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Week 6 · Chapter

Chapter 6 — Random Variables

Introduction to Statistics Free sample edition

The module's primary reading · ~40–50 minutes · Objective 4 (apply the basic rules of probability, including conditional probability and independence, and compute and interpret the expected value and standard deviation of a discrete random variable, including binomial settings), Objective 2 (summarize and display a single quantitative or categorical variable with appropriate tables and graphs, and describe its distribution's shape, center, and spread with the right numerical summaries) · Read it first this week; the lecture, the tutorial, the quiz, and the Data Lab all build on it.

In plain language

Many numbers in daily life stay unknown until chance decides them, like the total on two dice or the weight of the next bird a field team catches. This week you treat such a number as a random variable (a number whose value a chance process sets). You list its possible values with their chances, which is its probability distribution (the full list of values and how likely each one is). Then you find its expected value (the average it settles toward over many tries) and its standard deviation (how far one result typically lands from that average). Last, you meet the density curve (a smooth curve whose area gives chances) for numbers such as weight that can take any value in a range.

Why you already care about this

Roll two dice in a board game and the total can be anything from 2 to 12. Pull a page at random from a wall calendar and the month might have 28 days, 30 days, or 31. Neither number is known until chance decides it. Yet you already have a feel for what is typical: a total somewhere near 7, a month of about 30 days.

That feel shows up at work, too. An emergency-room nurse planning staff thinks about how many patients typically arrive in an hour. A delivery driver plans around how long a route usually takes and how much that time swings from day to day. In each case there is a number you care about, chance sets it, and you want two things: its long-run average and its typical swing. Last week you learned to put a number on chance itself. This week you attach chance to numbers, and you compute with the pair.

By the end of this chapter you will be able to:

  • tell a discrete random variable from a continuous one;
  • check whether a table of values and probabilities is a legitimate probability distribution, and read probabilities from it;
  • compute the expected value of a discrete random variable and explain why it is a long-run average, not a prediction for one try;
  • compute the variance and the standard deviation of a discrete random variable, with the right units;
  • predict what adding a constant, or multiplying by one, does to the mean and the standard deviation;
  • read a density curve, where probability is area and a single exact value has probability 0.

Carry this sentence with you: E(X) is the average over many tries — not a forecast for the next one.

Where we are

  • Already covered: Weeks 1 through 4 described data you already have; Week 5 measured chance itself with the probability rules.
  • This week: random variables, discrete and continuous; probability distributions; expected value; variance and standard deviation; shifting and rescaling; and a first look at density curves.
  • How they connect: Week 3's mean and standard deviation come back, but now they describe a chance process instead of a list of data, and Week 5's probabilities become the weights.
  • Where we're going next: Week 7, Binomial & Normal Models, will hand you two ready-made distributions that you will lean on for the rest of the course.

Each earlier week feeds this one. Week 1 sorted quantitative variables into discrete and continuous, and that same split sorts random variables. Week 2 taught you to read the shape of a distribution from a histogram, and you will read the shape of a probability distribution the same way. Week 3 gave you the mean and the standard deviation of a list, including the population values of the six die faces: a mean of 3.5 and a standard deviation of about 1.71. Week 5 gave you the 36 equally likely ways two dice can land and the law of large numbers, which says that a long run of tries settles down. This chapter puts those pieces together.

1. Random variables: put a number on a chance outcome

Hold a fair die before you roll it. You cannot say which face will land on top, but you can say it will be 1, 2, 3, 4, 5, or 6. The face is a number, and chance picks it. That is the whole idea of this week in one picture.

  • A random variable is a number whose value is set by the outcome of a chance process, such as the face that lands on top when you roll a die.
  • A discrete random variable is a random variable whose possible values can be listed with gaps between them, such as the sum of two dice or the number of days in a month.
  • A continuous random variable is a random variable that can take any value in a range, such as the body mass of a penguin picked at random.

You name a random variable with a capital letter. Let X be the face that lands on top of one fair die, and write a single possible value with a small letter, x. So "X = 3" means "the die shows 3".

