Summarizing numerical data sets - HiSET

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Question

Consider the data set

$\left { 17, 21, 27, 29, 29, 29 ,35, 37, x, x \right }$ ,

where is a prime integer.

How many possible values of make the set bimodal?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 29 already occurs three times in the data set. For the set to be bimodal, must be equal to one of the values that occurs once - 17, 21, 27, 35, or 37. Since it is given that is prime - having only two factors, 1 and itself - can only be either of 17 and 37, the other three values having other factors.

The correct response is two.

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Question

Consider the following data set:

$\left { 1, 4, 5, 8, 8, 9, 10, 11, x, x \right }$

Which of the following gives the arithmetic mean of the set in terms of ?

Answer

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

$\mu = \frac{ 1+4+ 5+ 8+ 8+ 9+ 10+11+ x+ x }{10}$

Simplify the numerator by combining the like terms:

$\mu = \frac{56+2x}{10}$

Now, split the fraction, and reduce to lowest terms:

$\mu = \frac{56 }{10}+ \frac{ 2x}{10}$

$= \frac{28}{5}+ \frac{ x}{5}$

$= \frac{ 1}{5} x+ \frac{28}{5}$ ,

the correct response.

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Question

Consider the following data set:

$\left { 1, 4, 5, 8, 8, 9, 10, 11, x, x \right }$

If , which is true of the mode or modes of the data set?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two or more values tie for most frequently occurring value, then the set has multiple modes.

We show that the question of the modes of the data set cannot be answered with certainty as follows:

Suppose assumes a value in the data set - for example, let . the data set is

$\left { 1,1, 1, 4, 5, 8, 8, 9, 10, 11 \right }$ ,

in which the most frequently occurring value is 1, which appears . This makes 1, , the mode.

Now, suppose assumes a value not already in the data set; for example, let . The data set becomes

$\left { 1,2,2 , 4, 5, 8, 8, 9, 10, 11 \right }$ ,

in which two values, 2 (the ) and 8 each occur twice.

Therefore, from the information given, the mode(s) cannot be determined with any certainty.

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Question

Consider the data set

$\left { 10, 16, 16, 18, 21, 25, x, 2x \right }$ ,

where is a prime integer.

How many possible values of make the set bimodal?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

$2x = 10 \Rightarrow x= 5$

$2x = 18 \Rightarrow x= 9$

$2x = 21 \Rightarrow x= 10.5$

$2x = 25 \Rightarrow x= 12.5$

Only in one case does it hold that is a prime integer.

The correct response is one - .

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Question

Consider the following data set:

$\left { 4, 5, 5, 6, 8, 9, 10, 10, x \right }$

where is an integer from 1 to 10 inclusive.

How many possible values of make 8 the median of the set?

Answer

The median of a set of nine data values - an odd number - is the value that appears in the middle when the values are ranked. For 8 to be the median, 8 must be in the middle - that is, four values must appear before 8, and four values must appear after 8.

In the given data set, four values - 4, 5, 5, 6 - are known to be appear before 8, so must appear after 8. Since is an integer from 1 to 10, can only be 8, 9, or 10. This makes three the correct response.

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Question

Consider the following data set:

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

where is an integer from 1 to 10 inclusive.

How many possible values of make 5 the median of the set?

Answer

The median of a set of eleven data values - an odd number - is the value that appears in the middle when the values are ranked. For 5 to be the median, 5 must be in the middle - that is, five values must appear before 5, and five values must appear after 8.

We can answer this question by looking at three cases.

Case 1:

Without loss of generality, assume ; this reasoning holds for any lesser value of . The data set becomes

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

and the median is 4.

Case 2:

The data set becomes

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

The middle value - the median - is 5.

Case 3:

Without loss of generality, assume ; this reasoning holds for any greater value of . The data set becomes

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

Again, the median is 5.

Therefore, we can set equal to 5, 6, 7, 8, 9, or 10 - any of six different values - and make the median of the set 5.

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Question

Consider the data set:

$\left { 12, 15, 17, 19, 19, 19, 22, 23, 23 \right }$

Which is true of the arithmetic mean $\mu$ , the median , and the midrange ?

Answer

The mean of a data set is equal to the sum of the entries divided by the number of entries. There are nine entries in the set, so

$\mu =( 12+15+ 17+ 19+ 19+ 19+ 22+ 23+ 23 ) \div 9 = 169 \div 9 \approx 18.8$

The median of a data set with an odd number of values is the value of the entry in the middle when the values are arranged in ascending order:

$\left { 12, 15, 17, 19, \textbf{19}, 19, 22, 23, 23 \right }$

This value is .

