Statistics

Combining random variables

A bottle is filled and then capped, and each step takes a random amount of time. If you know the mean and the spread of each step, what are the mean and the spread of the total time, and of the gap between the two?

You need

Normal random variablesGrade 12 Compound eventsGrades 7 to 11

Where you are 11 in Random variables

Probability4

Random variables11

Combining random variables

Continuous random variables and density curves Cumulative probability Expected value Linear transformations of a random variable ▾ Show all 11 (6 more)

Why this exists

A random variable X has a mean \mu_X and a variance \operatorname{Var}(X) = \sigma_X^2, with standard deviation \sigma_X. Linear transformations of a random variable showed that a changed variable a + bX has mean a + b\mu_X and standard deviation |b|\sigma_X. That handles one variable.

Many quantities are built from two or more random variables. At Teuila’s small guava juice shop, each bottle is filled by Teuila’s filler in X seconds and then capped by Losa’s capper in Y seconds, so a bottle takes X + Y seconds. Whether the filler runs ahead of the capper depends on the gap X - Y. When Sefo loads four crates into the van, the load is a sum of four weights. Two steps in sequence are a compound event, and a small table of pairs can be listed, but the list grows fast and cannot be written at all for a continuous variable. This lesson gives the rules that skip the list.

The lesson runs over three days: Day 1 is sums, Day 2 is differences and what happens without independence, Day 3 is normal variables and totals of several. The shop, its people and every probability are made up.

Definition

Let X and Y be random variables attached to the same chance process. Their sum X + Y is the random variable whose value, for each outcome, is the value of X plus the value of Y. Their difference X - Y is defined the same way with subtraction.

X and Y are independent when knowing the value of one does not change the probabilities for the other. For discrete variables this means that for every pair of values,

\begin{aligned} &P(X = x \text{ and } Y = y) \\ &\quad = P(X = x)\,P(Y = y) \end{aligned}

which is the multiplication rule for independent events applied to every pair. Several random variables X_1, X_2, \ldots, X_n are independent when knowing any of them does not change the probabilities for the rest.

Intuition

Means. If filling averages 5 seconds and capping 2.3 seconds, a bottle averages 7.3 seconds. Each bottle’s total is its fill time plus its cap time, and averaging over many bottles keeps that sum. How the two times are related cannot change this.

Spread. A total lands far from its mean most easily when both parts miss in the same direction. If the parts are independent, a slow fill is just as likely to meet a fast cap as a slow one, so the misses often cancel in part. The total is more spread out than either part, but less than the two standard deviations added.

The picture is a right triangle. Independent deviations behave like movements in two perpendicular directions: \sigma_X and \sigma_Y are the legs, and the total’s standard deviation is the hypotenuse, so \sigma^2_{X+Y} = \sigma_X^2 + \sigma_Y^2. Adding the standard deviations would be walking along both legs; the spread of the sum is the straight-line distance.

Figure 1Standard deviations of independent variables combine like the legs of a right triangle. The total’s spread is the hypotenuse, not the two legs laid end to end.

Differences. X - Y is X + (-Y), and -Y flips every value of Y without changing how spread out they are: \sigma_{-Y} = |-1|\,\sigma_Y = \sigma_Y. So a gap between two independent times is exactly as uncertain as their total. Each variable brings its own uncertainty, whichever sign it carries.

Results

Day 1: sums

For any two random variables, independent or not, means add:

\mu_{X+Y} = \mu_X + \mu_Y

For independent random variables, variances add:

\sigma^2_{X+Y} = \sigma_X^2 + \sigma_Y^2

\sigma_{X+Y} = \sqrt{\sigma_X^2 + \sigma_Y^2}

Standard deviations do not add: for independent X and Y whose standard deviations are both positive, \sigma_{X+Y} < \sigma_X + \sigma_Y.

Day 2: differences

For any two random variables, the mean of a difference is the difference of the means:

\mu_{X-Y} = \mu_X - \mu_Y

For independent random variables the variances still add:

\sigma^2_{X-Y} = \sigma_X^2 + \sigma_Y^2

With constants a and b, the mean of aX + bY is a\mu_X + b\mu_Y always, and if X and Y are independent its variance is a^2\sigma_X^2 + b^2\sigma_Y^2.

