How to find median - ISEE Upper Level: Quantitative Reasoning

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Question

Consider the data set

$\left \{ 10, 8, 12, 15, 16, 17, 19, 30, 22, 21\right \}$

Which is the greater quantity?

(a) The mean of this data set

(b) The median of this data set

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Answer

(a) The mean of this data set is the sum divided by 10:

$(10+8+12+15+16+17+19+30+22+21) \div 10$

$= 170 \div 10 = 17$

(b) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements:

The set, arranged, is $\left \{ 8, 10, 12, 15, 16, 17, 19, 21, 22, 30 \right \}$

The median is $(16 + 17) \div 2 = 16.5$

This makes (a) the greater quantity

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Question

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

$\left \{ 15,28,14,11,10,5 \right \}$

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Answer

If there are an even number of values, the median is the average of the two middle values of a set of data. In this question in order to find the median we should first put the numbers in order:

$\left \{ 5,10,11,14,15,28 \right \}$

So the median is:

$Median=\frac{11+14}{2}=\frac{25}{2}=12.5$

The 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:

$Mean = \frac{5+10+11+14+15+28}{6}=\frac{83}{6}\approx 13.83$

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:

$\left \{ 7,3,2,-1,14 \right \}$

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Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ -1,2,3,7,14 \right \}$

So the median is .

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:

$Mean = \frac{-1+2+3+7+14}{5}=\frac{25}{5}=5$

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

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Question

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

$\left \{ 28,25,31,34,37,43,40 \right \}$

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Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ 25,28,31,34,37,40,43 \right \}$

So the median is .

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.

$Mean = \frac{25+28+31+34+37+40+43}{7}=\frac{238}{7}=34$

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

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Question

Consider the following set of data:

$\left \{ 0,0.0001,0.001,0.01,0.1 \right \}$

$A=Median\ of\ the\ set$

$B=Mean\ of\ the\ set$

Compare and

Tap to reveal answer

Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ 0, 0.1, 0.01, 0.001, 0.0001 \right \}$

So the median is .

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.

$Mean = \frac{0+0.1+0.01+0.001+0.0001}{5}=\frac{0.1111}{5}=0.02222\approx 0.02$

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

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Question

In the following set of data the mean is . Find the median.

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

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Answer

First we need to find . We know that the mean of a set of data is given by the sum of the data, divided by the total number of values in the set. So we can write:

$Mean = \frac{2+4+N+8+10}{5}=\frac{24+N}{5}=10$

$\Rightarrow 24+N=50\Rightarrow N=50-24=26$

Now we have:

$\left \{ 2,4,26,8,10 \right \}$

If there are an odd number of values in a data set, the median is the middle value. First we need to put the numbers in order:

$\left \{ 2,4,8,10,26 \right \}$

So the median is .

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Question

In the following set of data the mean and the mode are equal. Find the median.

$\left \{ 1,3,3,3,N,4,8 \right \}$

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Answer

The mode of a set of data is the value which occurs most frequently which is in this problem (it is not dependent on in this problem). So the mean is also equal to .

We know that the mean of a set of data is given by the sum of the data, divided by the total number of values in the set. So we can write:

$Mean = \frac{1+3+3+3+N+4+8}{7}=\frac{22+N}{7}=3$

$\Rightarrow 22+N=21\Rightarrow N=21-22=-1$

So we have:

$\left \{ 1,3,3,3,-1,4,8 \right \}$

If there are an odd number of values in a data set, the median is the middle value. First we need to put the numbers in order:

$\left \{ -1,1,3,3,3,4,8 \right \}$

So the median is also .

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Question

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

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

Which is the greater quantity?

(A) The median of the set if and

(B) The median of the set if and

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Answer

The median of a data set with an even number of elements is the mean of the middle two numbers, assuming the elements are arranged in order; since the data set has eight elements, the median will be the mean of the fourth-lowest element and the fourth-highest element.

If and , the data set becomes $\left \{ 10, 15, 25, 25, 35, 35, 40, 50 \right \}$ ; if and , the data set becomes $\left \{ 10, 15, 25, 25, 35, 35, 40, 60 \right \}$ . In both data sets, the middle two elements are 25 and 35, making both medians equal to $\frac{25+35}{2} = 30$ . The quantities are equal.

Note: You don't need to do the last step and actually find the median. As long as you know that the two medians are the same, you've answered the question!

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Question

The following are the scores from a math test in a given classroom. What is the median score?

$\small \left \{ 70,89,67,77,92,83,68,75\right \}$

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Answer

To find the median you need to arrange the values in numerical order.

Starting with this:

Rearrange to look like this:

$\small \small \left \{ 67,68,70,75,77,83,89,92\right \}$

If there are an odd number of values, the median is the middle value. In this case there are 8 values so the median is the average (or mean) of the two middle values.

$\small (75+77)\div2=76$

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Question

Scores from a math test in a given classroom are as follows:

$\left \{ 80, 84, 86, 66, 57, 69, 92, 90 \right \}$

What is the median score?

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Answer

In order to find the median the data must first be ordered. So we have:

$\left \{ 57,66,69,80,84,86,90,92 \right \}$

In this problem the number of values is even. We know that when the number of values is even, the median is the mean of the two middle values. So we get:

$Median=\frac{80+84}{2}=82$

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Question

Heights of a group of students in a high school are as follows (heights are given in ):

$\left \{ 176,180,181,182,168,174,169,188,165 \right \}$

Find the median height.

