \(Y\) is travel time in minutes, and \(X+Y\) is total delivery time. The model states that \(X\) and \(Y\) are independent. We want \(\sigma_{X+Y}\).
Plan. Because the two component times are independent and measured in minutes, add their variances and take the square root. The provided means are not needed for this standard-deviation calculation.
Do. The variances are \(\sigma_X^2=4^2=16\) and \(\sigma_Y^2=3^2=9\), in squared minutes. Therefore:
As an arithmetic check, the combined variance is \(16+9=25\) square minutes, and \(5^2=25\). So taking the square root returns 5 minutes, as required. Adding the component standard deviations would instead give \(4+3=7\) minutes, which is not the independent-sum rule.
Conclude. Under this model, the standard deviation of total delivery time is 5 minutes. Total delivery times typically differ from their mean by about 5 minutes. This describes the spread of the total, not the mean or a guaranteed time for a particular delivery.
Worked Example: Combining Three Independent Components
Worked Example: Combining Three Independent Components
A fictional repair service models a technician’s travel time, setup time, and testing time as mutually independent random variables. Their standard deviations are 3 minutes, 4 minutes, and 12 minutes, respectively. Find the standard deviation of the total time for these three components.
State. Let \(X\), \(Y\), and \(Z\) be the travel, setup, and testing times, all measured in minutes. The total time is \(X+Y+Z\), and the model states that the three variables are mutually independent.
Plan. For mutually independent components, their variances add. Square each component standard deviation, add the three variances, and take the square root.
Do. The component variances are \(3^2=9\), \(4^2=16\), and \(12^2=144\) square minutes. Thus:
Check the addition: \(9+16=25\), and \(25+144=169\). Since \(13^2=169\), the square root is 13. By contrast, adding the three standard deviations would give \(3+4+12=19\) minutes; that is not the standard deviation under the independent-components model.
Conclude. The standard deviation of the technician’s total time is 13 minutes. Under this model, total times typically vary by about 13 minutes from their mean. This calculation depends on the stated mutual independence of all three component times.
Worked Example: Finding Component Standard Deviations First
Worked Example: Finding Component Standard Deviations First
A fictional delivery service models two independent parts of a wait. The random variable \(X\) is the time, in minutes, that an order waits before preparation begins. The random variable \(Y\) is a separate scanning delay, also in minutes. Find the standard deviation of \(X+Y\) from the distributions below.
| \(X\), minutes | \(P(X=x)\) | \(Y\), minutes | \(P(Y=y)\) |
|---|---|---|---|
| 0 | 0.25 | 2 | 0.25 |
| 5 | 0.50 | 6 | 0.50 |
| 10 | 0.25 | 10 | 0.25 |
State. The variables \(X\) and \(Y\) are independent according to the model, and both are measured in minutes. Their probabilities are between 0 and 1 and sum to 1 in each distribution: \(0.25+0.50+0.25=1\). The total wait is \(X+Y\).
Plan. First use each distribution to find its mean and variance, following the earlier tutorials on variance and standard deviation. Then add the variances and take the square root. Keep the unrounded variances until the final calculation.
Do. For \(X\), the mean is:
Its variance, found by weighting the squared deviations from 5, is:
As a check, \(E(X^2)=(0^2)(0.25)+(5^2)(0.50)+(10^2)(0.25)=0+12.5+25=37.5\). Subtracting \(\mu_X^2=25\) gives \(37.5-25=12.5\), the same variance.
For \(Y\), the mean is:
Its variance is:
For a second check, \(E(Y^2)=(2^2)(0.25)+(6^2)(0.50)+(10^2)(0.25)=1+18+25=44\). Subtracting \(\mu_Y^2=36\) gives \(44-36=8\), matching the variance above.
Now apply the independent-sum rule:
The variance addition checks as \(12.5+8=20.5\) square minutes. The reported standard deviation checks because \(4.5277^2\) is approximately \(20.5\), allowing for rounding.