Worked example — four random variables, sorted. Here are the four random variables you will use all week.

  1. One die. X (the face of one fair die) takes the values 1, 2, 3, 4, 5, 6. You can list them, with gaps between them, so X is discrete.
  2. Two dice. S (the sum of two fair dice) takes the values 2 through 12. Again you can list them, so S is discrete.
  3. A month. Pick a month at random from a common year, each month equally likely, and let D be its number of days. D takes only 28, 30, or 31, so D is discrete.
  4. A penguin. Pick a penguin at random from the 342 birds with a recorded body mass, and let B be its body mass in grams. Masses run from 2,700 g to 6,300 g and can fall anywhere in between, so B is a continuous random variable.

In words: three of the four can be listed value by value, and the fourth can only be described as a range. That difference decides which tools you use, as Section 6 shows.

Misconception check — whole numbers versus decimals. Students often think a variable is continuous whenever its values have decimals, and discrete whenever they are whole numbers. It is tempting because counts are whole numbers. But the test is not how a value is written. The penguin file records body mass in whole grams, yet a bird's true mass can be any value in a range, so B is continuous. Later this chapter, the die's long-run average works out to 3.5, a decimal, yet X itself only ever shows a whole face. The cure: ask whether you can list the possible values with gaps between them. If you can, the variable is discrete.

Honesty note: every measuring tool rounds. A scale that reads to the nearest gram still treats mass as continuous; it cannot report every digit.

Memory hook: list the values with gaps — discrete; any value in a range — continuous.

2. Probability distributions: check that the table is legitimate

A random variable is only half the story. You also need to know how likely each value is. Write the values in one column and their chances in the next, and you have the variable's full description.

  • A probability distribution is a list or rule that gives every possible value of a random variable together with its probability.
  • A legitimate probability distribution is a probability distribution in which every probability is between 0 and 1 and the probabilities of all the possible values add to exactly 1.

The two checks come straight from Week 5: a probability is never below 0 or above 1, and the chances of everything that can happen add to 1. You write P(X = x) for the probability that X takes the value x.

Worked example — the die table. Each face of a fair die is equally likely.

Probability distribution of X, the face of one fair die
Face x 1 2 3 4 5 6
P(X = x) 1 ⁄ 6 1 ⁄ 6 1 ⁄ 6 1 ⁄ 6 1 ⁄ 6 1 ⁄ 6
  1. Check each probability. Each one is 1 ⁄ 6, which sits between 0 and 1.
  2. Check the total. Six probabilities of 1 ⁄ 6 each add to 6 × 1 ⁄ 6 = 1.

In words: the die table passes both checks, so it is a legitimate probability distribution.

Worked example — the two-dice sum. In Week 5 you listed the 36 equally likely ways two dice can land. Group them by their total and you get the distribution of S.

Probability distribution of S, the sum of two fair dice, from 36 equally likely outcomes
Sum s 2 3 4 5 6 7 8 9 10 11 12
Ways 1 2 3 4 5 6 5 4 3 2 1
P(S = s) 1 ⁄ 36 2 ⁄ 36 3 ⁄ 36 4 ⁄ 36 5 ⁄ 36 6 ⁄ 36 5 ⁄ 36 4 ⁄ 36 3 ⁄ 36 2 ⁄ 36 1 ⁄ 36
  1. Count the ways. 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36 outcomes.
  2. Check the total. The eleven probabilities add to 36 ⁄ 36 = 1, and each one sits between 0 and 1.
  3. Read a probability. P(S ≥ 10) = (3 + 2 + 1) ⁄ 36 = 6 ⁄ 36 = 1 ⁄ 6.

In words: about one roll in 6 gives a total of 10 or more. A player who needs a big total should expect to wait several turns.

A distribution has a shape, like the data in Week 2. Picture a bar for each total, as tall as its probability. The bars climb from 2 to a peak at 7 and fall away evenly to 12, so the distribution of S is symmetric and unimodal. The die's bars, by contrast, are all the same height: a flat distribution.

