Making predictions

Question

An ice cream parlor serves ice cream cones, milkshakes, and waffles with ice cream. The cost of an ice cream cone is $1.20, the cost of a milkshake is $2.00, and the cost of a waffle with ice cream is $3.50. If 50% of the customers choose an ice cream cone, 30% choose a milkshake, and 20% choose a waffle with ice cream, what is the expected value of the cost of one item served at the parlor?

Answer 1

This is kind of like a weighted-average. You can't just average the prices (which would be (1.20+2.00+3.50)/3 or $2.23) because people get different quantities of each one. So you need to build the weighted average. That "expected value" is going to be made up of (probability of ice cream cone)(price ice cream cone) + (probability of milkshake)(price milkshake) + (probability of waffle/ice cream)(price waffle/ice cream). So this weighted average or "expected value" is made up of a portion of each but "weighted" more heavily toward the ice cream cone because more people order that.

So EV = (0.50)(1.20) + (0.30)(2.00) + (0.20)(3.50) = $1.90

Notice that $1.90 is not the price of any one item but it is more of an "average" price. It's better understood by thinking about what you would expect the store to make if it sold 100 items in a day. If they sold 100 items, the would expect to bring in 100•1.90 or $190. That $190 would be made up of 50% cones, 30% shakes and 20% waffle & ice cream.

Answer 2

Expected value is the weighted average

1.2(.5) + 2(.3) +3.5(.2) = .6 + .6 + .7 = 1.9 = $1.90 = expected value for one item

Since more are buying just the cone than the other two, best bet would be on a cone. The expected value is not equal to the price of any of the 3 items. It's more like the way they used to say the average family has 2.3 children, knowing full well there's no such thing as a part of a child, and no family with that number of children.