Day 3: normal variables and several variables

If X and Y are independent and each is normal, then X + Y and X - Y are also normal, with the means and standard deviations above. For independent X_1, \ldots, X_n, the mean of the total is the sum of the means and its variance is the sum of the variances; if each is normal, the total is normal.

For n independent variables that each have mean \mu and standard deviation \sigma:

\begin{aligned} X_1 + \cdots + X_n &: \ \text{mean } n\mu, \ \text{sd } \sigma\sqrt{n} \\ nX &: \ \text{mean } n\mu, \ \text{sd } n\sigma \end{aligned}

The normal facts are stated, not proved, here; their proof needs more than weighted sums.

Why it is true

The argument is written for discrete variables with finite lists of values; for continuous variables the sums become areas and the same steps go through. Write P(x, y) for P(X = x \text{ and } Y = y).

Means add. Each pair of values contributes x + y, weighted by its probability:

\begin{aligned} \mu_{X+Y} &= \sum (x + y)\,P(x, y) \\ &= \sum x\,P(x, y) + \sum y\,P(x, y) \end{aligned}

In the first sum, group the pairs that share the same x. Those pairs are the separate ways X = x can happen, so their probabilities add to P(X = x), and the first sum is \sum x\,P(X = x) = \mu_X. In the same way the second is \mu_Y. Independence was never used. A minus sign in place of the plus gives \mu_{X-Y} = \mu_X - \mu_Y.

Variances add. Write d = x - \mu_X and e = y - \mu_Y for the two deviations. The sum’s deviation from its mean \mu_X + \mu_Y is d + e, so

\begin{aligned} \sigma^2_{X+Y} &= \sum (d + e)^2 P(x, y) \\ &= \sum d^2 P(x, y) + \sum e^2 P(x, y) \\ &\quad + 2\sum d\,e\,P(x, y) \end{aligned}

By the same grouping, the first two sums are \sigma_X^2 and \sigma_Y^2. The last one is the cross term. If X and Y are independent, P(x, y) = P(X = x)P(Y = y), and the double sum factors:

\begin{aligned} &\sum_x \sum_y d\,e\,P(X = x)P(Y = y) \\ &= \Big(\sum_x d\,P(X = x)\Big)\Big(\sum_y e\,P(Y = y)\Big) \end{aligned}

Each factor is a probability-weighted sum of deviations from a mean, which is 0 (shown in Expected value). So the cross term vanishes and \sigma^2_{X+Y} = \sigma_X^2 + \sigma_Y^2. For the difference, the deviation is d - e, and (d - e)^2 = d^2 + e^2 - 2de: the same cross term, with a minus sign, still 0. That is why the variances add for a difference too.

Without independence the cross term need not vanish. It is positive when the two deviations tend to share a sign, making the sum more spread out, and negative when they tend to have opposite signs (Example 4). For several variables, apply the rule again and again: X_1 + X_2 + X_3 = (X_1 + X_2) + X_3, where X_1 + X_2 is independent of X_3. With constants, \operatorname{Var}(aX) = a^2\sigma_X^2 from Linear transformations of a random variable, and the sum rule does the rest.

Worked examples

Normal areas below use z rounded to two decimal places and \Phi(z) = P(Z < z), the area to the left under N(0, 1), as in Normal calculations: areas and percentiles. The values used:

z	\Phi(z)	z	\Phi(z)
0.40	0.6554	1.22	0.8888
0.67	0.7486	1.33	0.9082
0.80	0.7881	1.39	0.9177
1.11	0.8665	1.60	0.9452
1.20	0.8849	3.05	0.9989

For a negative z, \Phi(-z) = 1 - \Phi(z). Technology with an unrounded z may differ in the fourth decimal place.

Example 1

Day 1: listing every pair

Teuila’s filling time X is 4, 5 or 6 seconds with probabilities 0.2, 0.6 and 0.2. Losa’s capping time Y is 2 or 3 seconds with probabilities 0.7 and 0.3. The machines run separately, so take X and Y to be independent. List the distribution of the total T = X + Y, find \mu_T and \operatorname{Var}(T) from it, and compare with the rules.