Worked example — the month table. Seven of the 12 months have 31 days, four have 30, and February has 28 in a common year.

  1. Write the probabilities. P(D = 28) = 1 ⁄ 12, P(D = 30) = 4 ⁄ 12, and P(D = 31) = 7 ⁄ 12.
  2. Check the total. 1 ⁄ 12 + 4 ⁄ 12 + 7 ⁄ 12 = 12 ⁄ 12 = 1.

In words: the month table is legitimate too. Its tallest bar sits at 31, with a lone short bar out at 28, so its shape is left-skewed, the same shape Week 2 found in the month lengths.

Misconception check — a table that looks fine but fails. Students often accept any table whose numbers look like probabilities. Suppose someone miscounts and lists eight 31-day months. Then 1 ⁄ 12 + 4 ⁄ 12 + 8 ⁄ 12 = 13 ⁄ 12, which is more than 1. Every single entry sits between 0 and 1, so the first check passes, but the total fails. That table is not a legitimate distribution. The cure: run both checks every time, and always add the column.

In practice (your tools). In Google Sheets, type the sums 2 through 12 in cells A2 to A12 and the ways in B2 to B12. In C2, type =B2/36 and copy it down to C12 to get the probabilities. In an empty cell, =SUM(C2:C12) returns 1, which is your total check, and =MIN(C2:C12) and =MAX(C2:C12) confirm that every entry sits between 0 and 1. Excel uses the same formulas.

Memory hook: each one between 0 and 1; all of them add to exactly 1.

3. Expected value: the long-run average, not the next result

Imagine rolling one die thousands of times and keeping a running average of the faces. Early on, the average jumps around. Then, the way the share of sixes settled in Week 5's law of large numbers, the running average settles near one number. That settling point is the variable's expected value.

  • The expected value is the probability-weighted average of a random variable's values, found by multiplying each value by its probability and adding the products; it is the long-run average over many repetitions and need not be a possible value.

You write it E(X), read "E of X". It is also the mean of the distribution, written μ (read "mu"), the same symbol Week 3 used for a population mean. In symbols, E(X) = Σ x · P(X = x): multiply each value by its probability, then add them all up.

Worked example — one die. Each face has probability 1 ⁄ 6.

  1. Multiply and add. E(X) = 1 × 1 ⁄ 6 + 2 × 1 ⁄ 6 + 3 × 1 ⁄ 6 + 4 × 1 ⁄ 6 + 5 × 1 ⁄ 6 + 6 × 1 ⁄ 6.
  2. Collect the faces. E(X) = (1 + 2 + 3 + 4 + 5 + 6) × 1 ⁄ 6 = 21 ⁄ 6 = 3.5.

In words: over many rolls, the average face settles near 3.5. A player who rolls one die again and again will see the faces average out to about 3.5 per roll.

Worked example — two dice. Multiply each sum by its number of ways, add, and divide by the 36 outcomes.

  1. The products. 2 × 1, 3 × 2, 4 × 3, 5 × 4, 6 × 5, 7 × 6, 8 × 5, 9 × 4, 10 × 3, 11 × 2, 12 × 1 give 2, 6, 12, 20, 30, 42, 40, 36, 30, 22, 12.
  2. Add them. The products add to 252.
  3. Divide. E(S) = 252 ⁄ 36 = 7.

In words: over many rolls of two dice, the totals average out near 7, the same total that had the most ways in Week 5's grid. Adding a second die makes a 2-fold change in the long-run average, E(S) = 7 versus E(X) = 3.5, because each die contributes 3.5 on average.

Worked example — the month. A month picked at random has 28, 30, or 31 days with probabilities 1 ⁄ 12, 4 ⁄ 12, and 7 ⁄ 12.

  1. Multiply and add. E(D) = (28 × 1 + 30 × 4 + 31 × 7) ⁄ 12 = (28 + 120 + 217) ⁄ 12.
  2. Divide. E(D) = 365 ⁄ 12 ≈ 30.42 days, rounded at the end to two decimals.