The midrange of a data set is the arithmetic mean of the least and greatest elements in the set. This element is

$m = (12+23)\div 2 = 17.5$

The correct response is that

$m < \mu < M$ .

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Question

Consider the following data set:

$\left { 4, 5, 5, 6, 8, 9, 10, 10, x, x \right }$

where is an integer from 1 to 10 inclusive.

How many possible values of make the median of the set?

Answer

The median of a set of ten data values - and even number - is the arithmetic mean of the two values that appear in the middle when the values are ranked. For to be the median, it would have to hold that either the two middle values are both . The data set, in ascending order, would be as follows:

$\left { 4, 5, 5, 6, x, x, 8, 9, 10, 10 \right }$ .

must be between 6 and 8 inclusive. Since it is given that is an integer, there are three possibilities: 6, 7, and 8.

Three is the correct response.

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Question

Janice will take twelve tests in her political science class, worth one hundred points each. Her score for the term will be the arithmetic mean of the best ten.

She has taken eleven tests already; her scores, in order, are:

74, 79, 60, 77, 54, 80, 81, 60, 66, 68, 71

How high will Janice have to score on the twelfth test in order to get a "C" in the course, which is defined to be a mean of 70 points?

Note: Assume that no extra credit is given on any test.

Answer

First, we test to see if she is already assured of a 70 average. The worst-case scenario for Janice is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

$\frac{74+ 79+ 60+ 77 + 80+ 81+60+ 66+ 68+ 71}{10} = \frac{716}{10} = 71.6$

Janice is already assured of a mean of at least 70, since any score can only raise her grade.

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Question

Consider the data set:

$\left { 10, 12,13, 15, 15, 19, 19, 19, 20 , 20\right }$

Which is true of the arithmetic mean $\mu$ , the median , and the midrange ?

Answer

The mean of a data set is equal to the sum of the entries divided by the number of entries. There are ten entries in the set, so

$\mu = \frac{10+12+13+ 15+ 15+ 19+ 19+ 19+ 20 + 20 }{10 } = \frac{162}{10}= 16.2$

The median of a data set with an even number of values is the arithmetic mean of the values of the two entries in the middle when the values are arranged in ascending order:

$\left { 10, 12,13, 15, \textbf{15, 19}, 19, 19, 20 , 20\right }$

This value is $M = \frac{15+19}{2} = \frac{34}{2}= 17$ .

The midrange of a data set is the arithmetic mean of the least and greatest elements in the set. This element is

$m = \frac{10 + 20}{2} = 15$

The correct response is that

$m < \mu < M$ .

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Question

Donna will take twelve tests in her biology class, worth one hundred points each. Her score for the term will be the arithmetic mean of the best ten.

She has taken eleven tests already; her scores, in order, are:

74, 79, 60, 77, 54, 80, 81, 60, 66, 68, 71

How high will Donna have to score on the twelfth test in order to get a "B" in the course, which is defined to be a mean of 80 points?

Note: Assume that no extra credit is given on any test.

Answer

First, we test to see if she is already assured of a 80 average. The worst-case scenario for Donna is that she will score 0 points on the twelfth test,. If this happens, her grade will be the mean of the ten best tests so far. She will drop the 0 and the fifth score, 54, so her average will be the sum of the other ten tests divided by 10:

$\frac{74+ 79+ 60+ 77 + 80+ 81+60+ 66+ 68+ 71}{10} = \frac{716}{10} = 71.6$

Donna has not yet attained her 80. Now, we test to determine what her twelfth test score must be. If she scores 60 or less she will drop this score as well as the 54, so we will assume that she scores more than 60. Calling this score , the mean of her scores will be the the sum of her best nine scores thus far and this unknown score , divided by 10. The mean should be greater than or equal to 80, so we can set up and solve for in this inequality:

$\frac{74+ 79 + 77 + 80+ 81+60+ 66+ 68+ 71+x }{10} \ge 80$

Note that the third (60) and fifth (54) scores have been omitted. Add the known scores to get

$\frac{ x+ 656 }{10} \ge 80$

Multiply both sides by 10:

$\frac{ x+ 656 }{10} \cdot 10 \ge 80 \cdot 10$

$x+ 656 \ge 800$

Subtract 656 from both sides:

$x+ 656 - 656 \ge 800 - 656$

$x \ge 144$

Donna would have to score 144 or more on the twelfth test - an impossible feat. She cannot attain an average of 80 or greater.

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Question

Which of the following is an example of a set with no mode?

Answer

The mode of a data set is the value that appears in the set the most frequently (or a mode is one of several such values). A set is considered to have no mode if and only if no value is repeated. Of the four data sets given, $\left { 70, 72, 74, 76, 77, 81, 83, 85, 89, 92 \right }$ is the only one that fits this definition.

