Data Analysis and Probability - 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) $\frac{1}{4}$

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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) $\frac{3}{4}$

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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:

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

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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

Consider the following data set:

$\left \{ 67, 68, 70,70, 70, 72, 74, 74, 79\right \}$

Which of these numbers is greater than the others?

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Answer

The median of the set is the fifth-highest value, which is 70; this is also the mode, being the most commonly occurring element.

The mean is the sum of the elements divided by the number of them. This is

$\frac{67 +68+ 70+70+ 70+ 72+ 74+ 74+ 79}{9} \approx 71.6$

The midrange is the mean of the least and greatest elements, This is $\frac{67+79}{2} =73$

The midrange is the greatest of the four.

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Question

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

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

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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

Which is true about the mean and the median of the following data set:

$\left \{ 78, 92, 80, 79, 74\right \}$

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Answer

The mean of the data set is the sum of the elements divided by 5:

$\frac{78 + 92 + 80 + 79 + 74 }{5} = \frac{403 }{5} = 80.6$

The median of the data set is the middle value when the values are arranged in ascending order, which here is the third-highest value. This is 79.

The mean exceeds the median by

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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

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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 following set of numbers:

$\left \{ 20,35,7,12,73,12,18,31\right \}$

Quantity A: Median of the set

Quantity B: Mean of the set

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Answer

The median of the set of numbers is determined by arranging the numbers in numerical order and finding the middle number. In this case there are two middle numbers, and , so we find the average of those numbers, which gives us .

The mean is found by dividing the sum of elements by the number of elements in the set:

$\frac{20+35+7+12+73+12+18+31}{8}=26$

Quantity B is larger.

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Question

Eight judges scored Donna's gymnastics routine as follows:

$\left \{ 8, 6, 7, 7, 9, 10, 5, 9\right \}$

The highest and lowest scores are disregarded to account for possible bias on the part of the judges. Donna's score is the mean of the scores that remain.

What is Donna's score, to the nearest tenth?

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Answer

The highest and lowest scores are 10 and 5, so these are disregarded; the remaining six are averaged by dividing their sum by six:

$\left ( 8+ 6+ 7+7+ 9 + 9 \right ) \div 6 = 46 \div 6 \approx 7.7$

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Question

Consider the data set $\left \{10, 12, 13, 13, 13, 15, 16, 17, 19, 20\right \}$

Which is the greater quantity?

(a) The median of the data set

(b) The mode of the data set

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Answer

(a) The median of a data set with ten elements is the arithmetic mean of the fifth-highest and sixth-highest elements. These elements are 13 and 15, so the median is $\frac{13 + 15}{2} = 14$

(b) The mode of a data set is the element that occurs most frequently. Since 13 is the only repeated element, it is the mode.

This makes (a) the greater quantity.

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Question

Compare the mean and the mode in the following set of data:

$\left \{ 5,8,11,14,17,20,28,14,14 \right \}$

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Answer

The mode of a set of data is the value which occurs most frequently, which in this case is .

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:

$Mean = \frac{5+8+11+14+17+20+28+14+14}{9}=\frac{131}{9}\approx 14.56$

So the mean of is greater than the mode.

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Question

Compare the mean and the mode in the following set of data:

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

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Answer

The mode of a set of data is the value which occurs most frequently, which in this case is $\frac{1}{2}=0.5$ .

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:

$Mean = \frac{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{5}{6}+\frac{1}{2}}{5}=\frac{\frac{6+8+9+10+6}{12}}{5}=\frac{39}{60}=\frac{13}{20}=0.65$

So the mean is greater than the mode.

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Question

Consider the data set $\left \{ 20, 30, 30, 40, 40, 40, 40, 50, 50, 60, \square \right \}$ .

Which of the following elements replaces the box to make the data set bimodal?

(A)

(B)

(C)

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Answer

At current, 40 appears four times, more than any other value. For a set to be bimodal, it needs to have two modes; that is, another value would have to appear four times as well. Regardless of whether 30, 50, 60, or any other value replaces the box, this is impossible. The set cannot be made bimodal by adding one element.

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Question

Consider the set of numbers: $\left \{ 2,4,5,5,6,6,6,10,12\right \}$

Quantity A: The sum of the median and mode of the set

Quantity B: The range of the set

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Answer

Quantity A: The median (middle number) is , and the mode (most common number) is , so the sum of the two numbers is .

Quantity B: The range is the smallest number subtracted from the largest number, which is .

Quantity A is larger.

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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 ?

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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?

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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

Give the mode(s) of the data set

$\left \{ 1, 3, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 10\right \}$

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Answer

The mode of a data set is the element that appears the most frequently. When two numbers both appear most frequently, there are two modes:

$5 \textrm{ and }7$

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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) $\frac{1}{4}$

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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) $\frac{3}{4}$

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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:

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

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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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