In words: the average month of a common year runs about 30.42 days, which matches the mean of the twelve month lengths from Week 3. A person planning a monthly budget around "about 30 days" is close to right on average, and wrong for every single month.

Misconception check — "the expected value is what I should expect on one try". Students often read E(X) = 3.5 as a forecast for the next roll. The name invites it. But no face shows 3.5, and no month has 30.42 days. The expected value is what results average over many tries; it need not be a value the variable can take. The cure: say "over many tries, the average settles near …" every time you report an expected value.

Pause and think, and choose one: (a) describe a decision in your own life where the long-run average is the right guide, because you face the same chance many times; or (b) describe a decision where one try matters so much that the average is not enough, and say what else you would want to know.

In practice (your tools). In Google Sheets, put the faces 1 to 6 in A2 to A7 and =1/6 in B2 to B7. Then =SUMPRODUCT(A2:A7,B2:B7) returns 3.5. For two dice, with the sums in A2 to A12 and their probabilities in C2 to C12, =SUMPRODUCT(A2:A12,C2:C12) returns 7. SUMPRODUCT does the multiply-then-add step in one formula. Excel uses the same formula.

Memory hook: weight each value by its chance, then add.

4. Variance and standard deviation: measure how far results typically land

Two games can share the same expected value and still feel very different. One pays close to its average every time; the other swings from nothing to a lot. The expected value cannot tell them apart. You need a measure of spread, and Week 3 already built one.

  • Var(X) is the variance of a random variable X: multiply each squared deviation from the mean by its probability and add the products; it is measured in squared units.
  • SD(X) is the standard deviation of a random variable X, the square root of Var(X); it describes how far the variable's value typically lands from its mean, in the variable's own units.

In symbols, Var(X) = Σ (x − μ)² · P(X = x) and SD(X) = √Var(X). The standard deviation of a distribution is also written σ (read "sigma").

Worked example — one die. The mean is 3.5.

  1. Deviations. Each face minus 3.5 gives −2.5, −1.5, −0.5, 0.5, 1.5, 2.5.
  2. Square them. 6.25, 2.25, 0.25, 0.25, 2.25, 6.25, which add to 17.5.
  3. Weight by the probabilities. Every face has probability 1 ⁄ 6, so Var(X) = 17.5 × 1 ⁄ 6 = 35 ⁄ 12 ≈ 2.9167 square units.
  4. Take the square root. SD(X) = √2.9167 ≈ 1.71.

In words: a single roll typically lands about 1.71 away from 3.5. These are exactly Week 3's population variance and standard deviation of the six faces. In Week 3 they described a list of six numbers; now they describe a process you can repeat forever.

Worked example — two dice. The mean is 7. For each sum, square its distance from 7 and multiply by its ways.

Squared deviations of each two-dice sum from 7, multiplied by the number of ways
Sum s 2 3 4 5 6 7 8 9 10 11 12
(s − 7)² × ways 25 32 27 16 5 0 5 16 27 32 25
  1. Add the row. The entries add to 210.
  2. Divide by the 36 outcomes. Var(S) = 210 ⁄ 36 = 35 ⁄ 6 ≈ 5.83 square units.
  3. Take the square root. SD(S) = √(35 ⁄ 6) ≈ 2.4152, rounded at the end to four decimals.

In words: a two-dice total typically lands about 2.4 away from 7. A player waiting on a 7 should not be surprised by a 5 or a 9, which are less than one standard deviation from 7.

Your month variable is quieter. Its standard deviation is SD(D) ≈ 0.86 days, Week 3's population standard deviation of the month lengths, because the twelve months are a whole population. A month picked at random rarely strays far from its average.

Misconception check — dividing by one less. Students often reach for the Week 3 sample formula here and divide the squared deviations by one less than the count. For the six faces that gives 17.5 ⁄ 5 = 3.5, the sample variance from Week 3. But a probability distribution describes the whole process, like a population, so each squared deviation is weighted by its probability and nothing is divided by one less. The cure: weight by P(X = x), and the die's variance is 35 ⁄ 12, not 3.5. Keep the units straight too: the variance is in squared units, and the standard deviation is back in the variable's own units.

