How to find median - ISEE Upper Level Quantitative Reasoning

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}, \frac{1}{7}\right \}$

(b) $\frac{1}{4}$

Answer

The median of a data set with seven elements is its fourth-greatest element, which here is $\frac{1}{4}$ .

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \right \}$

(b) $\frac{3}{4}$

Answer

The median of a data set with five elements is its third-greatest element, which here is $\frac{3}{4}$ .

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Question

In the following set of numbers compare the mean and the median of the set:

$\left \{ 1,2,4,8,10 \right \}$

Answer

The median is the middle value of a set of data containing an odd number of values which is in this set of numbers.

Mean of a data set $x_{1},x_{2}, ..., x_{n}$ is the sum of the data set values divided by the number of data:

$Mean=\frac{x_{1}+x_{2}+...+x_{n}}{n}$

So we have:

$Mean = \frac{1+2+4+8+10}{5}=\frac{25}{5}=5$

So the mean of the set is greater than the median of the set.

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Question

In the following set of numbers compare the mean and the median of the set:

$\left \{ 1,3,4,115,120,185,224 \right \}$

Answer

The median is the middle value of a set of data containing an odd number of values which is in this problem.

Mean of a data set $x_{1},x_{2}, ..., x_{n}$ is the sum of the data set values divided by the number of data:

$Mean=\frac{x_{1}+x_{2}+...+x_{n}}{n}$

So the mean of the set is equal to the median of the set.

$Mean = \frac{1+3+4+115+120+185+224}{7}=\frac{652}{7}\approx 93.14$

So the median of the set is greater than the mean of that.

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Question

A data set has six known quantities and two unknown positive quantities, as follows:

$\left \{ 15, 25, 25, 35, 35, 40, M, N \right \}$

It is known, however, that

Which is the greater quantity?

(A) The mean of the data set

(B) The median of the data set

Answer

If , then the mean of the data set is the sum of the eight elements divided by eight:

$\left ( 15 + 25+ 25+ 35+ 35+40+ M + N \right ) \div 8$

$=\left ( 175+ 70 \right ) \div 8 = 245\div 8 = 30.625$

The median of the data set, however, depends on the values of and , as can be demonstrated using two cases.

Case 1:

The data set is then

$\left \{ 15, 25, 25, 35, 35, 35, 35, 40 \right \}$

and the median is the mean of the two middle values. Since both middle values are 35, the median is 35.

Case 2: and

The data set is then

$\left \{ 15,15, 25, 25, 35, 35, 40,55 \right \}$

and the median is the mean of the two middle values. They are 25 and 35, so the median is $\left (25+35 \right ) \div 2 = 30$

In the first case, the median is greater than the mean; in the second case, the mean is greater than the median. Therefore, the information is insufficient.

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Question

Consider the data set

$\left \{ 24,61, 27, 57, 34, 37, 63, 50, N\right \}$ .

For what value(s) of would this set have median ?

Answer

Arrange the eight known values from least to greatest.

$\left \{ 24, 27, 34, 37, 50, 57,61, 63\right \}$

For to be the median of the nine elements, it muct be the fifth-greatest, This happens if $N \geq 50$ .

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Question

Consider the data set: $\left \{ 12, 15, 18, 20, 20, 21, 23, 26, N\right \}$

where is not known.

What are the possible values of the median of this set?

Answer

The median of this nine-element set is its fifth-highest element. Of the eight known elements, the fourth-highest and fifth-highest elements are both 20. Regardless of the value of , 20 is the fifth-highest element of the nine.

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}, \frac{1}{7}\right \}$

(b) $\frac{1}{4}$

Answer

The median of a data set with seven elements is its fourth-greatest element, which here is $\frac{1}{4}$ .

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \right \}$

(b) $\frac{3}{4}$

Answer

The median of a data set with five elements is its third-greatest element, which here is $\frac{3}{4}$ .

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Question

In the following set of numbers compare the mean and the median of the set:

$\left \{ 1,2,4,8,10 \right \}$

Answer

The median is the middle value of a set of data containing an odd number of values which is in this set of numbers.

Mean of a data set $x_{1},x_{2}, ..., x_{n}$ is the sum of the data set values divided by the number of data:

$Mean=\frac{x_{1}+x_{2}+...+x_{n}}{n}$

So we have:

$Mean = \frac{1+2+4+8+10}{5}=\frac{25}{5}=5$

So the mean of the set is greater than the median of the set.

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Question

In the following set of numbers compare the mean and the median of the set:

$\left \{ 1,3,4,115,120,185,224 \right \}$

Answer

The median is the middle value of a set of data containing an odd number of values which is in this problem.

