- When you add or subtract random variables, the means behave simply: \(E(X \pm Y) = E(X) \pm E(Y)\).
- Spread does not behave simply. For independent variables the variances add, for sums and for differences alike, so \(\sigma_{X \pm Y} = \sqrt{\sigma_X^2 + \sigma_Y^2}\).
- Standard deviations never add directly. Adding them overstates the spread and gives the wrong probabilities.
- The case answer: a student whose walk and bus ride add to a mean of 36 minutes, with SD 5.39 minutes, is late for a 46-minute window only about 3% of the time, not the 8% you get by adding SDs.
- Exam point: write the variance line first, state “independent”, and only then take the square root. Mark schemes typically credit that working.
The case: will Sara make her 9:00 lecture?
Sara is a first-year student. Every morning she walks from her flat to the bus stop and then takes the bus to campus. She has timed both parts of the journey for a full term. (The figures are illustrative.)
- The walk to the stop, including the wait for the bus, takes on average 12 minutes, with a standard deviation of 2 minutes.
- The bus ride takes on average 24 minutes, with a standard deviation of 5 minutes, because traffic varies a lot.
Her first lecture starts at 9:00. She leaves home at 8:14, which gives her a 46-minute window. Both parts of the trip are roughly normal, and a slow walk does not make the bus slower, so the two times are independent.
Sara wants one number: how often will she be late? To answer that she needs the distribution of her total journey time, \(T = W + B\). The mean is easy. The standard deviation is where most people go wrong, and the size of the error matters. One method says she is late about 1 day in 13. The correct method says about 1 day in 32. Over a semester of 60 lectures that is the difference between roughly 5 late arrivals and roughly 2.
The concept: combining random variables
A random variable is a number whose value depends on chance: the time of a bus ride, the weight of a bag of rice, the score on a test. Very often the quantity we care about is not one random variable but a combination of several. A total journey is a walk plus a ride. A profit is revenue minus cost. A gap between two runners is one time minus another. A week of sales is seven days added together.
So we need rules that turn the means and standard deviations of the parts into the mean and standard deviation of the whole. There are only two rules to learn, but they work in very different ways.[1]
The first rule is about centre. Means combine exactly as the variables do. If you add the variables, add the means. If you subtract, subtract. This works whether or not the variables are independent.
The second rule is about spread. Here the natural quantity is the variance, not the standard deviation. When the variables are independent, their variances add. They add for a sum, and, surprisingly to most students, they also add for a difference. Subtracting a random quantity does not cancel uncertainty. It brings in more.
Here \(X\) and \(Y\) are random variables and \(a\) and \(b\) are fixed numbers. \(E(\cdot)\) is the expected value, or long-run mean. \(\operatorname{Var}(\cdot)\) is the variance, the square of the standard deviation \(\sigma\). Put \(a = b = 1\) and you get a sum: \(\operatorname{Var}(X + Y) = \sigma_X^2 + \sigma_Y^2\). Put \(a = 1, b = -1\) and you get a difference. Because \(b^2 = (-1)^2 = 1\), the variance is again \(\sigma_X^2 + \sigma_Y^2\). The constants are squared in the variance rule because variance is measured in squared units. If you double every value, the spread in squared units goes up by a factor of four.
Statistics Made Simple explains combining random variables with worked examples, diagrams and tables.Get the textbook · $12
Why this method fits the case
Sara’s total time is a sum of two random parts, and she knows the mean and SD of each part. That is exactly the input the rules need. The rules also need independence for the variance step, and the case gives us a good reason to believe it: how long she waits at the stop has nothing to do with how bad the traffic is once she is on board.
The rules give the mean and SD of the total. To turn those into a probability we need a shape. A useful fact helps here: if \(X\) and \(Y\) are independent and each is normally distributed, then any linear combination \(aX + bY\) is also normal. Since both parts of Sara’s trip are roughly normal, \(T\) is roughly normal too, and we can use a z-score.[2]
When is this not the right tool?
- When the parts are dependent. If the walk and the bus were both slowed by the same heavy rain, they would be positively correlated. Then \(\operatorname{Var}(X + Y) = \sigma_X^2 + \sigma_Y^2 + 2\operatorname{Cov}(X, Y)\), and the simple rule understates the spread. Height and weight, or a student’s maths and physics marks, are classic dependent pairs.
