Descriptive Statistics - GMAT Quantitative

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.

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.

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.

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.

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.

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.

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.

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.

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.

Knowing the mean of the set is enough to calculate :

Knowing the median is enough to deduce . Since there are ten elements - an even number - the median is the arithmetic mean of the fifth and sixth highest elements. If , those two elements are 47 and 49, making the median 48. If , those two elements are both 49, making the median 49. This forces to be 48, making the median 48.5.

Therefore, either statement alone is sufficient to answer the question.

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.

Knowing the mean of the set is enough to calculate :

Knowing the median is enough to deduce . Since there are ten elements - an even number - the median is the arithmetic mean of the fifth and sixth highest elements. If , those two elements are 47 and 49, making the median 48. If , those two elements are both 49, making the median 49. This forces to be 48, making the median 48.5.

Therefore, either statement alone is sufficient to answer the question.

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.

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.

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.

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 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: