If the mean of list A is 6.8 and the standard deviation is 3.6, then how many elements of list A are within 1 unit of standard deviation of the mean?

A = {2, 9, 2, 6, 9, 10, 7, 4, 5, 14} q8

FAQ: Why do we divide by R? Why not divide by T?

First, let's think briefly about what exactly tax is and how it's calculated. At the start, we have the retail price--this is the price of something as given by the seller. Then, there is a tax imposed by whatever government. This tax, by nature, is almost always calculated as a percent of the retail price.

So, for example, if you have a $10 t-shirt, tax might be 15%, which would mean you'd pay $1.50 in tax. It's not possible, though, to have a tax that's calculated as a percentage of the cost after tax. Back to the $10 t-shirt, it would be paradoxical to try to tax the shirt based on the after-tax cost ($11.50 total, which is T). How can you calculate a tax based on a number that already incorporates the tax? We have to start the tax as a percentage of the retail cost, R, or else we would be stuck pretty quickly.

If we reword this question a bit, we get this: If the original price of a shirt is R dollars, and the original price + tax is T dollars, then the tax is __________ %. Given an understanding of what tax generally refers to, then we know we're looking for the % of R dollars, not the % of R dollars + tax.

That being said, although the algebraic solution seems more direct and simpler, it's actually a bit difficult to deal with those abstract variables, as we're seeing here! That's why we also show the method of plugging in concrete, easy numbers in the answer video. It may seem like a more complicated solution than using just the variables, but it can be much easier to deal with real numbers than it is to understand the relationships between variables. q10

FAQ: Why do we multiply the two values? Why not add 5C2 and 4C2?

In a way, this is just a question of "and" or "or." Make sure that we don't confuse it with probability, though, which has very different rules for "and" and "or."

In this question, we're looking for the number of different ways we can choose both 2 boys and 2 girls. In the end, we're really choosing people to go in one group—we're not considering the two boys as a separate group from the 2 girls.

Here's a couple of simple examples to illustrate what I mean by this:

If I have an a peach, a plum, an apricot, an orange, a lime, and a lemon, how many ways can I pick two pieces of fruit with pits OR two pieces of citrus fruit? Well, there are 3C2 possible groups of stone fruit and 3C2 possible groups of citrus fruit. That's 3 + 3 = 6.

peach, plum

peach, apricot

plum, apricot

orange, lemon

orange, lime

lemon, lime

From that same group, how many ways can I pick two pieces of fruit with pits AND two pieces of citrus fruit? This gives more possibilities.

peach, plum, orange, lemon

peach, plum, orange, lime

peach, plum, lemon, lime

peach, apricot, orange, lemon

peach, apricot, orange, lime

peach, apricot, lemon, lime

plum, apricot, orange, lemon

plum, apricot, orange, lime

plum, apricot, lemon, lime

That makes 3 * 3 = 9 possible groups.

Really, we found that there are 3 ways to choose 2 stone fruits and 3 ways to choose two citrus fruits. If we're we're counting the possibilities of both together in one group, then we have to multiply the results, as per the fundamental counting principal (in the related lessons below).

This is the same as the boys and girls situation. If we asked how many ways you can make a group of 2 boys or a group of 2 girls, then we would add the two. But since we're combining these into one group, we're looking at boys and girls, and so we multiply. ;