Solution

Each part first. \mu_X = 4(0.2) + 5(0.6) + 6(0.2) = 5 seconds and \sigma_X^2 = (-1)^2(0.2) + 0^2(0.6) + 1^2(0.2), which is 0.4, so \sigma_X \approx 0.63 seconds. \mu_Y = 2(0.7) + 3(0.3) = 2.3 seconds and \sigma_Y^2 = (-0.3)^2(0.7) + 0.7^2(0.3), which is 0.063 + 0.147 = 0.21, so \sigma_Y \approx 0.46 seconds.

There are six pairs. By independence each pair’s probability is the product of the two:

fill x	cap y	total t	probability
4	2	6	0.2(0.7) = 0.14
4	3	7	0.2(0.3) = 0.06
5	2	7	0.6(0.7) = 0.42
5	3	8	0.6(0.3) = 0.18
6	2	8	0.2(0.7) = 0.14
6	3	9	0.2(0.3) = 0.06

Pairs with the same total are separate ways of reaching it, so their probabilities add:

total t	6	7	8	9
P(t)	0.14	0.48	0.32	0.06

From this table, \mu_T = 0.84 + 3.36 + 2.56 + 0.54 = 7.3 seconds, which is 5 + 2.3. With the shortcut, E(T^2) = 36(0.14) + 49(0.48) + 64(0.32) + 81(0.06) = 53.9 square seconds, so

\operatorname{Var}(T) = 53.9 - 7.3^2 = 53.9 - 53.29 = 0.61

and 0.61 = 0.4 + 0.21: the variances add. Then \displaystyle \sigma_T = \sqrt{0.61} \approx 0.78 seconds, while \sigma_X + \sigma_Y \approx 0.63 + 0.46 = 1.09. Adding standard deviations would overstate the spread by about 0.3 seconds. In the long run a bottle takes 7.3 seconds on average, typically about 0.78 seconds more or less (Figure 2).

Figure 2The total’s mean is 5 + 2.3 = 7.3, but its standard deviation, about 0.78, is less than 0.63 + 0.46 (made-up probabilities).

Example 2

Day 1: standard deviations do not add

Before the line starts each morning, Teuila washes the fruit, taking X minutes with \mu_X = 12 and \sigma_X = 2, and then Losa sterilizes the bottles, taking Y minutes with \mu_Y = 9 and \sigma_Y = 1.5, independently of the washing. Find the mean and standard deviation of the total time. Sefo says the standard deviation is 3.5 minutes. Is he right?

Solution

Means add: \mu_{X+Y} = 12 + 9 = 21 minutes. The parts are independent, so variances add:

\sigma^2_{X+Y} = 2^2 + 1.5^2 = 4 + 2.25 = 6.25

in square minutes, and \displaystyle \sigma_{X+Y} = \sqrt{6.25} = 2.5 minutes. Sefo added the standard deviations, 2 + 1.5 = 3.5. That would be right only if every slow wash came with a slow sterilizing by a matching amount, so that the misses never cancel. With independent parts they partly cancel, and the spread is the hypotenuse of a right triangle with legs 2 and 1.5. The morning setup takes 21 minutes on average, typically about 2.5 minutes more or less.

Example 3

Day 2: a difference, listed

With X and Y from Example 1, the gap D = X - Y is how many seconds longer the fill takes than the cap. List its distribution, find \mu_D and \operatorname{Var}(D), and compare with the rules.

Solution

The same six pairs, with the same probabilities, now give x - y:

fill x	cap y	gap d	probability
4	2	2	0.14
4	3	1	0.06
5	2	3	0.42
5	3	2	0.18
6	2	4	0.14
6	3	3	0.06

Combining equal gaps:

gap d	1	2	3	4
P(d)	0.06	0.32	0.48	0.14

\mu_D = 0.06 + 0.64 + 1.44 + 0.56 = 2.7 seconds, which is 5 - 2.3. Next, E(D^2) = 1(0.06) + 4(0.32) + 9(0.48) + 16(0.14) = 7.9 square seconds, so

\operatorname{Var}(D) = 7.9 - 2.7^2 = 7.9 - 7.29 = 0.61

the same as \operatorname{Var}(T) and equal to 0.4 + 0.21, not 0.4 - 0.21 = 0.19. The reason shows in the tables: the probabilities for D are those for T in reverse order, a mirror image with the same spread. On average the fill runs 2.7 seconds longer than the cap, so the capper waits about that long for each bottle.