Conclude. The standard deviation of the combined wait is approximately 4.5277 minutes. According to the model, the total wait typically differs from its mean by about 4.5277 minutes. The calculation uses the stated independence and the distributions’ exact variances, rather than rounded component standard deviations.
Why Independence Changes the Calculation
It can be tempting to use the Pythagorean formula whenever two random variables are added. The formula is specifically for independent variables. If components tend to be high together, the total can spread out more than the independent rule predicts; if one tends to be high when the other is low, the total can spread out less.
Worked Example: When Two Times Are Dependent
Suppose a fictional delivery model has two delays, \(X\) and \(Y\). Each delay is 4 minutes or 10 minutes, with probability 0.50 for each value. However, the delays always match: both are 4 minutes or both are 10 minutes. Find the standard deviation of the total and compare it with the result from incorrectly treating the delays as independent.
State. \(X\) and \(Y\) are the two delays, in minutes. They are dependent because their values always match. Each has mean \((4)(0.50)+(10)(0.50)=7\) minutes and standard deviation 3 minutes: the deviations from 7 are \(-3\) and 3, so the variance is \(9(0.50)+9(0.50)=9\) square minutes.
Plan. Since \(X\) and \(Y\) are dependent, do not use the independent-sum formula. Instead, find the distribution of the total from the specified paired outcomes.
Do. The total is 8 minutes with probability 0.50 and 20 minutes with probability 0.50. Its mean is \((8)(0.50)+(20)(0.50)=14\) minutes. Its variance is:
Therefore, \(\sigma_{X+Y}=\sqrt{36}=6\) minutes. If we incorrectly used the independent formula, we would get \(\sqrt{3^2+3^2}=\sqrt{18}\approx4.2426\) minutes, which does not match the actual standard deviation of 6 minutes.
Conclude. The standard deviation of the total is 6 minutes under this dependent model. The example shows why the independence check is not optional: the individual standard deviations alone do not determine the spread of a sum when the variables are dependent.
Common Mistakes and AP Exam Tips
- Adding standard deviations. The independent-sum rule adds variances, not standard deviations. A full-credit calculation shows \(\sqrt{\sigma_X^2+\sigma_Y^2}\).
- Forgetting to take the square root. The sum \(\sigma_X^2+\sigma_Y^2\) is the variance of the total, in squared units. Take its square root to report the standard deviation in the original units.
- Skipping the independence condition. State that the model makes the variables independent, or explain why that assumption is justified. Without independence, this formula is not generally valid.
- Rounding component values too early. If variances or exact standard deviations are available, use them in the calculation and round the final result. Early rounding can slightly change the answer.
- Leaving out context and units. A complete interpretation identifies the total quantity and its units. For example, say that total delivery time typically differs from its mean by about 5 minutes—not merely that “the answer is 5.”
A strong AP response defines the variables, identifies their sum, justifies independence, displays the variance addition and square root, and interprets the result in context. Keep the units visible: variances use squared units during the calculation, but standard deviation returns to the original units.
Key Takeaway
For independent random variables, combine spread by adding variances and taking the square root. This differs from the mean rule: means add without requiring independence, while the standard deviation formula for a sum does require it.
Check Your Understanding
For each question, show the calculation and include a brief interpretation where appropriate.
- Independent preparation and travel times have standard deviations of 6 minutes and 8 minutes. Find the standard deviation of their total.
- Two independent parts of a task have standard deviations of 5 seconds and 12 seconds. What is the standard deviation of their combined time? Show why adding the standard deviations is not the correct method.
- Three mutually independent components have standard deviations of 2, 3, and 6 minutes. Find the standard deviation of their total.
- Explain why the Pythagorean formula for the standard deviation of a sum cannot automatically be used when the variables are dependent.
- Two independent random variables have variances of 7 and 18 square meters. Find the standard deviation of their sum, including units.