Descriptive Statistics - GMAT Quantitative

We are looking for an average here. Statement 1 tells us we are looking for even consecutive integers such as 2, 4, 6, 8, and 10. Statement 2 tells us the difference between the smallest and largest integer; however, the difference between the largest and smallest of five consecutive even (or odd) integers will ALWAYS be 8, regardless of what 5 consecutive integers we choose; therefore the two statements don't give us enough information to solve for the average.

Statement 2 does not provide enough information about the mean as it can vary greatly from the median. Statement 1 is sufficient to calculate the mean, because even though it is impossible to calculate the set of numbers, the mean is calculated by dividing the sum by the total number of incidences in the set.

In this case, average is also the middle value.

The arithmetic mean of and is equal to .

Statement 1 alone gives us this value directly.

From Statement 2 alone, the value can be determined by dividing both sides by 4:

Assume Statement 1 alone. The sequences

and

are both arithmetic, each term being the previous term plus the same number (in the first case, this common difference is 40; in the second, it is 20). The fourth term is 120 in both cases, The second and third terms of the first sequence have arithmetic mean ; the second and third terms of the second sequence have arithmetic mean . Therefore, the mean of those two terms cannot be determined for certain. A similar argument holds for Statement 2 alone being insufficient.

Now assume both statements. Let be the common difference of the sequences mentioned in Statement 2. By Statement 2, 0 is the first term, so the sequence will be

By the first statement, the fourth term is 120, so

The second terms is and the third term is , and their arithmetic mean is .

Statement 1 alone would not be helpful.

Example 1: If and , the list, in descending order, is and the median would be .

Example 2: If and , the list, in descending order, is and the median would be .

In contrast, if Statement 2 is true, since and , and . Regardless of their relationship, this makes the fourth-highest number, and, therefore, the median.

Looking at statements (1) and (2) separately, we cannot get the median number since we don’t know the 50th percentile. However, putting the two statements together, we know that the 50th percentile is 5. So the median is 5.

Median of A = 3

Mean of A = 3.2

Median of B = 5

Mean of B = 4.8

The four numbers appear in ascending order, so their median must be the arithmetic mean of the middle two numbers. Therefore,

If , then .

If , then

The two statements are equivalent. This is enough to allow the mean to be found:

The answer is that either statement alone is sufficient to answer the question.

In a data set of 19 elements, 19 being odd, the median is the element that occurs in the position when they are arranged in ascending (or descending) order. Neither statement alone tells us what that middle element is. But put together, they tell us what the ninth and eleventh elements would be when arranged in ascending (or descending) order. Since both are 72, the element between them - the tenth element, and, thus, the median - must be 72.

The answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.

The range tells us the difference between the maximum and the minimum values, but provides no information about the median. The mode indicates the number that appears most frequently in the data set. While it is possible that the median is 17, it is impossible to determine the median from the data provided.

If we only assume Statement 1, then by examining these two cases, we show that we do not know whether the median changes.

Suppose the original set is

If we only know that one of the elements added is 50, then it is possible that the median does not change - this happens if the other element added is 70:

has median 60.

It is also possible that the median does change - this happens if the other element added is 50.

has median 55.

A similar argument can be used to demonstrate that Statement 2 is insufficient.

However, suppose we know both. 60 is the sixth-highest element of the old data set, and, since one element greater than 60 and one less than 60 are added, 60 is the seventh-highest element of the new thirteen-element set. We therefore know the median did not change.

The median of a data set with thirteen elements is the seventh-highest element; the median of a data set with fifteen elements is the eighth-highest.

The two statements together provide insufficient information, as we show with two examples.

The data set

has thirteen elements and median 75; after 30 and 40 are added, the set is

,

which has median 75.

By contrast, the data set

has thirteen elements and median 75; after 30 and 40 are added, the set is

which has median 74.

Both sets started out with median 75, but the addition of 30 and 40 changed one median and not the other.

Statement 1 is sufficient to prove that . If , then each of 6 and 7 occurs four times, more than any other element, making the set bimodal. If , then no element other than 6 occurs more than three times, giving the set only one mode.

Statement 2 is insufficient. The median of this data set, which has fifteen elements, is its eighth-greatest element. This happens if .

For example, if , the set becomes

If , the set becomes

The median of both sets is 7.

Assume Statement 1 alone. The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. By Statement 1, this number is , so the question is answered.

Statement 2 alone, however, gives that the mean is . It is possible that the mean and the median can be one and the same or two different numbers.

Case 1:

The mean is

making this consistent with Statement 2.

The median is the middle element, .

Case 2:

again, making this consistent with Statement 2.

The median is the middle element, .

Assume Statement 1 alone. By the Addition Property of Inequality, if

,

then can be added to each quantity to give an equivalent inequality:

.

By extension, the continued inequality in Statement 1 is equivalent to

.

A similar argument holds for Statement 2. By the Multiplication Property of Inequality, if

, then

and .

By extension, the continued inequality in Statement 2 is also equivalent to

.

The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. Therefore, it can be established from either statement alone that of the five, the median is .

Both statements can be shown to be equivalent to the continued inequality

by way of the Multiplication Property of Inequality. We will demonstrate this as follows:

Multiply each expression in

(Statement 1)

by :

.

Multiply each expression in

(Statement 2)

by :

The median of five numbers (an odd number) is the number in the middle when they are arranged in ascending order. Therefore, it can be established from either statement alone that of the five, the median is .

If we are given only that 15 occurs four times in the data set, it is possible that another number can occur up to six times; similarly, if we are given only that 16 occurs three times, it is possible that another number can occur up to seven times. Either way, the mode - the most frequently occurring data value - cannot be determined.

However, if we know both facts, then no other data value can occur more than three times, so 15 must be the mode.

Therefore, the answer is that both statements are sufficient, but not one alone.

Suppose but is unknown. The above set, with known elements ordered, is

The median is the arithmetic mean of the two middle elements, when the elements are ordered.

Regardless of the value of , the two middle elements must both be 84, making the median 84.

Now, suppose but is unknown. The above set, with known elements ordered, is

The median cannot be determined with certainty. For example, if , as stated before, the median is 84. But if , the middle elements are 84 and 87, making the median 85.5.

The answer is that Statement 1 alone is sufficent to answer the question, but Statement 2 alone is not.