Example 4

Day 2: when independence fails

Sefo packs mixed crates of 6 bottles, each guava or passion fruit. The number of guava bottles X is 2, 3 or 4 with probabilities 0.3, 0.4 and 0.3, and Y is the number of passion fruit bottles. Find the mean and variance of X + Y and of X - Y, and compare with what the independent rules would give.

Solution

\mu_X = 0.6 + 1.2 + 1.2 = 3 and \sigma_X^2 = (-1)^2(0.3) + 0^2(0.4) + 1^2(0.3), which is 0.6. Every crate holds 6 bottles, so Y = 6 - X, a linear transformation of X: \mu_Y = 6 - 3 = 3 and \sigma_Y^2 = (-1)^2(0.6) = 0.6.

The sum X + Y = 6 for every crate. Its mean is 6 = 3 + 3: means add even here. Its variance is 0, not 0.6 + 0.6 = 1.2.

The difference is X - Y = X - (6 - X) = 2X - 6, with mean 2(3) - 6 = 0 = 3 - 3 and variance 2^2(0.6) = 2.4, not 1.2. Directly: X - Y is -2, 0 or 2 with probabilities 0.3, 0.4, 0.3, and (-2)^2(0.3) + 0^2(0.4) + 2^2(0.3) is 2.4. So \displaystyle \sigma_{X-Y} = \sqrt{2.4} \approx 1.55 bottles, where the independent rule would claim \displaystyle \sqrt{1.2} \approx 1.10.

X and Y are as dependent as two variables can be: one guava bottle more always means one passion fruit bottle fewer. Their deviations always have opposite signs, so they cancel completely in the sum and double up in the difference. The means still behave; only the variance rule needed independence.

Example 5

Day 3: the time for a batch

Filling one batch of bottles takes Teuila X minutes and capping and labeling it takes Losa Y minutes, independently. A normal random variable is described by its mean and standard deviation, written N(\mu, \sigma); take X to be N(18, 3) and Y to be N(7, 2). Find the distribution of the batch time T = X + Y and the probability that a batch takes more than 30 minutes.

Solution

\mu_T = 18 + 7 = 25 minutes. The parts are independent, so \sigma_T^2 = 3^2 + 2^2 = 13 and \displaystyle \sigma_T = \sqrt{13} \approx 3.61 minutes. A sum of independent normal variables is normal, so T is N(25, 3.61). The z-score of 30 is

z = \frac{30 - 25}{\sqrt{13}} \approx 1.39

so P(T > 30) = 1 - \Phi(1.39) = 1 - 0.9177 = 0.0823. About 8 batches in 100 take more than half an hour.

Example 6

Day 3: which step takes longer

With X and Y from Example 5, find (a) the probability that capping and labeling a batch takes longer than filling it, and (b) the probability that filling takes at least 15 minutes longer than capping and labeling.

Solution

Both questions are about the gap D = X - Y. Its mean is 18 - 7 = 11 minutes, and because the variances add, \displaystyle \sigma_D = \sqrt{9 + 4} = \sqrt{13} \approx 3.61 minutes, the same as for the total. As a difference of independent normal variables, D is N(11, 3.61).

(a) Capping takes longer when Y > X, that is, when D < 0:

z = \frac{0 - 11}{\sqrt{13}} \approx -3.05

and P(D < 0) = \Phi(-3.05) = 1 - 0.9989 = 0.0011. About 1 batch in 1{,}000.

(b) D \ge 15 has \displaystyle z = \dfrac{15 - 11}{\sqrt{13}} \approx 1.11, so P(D \ge 15) = 1 - 0.8665 = 0.1335.

Without the rule, the gap could not be found from the two separate distributions; with it, every question about “which is longer, and by how much” is one normal calculation.

Example 7

Day 3: four crates, or one crate four times

A full crate weighs W pounds, N(22, 1.5), and crates vary independently. Sefo loads four. (a) Find the mean and standard deviation of the total weight, and the probability that it exceeds 92 pounds. (b) Compare with 4W, four times the weight of a single crate.

Solution

(a) The total W_1 + W_2 + W_3 + W_4 has mean 4(22) = 88 pounds and variance 4(1.5^2) = 9, so its standard deviation is \displaystyle \sqrt{9} = 3 pounds, which is \displaystyle 1.5\sqrt{4}. It is a sum of independent normal variables, so it is normal, and z = \dfrac{92 - 88}{3} \approx 1.33: the probability is 1 - 0.9082 = 0.0918.