- When you need a probability but the shape is unknown and far from normal. The mean and SD rules still hold, but a z-score will not give an accurate probability. With many parts, the central limit theorem often rescues you; with two very skewed parts it may not.
- When the combination is not linear. Products and ratios, such as price times quantity when both are random, need other methods.
Setting it up
Write down every quantity before you calculate. This habit alone prevents most errors.
| Quantity | Symbol | Mean (min) | SD (min) | Variance (min²) |
|---|---|---|---|---|
| Walk and wait | \(W\) | 12 | 2 | 4 |
| Bus ride | \(B\) | 24 | 5 | 25 |
| Total journey | \(T = W + B\) | 36 | \(\sqrt{29} \approx 5.39\) | 29 |
| Time window | — | 46 minutes (fixed, not random) | ||
Notice the extra column. Converting each SD into a variance before you combine anything makes the right method almost automatic: you add the column of variances and take one square root at the end.
Solving it
Sara’s walk \(W\) has mean 12 min and SD 2 min; her bus ride \(B\) has mean 24 min and SD 5 min. Both are normal and independent. She allows 46 minutes. Find the probability that she is late.
Sara is late on about 3.1% of mornings, roughly 2 lectures in a 60-lecture semester.
Now a difference. Differences turn up whenever two random amounts compete: a filling against a container, a deadline against a finishing time, one runner against another.
A café machine pours drinks into cups. The volume poured, \(F\), is normal with mean 230 ml and SD 20 ml. The usable capacity of a cup, \(C\), is normal with mean 250 ml and SD 2 ml, independent of \(F\). A cup overflows if \(F > C\). (The figures are illustrative.) Find the probability that a cup overflows.
About 16% of cups overflow. Almost all of the problem comes from the machine (SD 20), not the cups (SD 2).
The third example shows the difference between adding several separate values and multiplying one value by a constant. These look alike on paper but behave very differently.
A market stall’s daily takings \(X\) are normal with mean $300 and SD $15, and days are independent. (The figures are illustrative.) (a) Find the mean and SD of the total takings over 6 days. (b) Find the probability that the 6-day total exceeds $1,850. (c) Compare with the random variable \(6X\).
Summing six independent days gives SD 36.74, far less than 90, because good and bad days partly cancel out.
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What the answer tells us
Sara’s total journey has a mean of 36 minutes and a standard deviation of about 5.4 minutes. Her 46-minute window is just under two standard deviations above the mean, so she is late on about 3% of days. That is the honest answer, and it is useful: it tells her how much “insurance” her 10-minute buffer is buying.
Reasonableness check. The total SD must be at least as large as the larger part (5 minutes) and no larger than the sum of the parts (7 minutes). Our answer, 5.39, sits between those limits and is close to 5. That makes sense: the bus is by far the more variable part, so it dominates. The walk adds only a little extra uncertainty. This is the “Pythagoras” picture of spread: the SDs combine like the two shorter sides of a right-angled triangle, and here the sides are 2 and 5.
Why adding SDs is wrong. If you add the SDs you get 7 minutes, and \(P(T > 46) = P(Z > 1.43) \approx 0.076\). That more than doubles the true risk. Adding SDs assumes the walk is always slow on exactly the days the bus is slow, which is the opposite of independence. In reality, a slow walk is often offset by a quick ride, so the total varies less than the parts suggest. Figure 1 shows the two curves.
Sensitivity. How much does the buffer matter? With a 40-minute window Sara would be late about 23% of the time. With 44 minutes, about 7%. With 48 minutes, about 1.3%. If she wants to be late on no more than 5% of days, she needs a window of at least \(36 + 1.645 \times 5.39 \approx 44.9\) minutes. For 1%, she needs about 48.5 minutes. The other lever is the variability itself. If she switched to a train with ride SD 2 minutes and the same mean, the total SD would fall to \(\sqrt{8} \approx 2.83\), and a 46-minute window would make lateness very rare. Cutting the biggest source of variance helps far more than trimming the smallest.
If you want a full chapter on random variables, with more diagrams and tables, the Statistics Made Simple textbook builds the rules step by step from expected value to linear combinations. For drill, the Statistics Made Simple Practice Questions book has graded sums-and-differences problems with full working.
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Practice questions
Try each question before opening the worked answer. Use \(z\) tables or a calculator for normal probabilities.