In practice (your tools). With the faces in A2 to A7 and their probabilities in B2 to B7, =SUMPRODUCT((A2:A7-3.5)^2,B2:B7) returns the variance, and wrapping it as =SQRT(SUMPRODUCT((A2:A7-3.5)^2,B2:B7)) returns the standard deviation. Use Format, then Number, to show two decimals. Do not use STDEV.S on the six faces: it divides by one less, which is the sample rule. Excel uses the same formulas.

Memory hook: square the distances, weight by the chances, then take the root.

5. Shifting and rescaling: predict the new mean and SD

Sometimes you measure the same random thing on a new scale, and you want its new mean and spread fast. You might count from a new starting point, or switch units. You do not need to rebuild the whole distribution; two short rules do the work.

  • Shifting is adding the same constant to every value of a random variable; it moves the mean by that constant and leaves the standard deviation unchanged.
  • Rescaling is multiplying every value of a random variable by the same positive constant; it multiplies both the mean and the standard deviation by that constant.

Put together: if a new variable is a + b × X, with b positive, its mean is a + b × E(X) and its standard deviation is b × SD(X). The a drops out of the spread because sliding every value by the same amount does not change the distances between them.

Worked example — shift the sum. Let S − 7 be how far a two-dice total lands from 7.

  1. The mean. Shifting by −7 moves the mean from 7 to 7 − 7 = 0.
  2. The SD. Every total slides by the same 7, so the gaps between totals do not change, and the SD is still ≈ 2.4152.

In words: on average a total lands right on 7, and it typically misses by about 2.4 in one direction or the other.

Worked example — shift the month. Let D − 30 be how many days a month runs past 30.

  1. The new values. D − 30 takes the values −2, 0, and 1.
  2. The mean, directly. (−2 × 1 + 0 × 4 + 1 × 7) ⁄ 12 = 5 ⁄ 12 ≈ 0.42 days.
  3. The mean, by the rule. 365 ⁄ 12 − 30 = 5 ⁄ 12, the same answer.
  4. The SD. Still ≈ 0.86 days, because a shift leaves the spread alone.

In words: the average month runs about 0.42 days past 30. Both routes agree, which is a good habit to check.

Worked example — rescale the month. Let H = 24 × D be the month's length in hours.

  1. The new values. 24 × 28 = 672, 24 × 30 = 720, and 24 × 31 = 744 hours.
  2. The mean. E(H) = 24 × 365 ⁄ 12 = 730 hours.
  3. The SD. SD(H) = 24 × SD(D) ≈ 20.69 hours, using the unrounded SD of D.

In words: the average month has 730 hours, though no single month does, and month lengths typically stray about 20.69 hours from that average. The spread in hours is 24 times the spread in days, exactly as rescaling predicts.

Misconception check — "adding a constant spreads things out". Students often think that adding a constant changes the standard deviation too, because every number changed. It is tempting because the values all look different. But the standard deviation measures distances between values, and a shift moves every value by the same amount. The cure: a shift moves the center and leaves the spread; a rescale changes both. For the record, a rescale multiplies the variance by the square of the constant, which is why the standard deviation is the easier number to carry.

Memory hook: adding slides the center; multiplying stretches everything.

6. Density curves: probability becomes area

Pick a penguin at random and ask for its body mass. There is no list of values to write in a table, because the mass could be any number in a range. Week 2's histogram of the 342 body masses hints at the answer. Make the bins narrower and narrower, and the tops of the bars blur into a smooth curve.

  • A density curve is a smooth curve that models a continuous random variable: it never dips below the horizontal axis, the total area under it is 1, and the area above a range of values is the probability of landing in that range.

You get two facts from "probability is area". The whole area is 1, because the variable has to land somewhere. And any single exact value has probability 0, because a single point has no width and so no area.

Worked example — a flat spinner. Picture a spinner whose pointer is equally likely to stop anywhere on a dial marked from 0 to 6, and let Y be the stopping point. Its density curve is a flat line at height 1 ⁄ 6 from 0 to 6.

