The median is the middle value in a set of data, so for this set of 7 cities’ data you’re looking for the 4th value once the values are arranged in order.  The values in order are:

36.7, 39.8, 41.9, 42.1, 44.2, 49.4, 59.4

So the 4th/middle term is 42.1. Since the question asks for the difference between the median and 44.9, you subtract 44.9 - 42.1 to get the answer of 2.8.

With only three ages in play here, your best bet is likely to use the given information to calculate each age and then select the middle one as the median. You're given that:

1) Andrea is 24

2) Andrea is 2 years older than Rory

So you can calculate that Rory is 24 - 2 = 22 years old.

3) Rory is 3 years older than Joe, so Joe is 22 - 3 = 19 years old.

With ages of 19, 22, and 24, the median age is 22.

Average is calculated as , so your goal is to express those two quantities so that you can perform that operation and set it equal to the desired average of 75.

The number of terms will be 5, as you're told that it is a five-day work week. And the sum of the terms also involves the average calculation: if you multiply both sides of the average equation by "number of terms," you can find the sum:

Since you know Danielle's four-day average, you can calculate her current sum by multiplying 72 × 4 = 288.

This means that her total number of calls for the week will be , where  represents the number of calls she makes on the fifth day. That allows you to plug in to the average formula:

 becomes:

When you multiply both sides by 5 you get:

So you can then subtract 288 from both sides to get:

Alternatively (and this is much easier on this problem as you avoid the somewhat tedious sum calculations) you can simply consider how far the first four days are below the required average. Since 72 is 3 below the required average of 75 you know that Danielle has a deficit of 12 (4 × 3) going into the last day. She must therefore make the required average of 75 + the deficit of 12 or 87 calls.

While the rule for calculating an average is to take the sum of the terms and divide by the number of terms, in practice that can lead to cumbersome calculations in a case like this. You could calculate (192 + 188 + 195 + 197 + 205 + 208 + 203 + x), then say that that sum divided by 8 equals 200. But that's a lot of addition that you don't have to do.

What you really want to know is how far away Neil is from his goal of 200 per game. And with the numbers you're given, you can pair off values that net to an average of 200. For example, that list contains 195 and 205: the average of those two is 200, so you can pair those off and say that with those two scores he's on pace to his goal. If you do that with the available numbers, you'll pair off:

192 and 208

195 and 205

197 and 203

And then you're just left with 188. So going into his last game, Neil is 12 points short of being on track for his goal. He'll need to balance that out with a score 12 points above his goal, so he must get at least a 212.

Both lists have the same mean value of 14, a fact you can determine by adding each list (the sum of each is 70) and dividing by the number of terms (each list has five terms). Or you can find this by inspection: List X is a set of consecutive even numbers and List Y is a set of consecutive integers, and in any evenly-spaced set the mean is equal to the median, which for each list is 14.

The standard deviations are different, however, and you can determine this without having to calculate. Standard deviation is a measure of how far the terms stray from the mean. Here you can see that List X is farther spread from the mean while List Y is closer to the mean, so you know that List X has a larger standard deviation.

Both lists have the same mean value of 14, a fact you can determine by adding each list (the sum of each is 70) and dividing by the number of terms (each list has five terms). Or you can find this by inspection: List X is a set of consecutive even numbers and List Y is a set of consecutive integers, and in any evenly-spaced set the mean is equal to the median, which for each list is 14.

The standard deviations are different, however, and you can determine this without having to calculate. Standard deviation is a measure of how far the terms stray from the mean. Here you can see that List X is farther spread from the mean while List Y is closer to the mean, so you know that List X has a larger standard deviation.

This problem tests two important statistical concepts: mean and standard deviation. Note that you are expected to know how to calculate the mean of a set of numbers: it’s the sum of the values divided by the number of values.  Here, however, you may not need to go to all that trouble. You know that each list has five terms, and you also know that the sums are completely different: List A’s sum is positive and List B’s is negative. So just based on the the means are different, although you could also calculate and find that the means are 7 for List A and -6 for List B.

For standard deviation, the SAT won’t ask you to actually calculate the standard deviation, but you should understand it conceptually as a measure of how far the data points stray from the mean. For these lists, notice that the differences from the mean are all the same:

List A: 3 and 11 are each 4 away from the mean; 5 and 9 are each 2 away; and 7 is 0 away.

List B: -2 and -10 are each 6 away from the mean; -4 and -8 are each 2 away; and -6 is 0 away.

So the standard deviations for the two lists are exactly the same.