Median - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

The median of seven consecutive integers is 129. What is the least integer?

Answer

The median of seven (an odd number) integers is the one in the middle when the numbers are arranged in ascending order; in this case, it is the fourth lowest. Since the seven integers are consecutive, the lowest integer is three less than the median, or .

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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

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

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

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 mean of the first ten prime numbers

(b) The median of the first ten prime numbers

Answer

The first ten primes form the data set:

$\left { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29\right }$

(a) Add these primes, and divide by :

$\left ( 2 + 3+5+ 7+ 11+13+17+19+ 23+29 \right )\div 10 = 129\div 10= 12.9$

(b) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements. These are and , so the median is

$\left (11 + 13 \right ) \div 2 = 12$ .

(a) is the greater quantity.

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Question

A data set with nine elements has mean 10 and median 10. A new data set is formed with these nine elements, plus two new elements, 2 and 18.

Which is the greater quantity?

(a) The mean of the new data set

(b) The median of the new data set

Answer

(a) The mean of nine elements is , so, if is their sum, $\frac{S}{9} = 10$ and $S = 9 \times 10 = 90$ . The sum of the new data set is . Since the new set has elements, its mean is $\frac{110}{11} = 10$ .

(b) The median of the nine elements is , so, when they are ranked, the fifth-highest element is . Since is less than and is greater than , when they are added to the set, is the sixth-highest of eleven elements, which is the median.

Therefore, both are equal to .

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Question

A data set has twelve elements; the mean and the median of the set are both 50.

A new data set is formed by increasing each element by 5. Which is the greater quantity?

(a) The mean of the new data set

(b) The median of the new data set

Answer

(a) Since each element of the old set is increased by 5, the sum of the elements is increased by $5 \times 12 = 60$ . This increases the mean by $60 \div 12 = 5$ , to 55.

(b) The median of the old set is the mean of the sixth- and seventh-highest elements. Since each element of the old set is increased by 5, these elements remain the sixth- and seventh-highest elements; their sum is increased by 10, and their mean is increased by 5, to 55.

The mean and the median of the new set 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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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

A data set has nine elements. Four of the elements are greater than 50; four are less than 50.

Which is the greater quantity?

(a) The median of the data set

(b) 50

Answer

If four of the elements are greater than 50 and four are less than 50, then the fifth-highest element, which is the median of a nine-element set, must be 50.

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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

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:

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

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:

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

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:

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

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:

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

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

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 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

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

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

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.

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

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

Give the median of this data set:

Round to the nearest tenth, if applicable.

Answer

The seven elements are already arranged in ascending order, so we need to look at the value occurring in the middle - that is, the fourth position. This element is 26.

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