  1. Check the total area. The region is a rectangle 6 wide and 1 ⁄ 6 tall, so its area is 6 × 1 ⁄ 6 = 1.
  2. Find a probability. P(2 ≤ Y ≤ 3.5) is the rectangle from 2 to 3.5, which is 1.5 wide, so P(2 ≤ Y ≤ 3.5) = 1.5 × 1 ⁄ 6 = 0.25.
  3. A single point. P(Y = 3) = 0, because the line at exactly 3 has no width.

In words: about one spin in 4 stops between 2 and 3.5, and no spin has a real chance of stopping at exactly 3, though plenty stop near it.

Worked example — the penguin curve. Smooth the histogram of the 342 body masses into a curve. It keeps the mild right skew Week 2 found, with a longer tail toward the heavy birds.

  1. Read an area from the quartiles. About half of the area lies between Q1 = 3,550 g and Q3 = 4,750 g, because those two quartiles hold the middle half of the masses.
  2. Read another. About half of the area lies below the median, 4,050 g.
  3. A single value. P(B = 4,000 g exactly) = 0 for the curve.

In words: a penguin picked at random has about an even chance of weighing between 3,550 g and 4,750 g. To ask a useful question about one mass, ask about a range around it.

You can now describe any random variable's distribution the Week 2 and Week 3 way: its shape (the die is flat, the two-dice sum is symmetric, the months are left-skewed, the penguin curve is right-skewed), its center (the expected value), and its spread (the standard deviation).

Misconception check — "probability 0 means it cannot happen". Students often read P(B = 4,000 g exactly) = 0 as "no penguin weighs 4,000 g". It sounds like a ban. But the zero belongs to the curve's single point, which has no width. Ranges near 4,000 g have real area, so masses near 4,000 g are common. A second trap: the height of the curve is not a probability; only area is. The cure: for a continuous random variable, always ask about a range, and read probability as area.

Limitation: the curve is a model. The real file records mass in whole grams, so two birds can share the same recorded value; the zero rule describes the smooth model of true mass, not the rounded record.

Memory hook: for a continuous variable, probability is area — never height, never a single point.

Key terms

  • random variable — A number whose value is set by the outcome of a chance process, such as the face that lands on top when you roll a die. (first used: 1. Random variables: put a number on a chance outcome)
  • discrete random variable — A random variable whose possible values can be listed with gaps between them, such as the sum of two dice or the number of days in a month. (first used: 1. Random variables: put a number on a chance outcome)
  • continuous random variable — A random variable that can take any value in a range, such as the body mass of a penguin picked at random. (first used: 1. Random variables: put a number on a chance outcome)
  • probability distribution — A list or rule that gives every possible value of a random variable together with its probability. (first used: 2. Probability distributions: check that the table is legitimate)
  • legitimate probability distribution — A probability distribution in which every probability is between 0 and 1 and the probabilities of all the possible values add to exactly 1. (first used: 2. Probability distributions: check that the table is legitimate)
  • expected value — The probability-weighted average of a random variable's values, found by multiplying each value by its probability and adding the products; it is the long-run average over many repetitions and need not be a possible value. (first used: 3. Expected value: the long-run average, not the next result)
  • Var(X) — The variance of a random variable X: multiply each squared deviation from the mean by its probability and add the products; it is measured in squared units. (first used: 4. Variance and standard deviation: measure how far results typically land)
  • SD(X) — The standard deviation of a random variable X, the square root of Var(X); it describes how far the variable's value typically lands from its mean, in the variable's own units. (first used: 4. Variance and standard deviation: measure how far results typically land)
  • shifting — Adding the same constant to every value of a random variable; it moves the mean by that constant and leaves the standard deviation unchanged. (first used: 5. Shifting and rescaling: predict the new mean and SD)
  • rescaling — Multiplying every value of a random variable by the same positive constant; it multiplies both the mean and the standard deviation by that constant. (first used: 5. Shifting and rescaling: predict the new mean and SD)
  • density curve — A smooth curve that models a continuous random variable: it never dips below the horizontal axis, the total area under it is 1, and the area above a range of values is the probability of landing in that range. (first used: 6. Density curves: probability becomes area)