Mean of a data set $x_{1},x_{2}, ..., x_{n}$ is the sum of the data set values divided by the number of data:

$Mean=\frac{x_{1}+x_{2}+...+x_{n}}{n}$

So the mean of the set is equal to the median of the set.

$Mean = \frac{1+3+4+115+120+185+224}{7}=\frac{652}{7}\approx 93.14$

So the median of the set is greater than the mean of that.

Compare your answer with the correct one above

Question

A data set has six known quantities and two unknown positive quantities, as follows:

$\left \{ 15, 25, 25, 35, 35, 40, M, N \right \}$

It is known, however, that

Which is the greater quantity?

(A) The mean of the data set

(B) The median of the data set

Answer

If , then the mean of the data set is the sum of the eight elements divided by eight:

$\left ( 15 + 25+ 25+ 35+ 35+40+ M + N \right ) \div 8$

$=\left ( 175+ 70 \right ) \div 8 = 245\div 8 = 30.625$

The median of the data set, however, depends on the values of and , as can be demonstrated using two cases.

Case 1:

The data set is then

$\left \{ 15, 25, 25, 35, 35, 35, 35, 40 \right \}$

and the median is the mean of the two middle values. Since both middle values are 35, the median is 35.

Case 2: and

The data set is then

$\left \{ 15,15, 25, 25, 35, 35, 40,55 \right \}$

and the median is the mean of the two middle values. They are 25 and 35, so the median is $\left (25+35 \right ) \div 2 = 30$

In the first case, the median is greater than the mean; in the second case, the mean is greater than the median. Therefore, the information is insufficient.

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Question

Consider the data set

$\left \{ 24,61, 27, 57, 34, 37, 63, 50, N\right \}$ .

For what value(s) of would this set have median ?

Answer

Arrange the eight known values from least to greatest.

$\left \{ 24, 27, 34, 37, 50, 57,61, 63\right \}$

For to be the median of the nine elements, it muct be the fifth-greatest, This happens if $N \geq 50$ .

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Question

Consider the data set: $\left \{ 12, 15, 18, 20, 20, 21, 23, 26, N\right \}$

where is not known.

What are the possible values of the median of this set?

Answer

The median of this nine-element set is its fifth-highest element. Of the eight known elements, the fourth-highest and fifth-highest elements are both 20. Regardless of the value of , 20 is the fifth-highest element of the nine.

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \right \}$

(b) $\frac{3}{4}$

Answer

The median of a data set with five elements is its third-greatest element, which here is $\frac{3}{4}$ .

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Question

In the following set of numbers compare the mean and the median of the set:

$\left \{ 1,3,4,115,120,185,224 \right \}$

Answer

The median is the middle value of a set of data containing an odd number of values which is in this problem.

Mean of a data set $x_{1},x_{2}, ..., x_{n}$ is the sum of the data set values divided by the number of data:

$Mean=\frac{x_{1}+x_{2}+...+x_{n}}{n}$

So the mean of the set is equal to the median of the set.

$Mean = \frac{1+3+4+115+120+185+224}{7}=\frac{652}{7}\approx 93.14$

So the median of the set is greater than the mean of that.

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Question

Consider the data set: $\left \{ 12, 15, 18, 20, 20, 21, 23, 26, N\right \}$

where is not known.

What are the possible values of the median of this set?

Answer

The median of this nine-element set is its fifth-highest element. Of the eight known elements, the fourth-highest and fifth-highest elements are both 20. Regardless of the value of , 20 is the fifth-highest element of the nine.

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Question

Consider the data set $\left \{ 1, 3, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 10\right \}$ .

Which is the greater quantity?

(a) The mean of this set

(b) The median of this set

Answer

(a) The mean of this set is $(1+3+4+5+5+5+6+7+7+7+8+9+10) \div 13 = 77 \div 13 \approx 5.92$ .

(b) Since there are elements, the median of this set is the seventh-highest number, which is .

Therefore (b) is the greater quantity.

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{8, 8, 9, 9, 10, 10, 11, 11,12, 12\right \}$

(b) The median of the data set $\left \{1,1, 2,2, 10, 10, 11, 11,12, 12\right \}$

Answer

Each data set has ten elements, so the median in each case is the arithmetic mean of the fifth-highest and sixth-highest elements. In each data set, these elements are 10 and 10, so the median of each set is 10. Therefore, both quantities are equal.

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Question

Which is the greater quantity?

(a) The median of the data set $\left \{ 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}, \frac{1}{7}\right \}$

(b) $\frac{1}{4}$

Answer

The median of a data set with seven elements is its fourth-greatest element, which here is $\frac{1}{4}$ .

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