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Answer

In order to find the median the data must first be ordered. So we have:

$\left \{ 165,168,169,174,176,180,181,182,188 \right \}$

When the number of values is odd, the median is the single middle value. In this problem we have nine values. So the median is th value which is .

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Question

Find the median in the following set of data:

$\left \{ 2,3,4,1,2,4,3,7,3,4 \right \}$

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Answer

In order to find the median, the data must first be ordered. So we should write:

$\left \{ 1,2,2,3,3,3,4,4,4,7 \right \}$

When the number of values is even, the median is the mean of the two middle values. In this problem we have values, so the median would be the mean of the and values:

$Median=\frac{3+3}{2}=3$

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Question

Consider the data set

$\left \{ 10, 8, 12, 15, 16, 17, 19, 30, 22, 21\right \}$

Which is the greater quantity?

(a) The mean of this data set

(b) The median of this data set

Tap to reveal answer

Answer

(a) The mean of this data set is the sum divided by 10:

$(10+8+12+15+16+17+19+30+22+21) \div 10$

$= 170 \div 10 = 17$

(b) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements:

The set, arranged, is $\left \{ 8, 10, 12, 15, 16, 17, 19, 21, 22, 30 \right \}$

The median is $(16 + 17) \div 2 = 16.5$

This makes (a) the greater quantity

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Question

Consider the data set $\left \{ -1, 2, -3, 4, -5, 6, -7\right \}$

Which is the greater quantity?

(a) The median of the data set

(b) The mean of the data set

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Answer

(a) Arrange the elements in ascending order:

$\left \{ -7, -5, -3, -1, 2,4,6 \right \}$

The median is the middle element, which is .

(b) Add the elements and divide by 7:

$[ (-7) + (-5) + (-3) + (-1) + 2 + 4 + 6 ] \div 7 = -4 \div 7 = - \frac{4}{7}$

$- \frac{4}{7} > -1$ , so the mean is greater.

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Question

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

$\left \{ 15,28,14,11,10,5 \right \}$

Tap to reveal answer

Answer

If there are an even number of values, the median is the average of the two middle values of a set of data. In this question in order to find the median we should first put the numbers in order:

$\left \{ 5,10,11,14,15,28 \right \}$

So the median is:

$Median=\frac{11+14}{2}=\frac{25}{2}=12.5$

The 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:

$Mean = \frac{5+10+11+14+15+28}{6}=\frac{83}{6}\approx 13.83$

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:

$\left \{ 7,3,2,-1,14 \right \}$

Tap to reveal answer

Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ -1,2,3,7,14 \right \}$

So the median is .

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:

$Mean = \frac{-1+2+3+7+14}{5}=\frac{25}{5}=5$

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

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Question

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

$\left \{ 28,25,31,34,37,43,40 \right \}$

Tap to reveal answer

Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ 25,28,31,34,37,40,43 \right \}$

So the median is .

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.

$Mean = \frac{25+28+31+34+37+40+43}{7}=\frac{238}{7}=34$

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

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Question

Consider the following set of data:

$\left \{ 0,0.0001,0.001,0.01,0.1 \right \}$

$A=Median\ of\ the\ set$

$B=Mean\ of\ the\ set$

Compare and

Tap to reveal answer

Answer

The median is the middle value of a set of data containing an odd number of values. In this question in order to find the median we should first put the numbers in order:

$\left \{ 0, 0.1, 0.01, 0.001, 0.0001 \right \}$

So the median is .

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.

$Mean = \frac{0+0.1+0.01+0.001+0.0001}{5}=\frac{0.1111}{5}=0.02222\approx 0.02$

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

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Question

In the following set of data the mean is . Find the median.

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

Tap to reveal answer

Answer

First we need to find . We know that the mean of a set of data is given by the sum of the data, divided by the total number of values in the set. So we can write:

$Mean = \frac{2+4+N+8+10}{5}=\frac{24+N}{5}=10$

$\Rightarrow 24+N=50\Rightarrow N=50-24=26$

Now we have:

$\left \{ 2,4,26,8,10 \right \}$

If there are an odd number of values in a data set, the median is the middle value. First we need to put the numbers in order:

$\left \{ 2,4,8,10,26 \right \}$

So the median is .

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Question

In the following set of data the mean and the mode are equal. Find the median.

$\left \{ 1,3,3,3,N,4,8 \right \}$

Tap to reveal answer

Answer

The mode of a set of data is the value which occurs most frequently which is in this problem (it is not dependent on in this problem). So the mean is also equal to .

We know that the mean of a set of data is given by the sum of the data, divided by the total number of values in the set. So we can write:

$Mean = \frac{1+3+3+3+N+4+8}{7}=\frac{22+N}{7}=3$

$\Rightarrow 22+N=21\Rightarrow N=21-22=-1$

So we have:

$\left \{ 1,3,3,3,-1,4,8 \right \}$

If there are an odd number of values in a data set, the median is the middle value. First we need to put the numbers in order:

$\left \{ -1,1,3,3,3,4,8 \right \}$

So the median is also .

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