(b) 4W also has mean 88 pounds, but by Linear transformations of a random variable its standard deviation is 4(1.5) = 6 pounds, twice as large. Then z = \dfrac{92 - 88}{6} \approx 0.67, and P(4W > 92) = 1 - 0.7486 = 0.2514.

Four separate crates are heavy and light in no particular pattern, so their deviations partly cancel. Multiplying one crate by 4 repeats a single deviation four times, always in the same direction. A total of n independent pieces spreads like \displaystyle \sqrt{n}; one piece scaled by n spreads like n.

Boundaries

Means add (and subtract) for any random variables with means, dependent or not.
The variance rules need the cross term to be 0, which independence guarantees. Without it the variance of a sum can be larger or smaller than \sigma_X^2 + \sigma_Y^2, even 0 (Example 4).
Independence is a modeling choice that has to fit the situation: separate machines and separate crates, yes; counts that share a fixed total, or quantities driven by the same cause (a hot day and sales of two flavors), no.
For independent variables, variances add; standard deviations do not (unless one of them is 0), and variances never subtract, not even for a difference. Standard deviations can add only when the parts always miss together, as in X + X = 2X.
The shape rule is only for normal variables. A sum or difference of independent non-normal variables has the mean and standard deviation above but need not be normal (Example 1’s total is a four-bar distribution).
A total of n independent copies is not nX: same mean, different spread (Example 7).

Where this goes next

The sample mean of n independent observations is their total divided by n, so its standard deviation is \displaystyle \dfrac{\sigma\sqrt{n}}{n} = \dfrac{\sigma}{\sqrt{n}}, the starting point of the Sampling distribution of a sample mean. A binomial count of successes is a sum of n independent variables that are each 1 or 0, so its mean and variance follow from this lesson by adding, as in Mean and standard deviation of a binomial. Comparing two groups, two means or two proportions, uses the difference rule: the means subtract and the variances add.

Exercises

All probabilities and models are made up. Give standard deviations to two decimal places unless they are exact.

Day 1: sums

The capper jams X times in an hour, 1 or 2 with probability 0.5 each. The filler jams Y times in the same hour, 0 or 1 with probabilities 0.6 and 0.4, independently. List the distribution of S = X + Y, find \mu_S and \operatorname{Var}(S) from your table, and check them with the rules.
Teuila’s shop sells X bottles a day with \mu_X = 60 and \sigma_X = 5. Losa’s market stall in another town sells Y bottles a day with \mu_Y = 45 and \sigma_Y = 12, independently. Find the mean and standard deviation of the day’s combined sales.
Filling takes X seconds with \mu_X = 9.2 and \sigma_X = 0.8; labeling takes Y seconds with \mu_Y = 4.6 and \sigma_Y = 1.5, independently. Find the mean, the variance and the standard deviation of the total time.
Two independent steps have standard deviations 2 and 5 minutes. Losa says the total time has a standard deviation of 7 minutes. Find the correct value and say why 7 is too large.
X and Y are independent, \sigma_X = 1 and \sigma_{X+Y} = 2.6. Find \sigma_Y.
On any day the shop sells X bottles of guava with \mu_X = 40 and Y bottles of passion fruit with \mu_Y = 25, and both rise on hot days. Find the mean total, and explain why its standard deviation cannot be found from \sigma_X and \sigma_Y alone.
Washing, filling and capping a batch take independent times with means 4, 6 and 10 minutes and standard deviations 1, 2 and 2 minutes. Find the mean and standard deviation of the total.

