Compute weighted averages (such as a course grade from weighted components) and the expected value of a random variable as a probability-weighted sum (Statistics and Probability).

What this topic is asking

A weighted average combines values that do not count equally, and an expected value is the long-run average outcome of a random process. Both are probability- or weight-weighted sums, and the ACT tests them in grade calculations, mixed-group averages, and simple games or random payouts.

Weighted averages

A weighted average accounts for unequal importance or group size.

When to weight

Use a weighted average whenever the items being combined are not equal. Two classes of different sizes: weight each class average by its number of students. A grade with categories of different importance: weight each by its percentage. A simple average of the category scores would treat a 40%-weighted final the same as a 30%-weighted quiz, which understates the final's effect. Reading whether the groups or components are equal (simple average) or unequal (weighted average) is the deciding step.

Expected value

Expected value is a weighted average where the weights are probabilities.

For a random process with outcomes and their probabilities, $E = \sum (\text{outcome} \times \text{probability})$. For a game paying \5$with probability$0.4$and$-\$2$ (a loss) with probability $0.6$, the expected value is $5(0.4) + (-2)(0.6) = 2 - 1.2 = 0.8$, so on average you gain \0.80$ per play. Because probabilities sum to 1, no division is needed; you just sum the products. Expected value tells you the long-run average, even though any single trial gives one of the actual outcomes.

Mixtures and average rates

A weighted average also handles mixture and average-rate problems. Mixing 3 liters of a 20% solution with 2 liters of a 70% solution gives a concentration of $\frac{3(0.20) + 2(0.70)}{3 + 2} = \frac{0.6 + 1.4}{5} = \frac{2.0}{5} = 0.40$, a 40% mixture, weighted by the volumes. Similarly, an average speed over a trip is the total distance divided by the total time, a weighted average of the leg speeds by their times, not the simple average of the speeds. Recognising these as weighted averages, where the weights are volumes or times, prevents the error of averaging the rates directly.

Fair games and decisions

Expected value lets you judge whether a game or choice is favorable. A game is fair if its expected value is 0 (you neither gain nor lose on average), favorable if positive, and unfavorable if negative. So a lottery ticket costing \1$with an expected payout of$\$0.60$ has an expected net value of 0.60 - 1 = -\0.40$, an unfavorable bet. Comparing the expected values of two options tells you which is better on average, which is exactly the kind of decision the ACT frames around expected value. Remember to subtract any cost to play from the expected payout.

Why both are the same idea

A weighted average and an expected value are the same calculation: each value times its weight, summed. The only difference is that expected value's weights are probabilities (summing to 1), so the division is built in. Recognising that "average grade with weighted categories", "average of unequal groups", and "average payout of a game" are all weighted sums lets you solve them with one method. The error to avoid is defaulting to a simple average when the items deserve different weights.

Try this

Q1. One class of 12 averages 85 and another of 8 averages 90. What is the combined average? [1 point]

• Cue. $\frac{12(85) + 8(90)}{20} = \frac{1020 + 720}{20} = \frac{1740}{20} = 87$.

Q2. A spinner pays $6 with probability 0.5 and$0 with probability 0.5. What is the expected payout? [1 point]

• Cue. $6(0.5) + 0(0.5) = 3$.