AP Statistics · Lesson 2

Discrete and binomial random variables

Know Know it

A discrete random variable X has a list of values xix_i with probabilities pip_i that are each between 0 and 1 and add to 1.

Its mean, or expected value, is μX=∑xipi\mu_X = \sum x_i p_i. Its variance is σX2=∑(xi−μX)2pi\sigma_X^2 = \sum (x_i - \mu_X)^2 p_i, and the standard deviation σX\sigma_X is the square root of the variance.

Linear changes. If Y=a+bXY = a + bX, then μY=a+bμX\mu_Y = a + b\mu_X and σY=∣b∣σX\sigma_Y = |b|\sigma_X. Adding a constant shifts the center but not the spread.

Sums and differences. For any X and Y, μX+Y=μX+μY\mu_{X+Y} = \mu_X + \mu_Y and μX−Y=μX−μY\mu_{X-Y} = \mu_X - \mu_Y. If X and Y are independent, σX±Y2=σX2+σY2\sigma_{X \pm Y}^2 = \sigma_X^2 + \sigma_Y^2: the variances add for a sum and for a difference.

A count X is binomial when there is a fixed number n of trials, each a success or failure, the trials are independent, and the success probability p is the same each time. Then P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, μX=np\mu_X = np and σX=np(1−p)\sigma_X = \sqrt{np(1-p)}. When you sample without replacement, you may treat the trials as independent when the sample is less than 10 percent of the population.

Worked example Easy

X = number of cars in a randomly chosen household in a fictional town

x0123
P(X = x)0.100.400.350.15

What is the mean of X?

  1. 1.50
  2. 1.55Answer
  3. 0.25
  4. 0.86

How to solve it

  1. Multiply each value by its probability: 0, 0.40, 0.70 and 0.45.
  2. Add them: μX=1.55\mu_X = 1.55 cars per household.
  3. Over many randomly chosen households, the average number of cars would be about 1.55.

Why each choice is right or wrong

  • A. Incorrect. 1.50 is the plain average of the values 0, 1, 2 and 3, which ignores that they have different probabilities.
  • B. Correct. 0(0.10)+1(0.40)+2(0.35)+3(0.15)=1.550(0.10) + 1(0.40) + 2(0.35) + 3(0.15) = 1.55 cars.
  • C. Incorrect. 0.25 is the average of the four probabilities, which is always 0.25 for four values and is not a mean of X.
  • D. Incorrect. 0.86 is the standard deviation of X, which measures spread around the mean, not the mean itself.

Trap Standard deviations do not add

For independent X and Y, σX+Y=σX2+σY2\sigma_{X+Y} = \sqrt{\sigma_X^2 + \sigma_Y^2}, not σX+σY\sigma_X + \sigma_Y. The same is true for X−YX - Y: subtracting does not subtract the spread, because two uncertain amounts are still both uncertain. Add the variances, then take the square root.

Explain Explain it

Task verbs: calculate, interpret

Before using the binomial formula, name the setting: what counts as a success, what n and p are, and why the trials are independent. Then write the formula with your numbers, or the calculator command with n, p and the values of k, so a reader can follow the work.

"At least" questions are often easiest with the complement: P(X≥2)=1−P(X≤1)P(X \geq 2) = 1 - P(X \leq 1).

To interpret an expected value, call it a long-run average: over many repetitions of the chance process, the values of X average out to about μX\mu_X. A single outcome need not equal it, or even be possible.

Worked example Medium

A fictional basketball player makes 70 percent of her free throws, and her shots are independent. She takes 10 free throws.

What is the probability that she makes exactly 8 of them?

  1. 0.0576
  2. 0.0052
  3. 0.3828
  4. 0.2335Answer

How to solve it

  1. Binomial setting: n = 10 shots, success = a make, p = 0.7, and the shots are independent.
  2. P(X=8)=(108)(0.7)8(0.3)2P(X = 8) = \binom{10}{8}(0.7)^8(0.3)^2.
  3. (108)=45\binom{10}{8} = 45, so the probability is 45⋅0.05765⋅0.09≈0.233545 \cdot 0.05765 \cdot 0.09 \approx 0.2335.

Why each choice is right or wrong

  • A. Incorrect. 0.0576 is 0.780.7^8, which ignores the 2 misses and the number of orders in which 8 makes can happen.
  • B. Incorrect. 0.0052 is 0.78⋅0.320.7^8 \cdot 0.3^2, the chance of one particular order; it leaves out the 45 possible orders.
  • C. Incorrect. 0.3828 is the probability of 8 or more makes, which adds the cases of 9 and 10 makes.
  • D. Correct. (108)(0.7)8(0.3)2=45⋅0.05765⋅0.09≈0.2335\binom{10}{8}(0.7)^8(0.3)^2 = 45 \cdot 0.05765 \cdot 0.09 \approx 0.2335.

Interpret Your turn

A fictional school raffle sells tickets for 2 dollars each. Let W be a buyer's net gain from one ticket, the prize minus the 2 dollars. The expected value of W is -0.80 dollars. Interpret this expected value in context.

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A full answer does this

  • Describes the expected value as a long-run average over many tickets
  • States the size and direction in context: a loss of about 0.80 dollars per ticket

Model answer

If someone bought a very large number of these raffle tickets, their net gain would average out to a loss of about 0.80 dollars per ticket. No single ticket loses exactly 0.80 dollars; it is the long-run average.

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