Day 2: differences and dependence

X and Y are independent with \mu_X = 30, \sigma_X = 7, \mu_Y = 12 and \sigma_Y = 2.4. Find the mean and standard deviation of X - Y.
X is 2 or 3 with probabilities 0.4 and 0.6; Y is 1 or 2 with probability 0.5 each, independently. List the distribution of X - Y, find its mean and variance, and check them with the rules.
Two filler machines each take a mean of 40 seconds with a standard deviation of 5 seconds, independently. Sefo says the difference of their times has standard deviation 0, since the two standard deviations are equal. Find the correct mean and standard deviation of the difference.
The capper uses one cap per bottle, so the caps used Y always equal the bottles filled X. If \sigma_X = 1.5, find the standard deviation of X + Y, and compare it with what the independent rule would give.
A crate of 12 bottles holds X guava and Y = 12 - X passion fruit, with \mu_X = 5 and \sigma_X = 1.2. Find the mean and standard deviation of X + Y and of X - Y.
Teuila sells G bottles at her shop for 4 dollars each, with \mu_G = 30 and \sigma_G = 5. Losa sells Q bottles at her stall for 3 dollars each, with \mu_Q = 20 and \sigma_Q = 4, independently. Find the mean and standard deviation of their combined revenue 4G + 3Q in dollars.
For each pair, say whether treating them as independent is reasonable, and why. (a) The fill time and the cap time of one bottle, on two machines that run separately. (b) The numbers of guava and passion fruit bottles in a 6-bottle crate. (c) A day’s high temperature and the bottles sold that day.

Day 3: normal variables and several

Sefo’s drive to the market takes N(35, 4) minutes and unloading takes N(15, 3) minutes, independently. (a) Give the distribution of the total time. (b) Find the probability that it exceeds 58 minutes. (c) How long should he allow so that he runs over only 5\% of the time? (\Phi(1.645) = 0.95.)
Losa labels a batch in N(11, 1.5) minutes and Teuila in N(10, 2) minutes, independently. Find the probability that Losa takes longer.
Each bottle receives N(16.1, 0.2) fluid ounces of juice, independently. Find the mean and standard deviation of the total in a six-pack, and the probability that the total is less than 96 fluid ounces.
A sugar scoop holds N(2, 0.15) ounces. (a) Three separate scoops are used: find the standard deviation of the total. (b) A scoop three times as large is used, so the amount is 3X for one X with that model: find its standard deviation.
Sefo’s route takes S minutes, N(50, 3); Losa’s route takes L minutes, N(46, 4), independently. Find (a) the probability that Sefo’s route takes more than 10 minutes longer than Losa’s, and (b) the probability that Losa’s takes longer than Sefo’s.

Reasoning

Explain why \mu_{X+Y} = \mu_X + \mu_Y needs no independence while \sigma^2_{X+Y} = \sigma_X^2 + \sigma_Y^2 does.
Explain why the variance of X - Y is \sigma_X^2 + \sigma_Y^2 and not \sigma_X^2 - \sigma_Y^2, and show that a subtracting rule would sometimes give a negative variance.
Show that for independent X and Y, \sigma_{X+Y} \le \sigma_X + \sigma_Y, with equality only when one of the standard deviations is 0.
Show that a total of n independent variables, each with standard deviation \sigma, has standard deviation \displaystyle \sigma\sqrt{n}, and explain in words why it is smaller than the standard deviation of nX.
In Example 1 the total T has the mean and standard deviation given by the rules but is not normal. Explain why the Day 3 normal result does not apply to it.

