Descriptive Statistics - GMAT Quantitative

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.

In this case, average is also the middle value.

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,

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 .

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.

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.

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 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, .

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.

If we know that , then the set is known to have one 24, five 26's, three 27's, two 28's, and one 29. The only way the set can have two modes is for and ; this makes 27 occur five times, just as frequently as 26.

If we know , however, the set is known to have one 24, four 26's, four 27's, two 28's, and one 29. There are two ways for the set to have two modes (26 and 27): for and , or for and .

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

The median of a data set with an even number of elements is the arithmetic mean of the two elements that fall in the middle when the elements are arranged in ascending order. These two elements are both 6, so 6 is the median.

The mode of a data set is the element that occurs most frequently. Since 6 appears thre times, 7 appears two times, and all other elements appear once each, the mode is 6.

Therefore, the median and the mode are equal.

The data set has four 6's and no more than two of any other element - and there cannot be more than four of any other element regardless of the value of - so 6 must be one of the modes. For the set to be bimodal, there must be four of another element. Since occurs twice, it must be set to a number known to occur exactly two other times. There are, however, two choices, 5 and 7, so Statement 1 is insufficient.

Statement 2 is sufficient, as seen below:

If , as given in Statement 1, the set can have one mode (the other four numbers are different from each other and from and ), two modes (for example, ) or three modes (for example, ) . Therefore, Statement 1 alone is not enough.

If none of , , , or are equal to each other, as given in Statement 2, the set can have one mode (for example, , the other numbers are different), two modes (, and are different), or no modes (all six different numbers).

If both statements are true, however, there are two possibilities - , with the other four elements being different, or , with one number being the same and the other three different. Either way, the set is known to have one mode.