Summary

  • A random variable is a number that chance sets; list its values with gaps and it is discrete, allow any value in a range and it is continuous.
  • A probability distribution lists every value with its probability; it is legitimate when each probability is between 0 and 1 and they add to exactly 1.
  • The expected value multiplies each value by its probability and adds; it is a long-run average, and it need not be a possible value.
  • The variance weights each squared deviation by its probability; the standard deviation is its square root, in the variable's own units.
  • Adding a constant moves the mean and leaves the standard deviation alone; multiplying by a positive constant multiplies both.
  • For a continuous random variable, probability is area under a density curve, the total area is 1, and a single exact value has probability 0.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. A campus help desk tracks how many laptop-repair requests arrive in its first hour of the day. Suppose the count has this distribution: 0 requests with probability 0.2, 1 with 0.5, 2 with 0.2, and 3 with 0.1. Is this a legitimate probability distribution? Find the expected number of requests and say what it means.
  2. A youth soccer coach writes two tables for the number of goals her team scores in a game. Table A: 0 goals with probability 0.3, 1 with 0.4, 2 with 0.4. Table B: 0 goals with probability 0.25, 1 with 0.5, 2 with 0.25. Which table is a legitimate probability distribution, and what is its expected value?
  3. A neighborhood bakery sells L loaves a day, with E(L) = 120 loaves and SD(L) = 15 loaves. Each loaf earns $3, and the bakery pays a fixed $50 a day for its oven. Daily profit is 3 × L − 50 dollars. Find the mean and the standard deviation of daily profit.
  4. A bus arrives at a stop at a time equally likely to fall anywhere in a 10-minute window, so your wait has a flat density curve at height 1 ⁄ 10 from 0 to 10 minutes. Find the probability that you wait between 3 and 7 minutes, and the probability that you wait exactly 5 minutes.

Answers (worked reasoning — read after attempting)

  1. Check each probability: 0.2, 0.5, 0.2, and 0.1 all sit between 0 and 1. Check the total: 0.2 + 0.5 + 0.2 + 0.1 = 1.0. Both checks pass, so the table is legitimate. Now weight each count by its probability: 0 × 0.2 + 1 × 0.5 + 2 × 0.2 + 3 × 0.1 = 0 + 0.5 + 0.4 + 0.3 = 1.2 requests. In words: over many mornings, the desk averages about 1.2 requests in its first hour. No single hour brings 1.2 requests, which is exactly why an expected value is a long-run average and not a forecast.
  2. Table A fails: 0.3 + 0.4 + 0.4 = 1.1, which is more than 1, even though each entry sits between 0 and 1. Table B passes: 0.25 + 0.5 + 0.25 = 1, and each entry is between 0 and 1. Its expected value is 0 × 0.25 + 1 × 0.5 + 2 × 0.25 = 0 + 0.5 + 0.5 = 1 goal per game. Over a long season the team averages about one goal a game.
  3. Profit is a rescale by 3 followed by a shift of −50. The mean: 3 × 120 − 50 = 360 − 50 = $310 a day. The standard deviation: the shift does not change the spread, and the rescale multiplies it by 3, so 3 × 15 = $45. In words: the bakery clears about $310 on an average day, and a typical day lands about $45 above or below that. Subtracting the $50 oven cost moved the center, not the swing.
  4. Probability is area. The region from 3 to 7 minutes is a rectangle 7 − 3 = 4 minutes wide and 1 ⁄ 10 tall, so the probability is 4 × 1 ⁄ 10 = 0.4. In words: about 4 waits in 10 last between 3 and 7 minutes. The probability of waiting exactly 5 minutes is 0, because a single point has no width; a wait of very nearly 5 minutes is still common.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, and the quiz — and next week the binomial model gives a ready-made distribution for counting successes, with its own rule for the mean and standard deviation.

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