Answers

Pairs: (1, 0) gives 1 with 0.3; (1, 1) gives 2 with 0.2; (2, 0) gives 2 with 0.3; (2, 1) gives 3 with 0.2. So S is 1, 2, 3 with 0.3, 0.5, 0.2. \mu_S = 0.3 + 1 + 0.6 = 1.9, E(S^2) = 0.3 + 2 + 1.8 = 4.1, so the variance is 4.1 - 3.61 = 0.49. Rules: \mu_X = 1.5, \mu_Y = 0.4, sum 1.9; \sigma_X^2 = 0.25, \sigma_Y^2 = 0.24, sum 0.49.
Mean 60 + 45 = 105 bottles; \displaystyle \sigma = \sqrt{25 + 144} = \sqrt{169} = 13 bottles.
Mean 9.2 + 4.6 = 13.8 seconds; variance 0.64 + 2.25 = 2.89 square seconds; \sigma = 1.7 seconds.
\displaystyle \sqrt{4 + 25} = \sqrt{29} \approx 5.39 minutes. Adding to 7 assumes the two steps always miss in the same direction by the full amount; independent misses partly cancel.
\sigma_Y^2 = 2.6^2 - 1^2 = 6.76 - 1 = 5.76, so \sigma_Y = 2.4.
Mean 40 + 25 = 65 bottles, since means add without independence. The two sales move together, so the cross term is not 0 (here likely positive); the variances alone do not give the total’s variance.
Mean 4 + 6 + 10 = 20 minutes; variance 1 + 4 + 4 = 9; \sigma = 3 minutes.
Mean 30 - 12 = 18; variance 49 + 5.76 = 54.76; \sigma = 7.4.
X - Y is 0 with 0.2, 1 with 0.2 + 0.3 = 0.5, 2 with 0.3. Mean 0.5 + 0.6 = 1.1, E(D^2) = 0.5 + 1.2 = 1.7, variance 1.7 - 1.21 = 0.49. Rules: 2.6 - 1.5 = 1.1 and 0.24 + 0.25 = 0.49.
Mean 40 - 40 = 0 seconds; \displaystyle \sigma = \sqrt{25 + 25} = \sqrt{50} \approx 7.07 seconds. The variances add, they do not cancel.
X + Y = 2X, so \sigma = 2(1.5) = 3. The independent rule would give \displaystyle \sqrt{2.25 + 2.25} = \sqrt{4.5} \approx 2.12, too small, because X and Y always miss together.
X + Y = 12: mean 12, standard deviation 0. X - Y = 2X - 12: mean 10 - 12 = -2, standard deviation 2(1.2) = 2.4.
Mean 4(30) + 3(20) = 180 dollars. Variance 16(25) + 9(16) = 400 + 144 = 544, so \displaystyle \sigma = \sqrt{544} \approx 23.32 dollars.
(a) Reasonable: neither machine’s time affects the other’s. (b) Not independent: the counts add to 6, so one fixes the other. (c) Not independent: hotter days bring more sales.
(a) N(50, 5), since \displaystyle \sqrt{16 + 9} = 5. (b) z = \dfrac{58 - 50}{5} = 1.6, probability 1 - 0.9452 = 0.0548. (c) 50 + 1.645(5) \approx 58.2 minutes.
With L for Losa’s time and M for Teuila’s, D = L - M is N(1, 2.5), since \displaystyle \sqrt{2.25 + 4} = 2.5. P(D > 0): z = \dfrac{0 - 1}{2.5} = -0.4, so 1 - \Phi(-0.4) = \Phi(0.4) = 0.6554.
Mean 6(16.1) = 96.6 fluid ounces; \displaystyle \sigma = 0.2\sqrt{6} \approx 0.49. The z-score of 96 is \displaystyle \dfrac{96 - 96.6}{0.2\sqrt{6}} \approx -1.22 and the probability is 1 - 0.8888 = 0.1112.
(a) \displaystyle 0.15\sqrt{3} \approx 0.26 ounces. (b) 3(0.15) = 0.45 ounces.
S - L is N(4, 5), since \displaystyle \sqrt{9 + 16} = 5. (a) z = \dfrac{10 - 4}{5} = 1.2, probability 1 - 0.8849 = 0.1151. (b) S - L < 0: z = -0.8, probability 1 - 0.7881 = 0.2119.
The mean of X + Y splits into \sum x\,P(x, y) + \sum y\,P(x, y), and grouping the pairs turns each part into \mu_X or \mu_Y for any joint probabilities. The variance has a cross term 2\sum de\,P(x, y), which factors into a product of two zero sums only when P(x, y) = P(X = x)P(Y = y).
X - Y = X + (-Y), and \operatorname{Var}(-Y) = (-1)^2\sigma_Y^2 = \sigma_Y^2, so the variances add. Subtracting would give, for \sigma_X = 1 and \sigma_Y = 2, the value 1 - 4 = -3, and no variance is negative.
(\sigma_X + \sigma_Y)^2 = \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y, which is at least \sigma_X^2 + \sigma_Y^2 = \sigma^2_{X+Y}, and both sides are positive, so taking square roots keeps the order. Equality needs 2\sigma_X\sigma_Y = 0, that is, one of them 0. (Geometrically: a hypotenuse is shorter than the two legs together.)
The variances add: n terms of \sigma^2 give n\sigma^2, with square root \displaystyle \sigma\sqrt{n}. nX has standard deviation n\sigma, larger once n > 1 because \displaystyle \sqrt{n} < n. Separate pieces miss in different directions and partly cancel; nX repeats one miss n times.
The result needs each variable to be normal. X and Y in Example 1 are discrete with three and two values, so their total is discrete with four values, not a normal curve.