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A high-level math concept likely to show up on an ACT exam is an expected value problem.

“Expected value” is another term for weighted average. It’s the “most common” result, though it may be a decimal and not a whole number.

One example:

On a normal six-sided die, the expected value is 3.5, since all six outcomes are equally likely to occur: (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5

Another example:

Let’s say there are 8 computers in a business office. 60% of the time, none of the computers are malfunctioning. 30% of the time, 1 is malfunctioning; 9% of the time, 2 are malfunctioning, and 1% of the time, 3 are malfunctioning.

On an “average” day, how many computers will be not be working?

The formula is: Find the sum of (each outcome * that particular outcome’s probability).

(.60 * 0) + (.30 * 1) + (.09 * 2) + (.01 * 3) = .51

So, the expected value is .51, about half a computer. Yes this is confusing, and there is no such thing as a 1/2 of a computer. More specifically, the number helps provide a value for how frequently we can “expect” a computer to malfunction each day.

More than half the time, no computer malfunctions. However, because of the small, but real chance that two or three computers are having problems, the expected value is dragged up from zero.


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Expected Value

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