EasyQ1. \(X\) has mean 10 and SD 3; \(Y\) has mean 6 and SD 4. They are independent. Find the mean and SD of \(X + Y\) and of \(X – Y\). Show worked answer
\(\operatorname{Var}(X + Y) = \operatorname{Var}(X – Y) = 9 + 16 = 25\).
Both SDs are \(\sqrt{25} = 5\).
EasyQ2. A fair six-sided die has mean 3.5 and variance \(35/12\). Two dice are rolled. Find the mean and SD of the total score, and the SD of twice the score on one die. Show worked answer
Twice one die: variance \(2^2 \times 35/12 = 35/3\); SD \(\approx 3.42\).
The total of two dice varies less than double one die.
MediumQ3. Bolt diameters \(B\) are normal with mean 10.00 mm and SD 0.03 mm. Hole diameters \(H\) are normal with mean 10.10 mm and SD 0.04 mm, independent of \(B\). A bolt fits if it is narrower than the hole. Find the probability that a randomly chosen bolt does not fit a randomly chosen hole. Show worked answer
\(\operatorname{Var}(D) = 0.03^2 + 0.04^2 = 0.0025\), so \(\sigma_D = 0.05\) mm.
\(P(D < 0) = P(Z < -0.10/0.05) = P(Z < -2) \approx 0.0228\).
About 2.3% of pairs do not fit.
MediumQ4. Ali’s 1,500 m times \(A\) are normal with mean 300 s and SD 20 s. Bilal’s times \(B\) are normal with mean 330 s and SD 5 s. Their times are independent. Find the probability that Bilal beats Ali in a race. Show worked answer
\(E(A – B) = 300 – 330 = -30\) s; \(\operatorname{Var}(A – B) = 400 + 25 = 425\); SD \(\approx 20.62\) s.
\(P(A – B > 0) = P(Z > 30/20.62) = P(Z > 1.455) \approx 0.073\).
Bilal wins about 7% of races, mainly because Ali is so inconsistent.
ExamQ5. The mass of a packet of biscuits is normal with mean 500 g and SD 12 g. A carton holds 8 packets, chosen at random. (a) Find the mean and SD of the total mass of the 8 packets. [3] (b) Find the probability that the total mass exceeds 4,050 g. [2] (c) State one assumption you have made. [1] Show worked answer
(b) \(z = (4050 – 4000)/33.94 \approx 1.47\) [M1]; \(P(Z > 1.47) \approx 0.0708\) [A1].
(c) The masses of the 8 packets are independent [B1].
Note: using \(8^2 \times 12^2\) (treating it as \(8X\)) scores M0.
Frequently asked questions
Why do variances add but standard deviations do not?
Variance is built from squared deviations, and for independent variables the cross-term (the covariance) is zero. That leaves a clean sum of squares. Standard deviation is the square root of that sum, and the square root of a sum is not the sum of square roots. It is the same reason a right-angled triangle with sides 3 and 4 has a hypotenuse of 5, not 7.
Do the rules for the mean need independence?
No. \(E(X + Y) = E(X) + E(Y)\) is always true, whatever the relationship between \(X\) and \(Y\). Only the variance rule needs independence, or more precisely zero correlation. If the variables are correlated, you add \(2\operatorname{Cov}(X, Y)\) for a sum and subtract it for a difference.
Is the sum of two normal random variables always normal?
If they are independent and each is normal, yes: any linear combination is normal. If the parts are not normal, the sum is generally not exactly normal, although sums of many independent parts tend towards a normal shape by the central limit theorem.
How is this connected to the standard error of a mean?
Directly. The sample mean is \((X_1 + \dots + X_n)/n\). The sum has variance \(n\sigma^2\), and dividing by \(n\) multiplies the variance by \(1/n^2\). So the variance of the mean is \(\sigma^2/n\) and its SD is \(\sigma/\sqrt{n}\). The same rule that answered Sara’s question is the foundation of every confidence interval you will meet later.[1]
References
- Illowsky, B. and Dean, S. (2023) Introductory Statistics 2e. Houston, TX: OpenStax. https://openstax.org/details/books/introductory-statistics-2e
- Starnes, D. S. and Tabor, J. The Practice of Statistics. New York: W. H. Freeman.
- Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
- College Board. AP Statistics Course and Exam Description. New York: College Board.