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Question

Consider the following data set:

$\left { 1, 4, 5, 8, 8, 9, 10, 11, x, x \right }$

Which of the following gives the arithmetic mean of the set in terms of ?

Answer

The arithmetic mean of a data set is the sum of the items in the set divided by the number of items. There are ten items, so the mean is

$\mu = \frac{ 1+4+ 5+ 8+ 8+ 9+ 10+11+ x+ x }{10}$

Simplify the numerator by combining the like terms:

$\mu = \frac{56+2x}{10}$

Now, split the fraction, and reduce to lowest terms:

$\mu = \frac{56 }{10}+ \frac{ 2x}{10}$

$= \frac{28}{5}+ \frac{ x}{5}$

$= \frac{ 1}{5} x+ \frac{28}{5}$ ,

the correct response.

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Question

Consider the following data set:

$\left { 1, 4, 5, 8, 8, 9, 10, 11, x, x \right }$

If , which is true of the mode or modes of the data set?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two or more values tie for most frequently occurring value, then the set has multiple modes.

We show that the question of the modes of the data set cannot be answered with certainty as follows:

Suppose assumes a value in the data set - for example, let . the data set is

$\left { 1,1, 1, 4, 5, 8, 8, 9, 10, 11 \right }$ ,

in which the most frequently occurring value is 1, which appears . This makes 1, , the mode.

Now, suppose assumes a value not already in the data set; for example, let . The data set becomes

$\left { 1,2,2 , 4, 5, 8, 8, 9, 10, 11 \right }$ ,

in which two values, 2 (the ) and 8 each occur twice.

Therefore, from the information given, the mode(s) cannot be determined with any certainty.

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Question

Consider the data set

$\left { 10, 16, 16, 18, 21, 25, x, 2x \right }$ ,

where is a prime integer.

How many possible values of make the set bimodal?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 16 already occurs twice in the data set. For the set to be bimodal, one of the following must happen:

Case 1: , which is not possible, since 0 is not considered prime or composite.

Case 2: must be equal to one of the other four known values, 10, 18, 21, or 25. however, is given to be prime - that is, to have only two factors, 1 and itself. Each of 10, 18, 21, and 25 is composite, having factors not equal to 1 or itself, so cannot assume any of these values.

Case 3: must be equal to one of the other four known values, 10, 18, 21, or 25. Then must be equal to half of the selected number:

$2x = 10 \Rightarrow x= 5$

$2x = 18 \Rightarrow x= 9$

$2x = 21 \Rightarrow x= 10.5$

$2x = 25 \Rightarrow x= 12.5$

Only in one case does it hold that is a prime integer.

The correct response is one - .

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Question

Consider the data set

$\left { 17, 21, 27, 29, 29, 29 ,35, 37, x, x \right }$ ,

where is a prime integer.

How many possible values of make the set bimodal?

Answer

The mode of a data set is the value that occurs most frequently in the set. If two values tie for most frequently occurring value, then the set has two modes - it is bimodal.

The value 29 already occurs three times in the data set. For the set to be bimodal, must be equal to one of the values that occurs once - 17, 21, 27, 35, or 37. Since it is given that is prime - having only two factors, 1 and itself - can only be either of 17 and 37, the other three values having other factors.

The correct response is two.

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Question

Consider the following data set:

$\left { 4, 5, 5, 6, 8, 9, 10, 10, x \right }$

where is an integer from 1 to 10 inclusive.

How many possible values of make 8 the median of the set?

Answer

The median of a set of nine data values - an odd number - is the value that appears in the middle when the values are ranked. For 8 to be the median, 8 must be in the middle - that is, four values must appear before 8, and four values must appear after 8.

In the given data set, four values - 4, 5, 5, 6 - are known to be appear before 8, so must appear after 8. Since is an integer from 1 to 10, can only be 8, 9, or 10. This makes three the correct response.

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Question

Consider the following data set:

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

where is an integer from 1 to 10 inclusive.

How many possible values of make 5 the median of the set?

Answer

The median of a set of eleven data values - an odd number - is the value that appears in the middle when the values are ranked. For 5 to be the median, 5 must be in the middle - that is, five values must appear before 5, and five values must appear after 8.

We can answer this question by looking at three cases.

Case 1:

Without loss of generality, assume ; this reasoning holds for any lesser value of . The data set becomes

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

and the median is 4.

Case 2:

The data set becomes

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

The middle value - the median - is 5.

Case 3:

Without loss of generality, assume ; this reasoning holds for any greater value of . The data set becomes

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

Again, the median is 5.

Therefore, we can set equal to 5, 6, 7, 8, 9, or 10 - any of six different values - and make the median of the set 5.

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Question

Consider the data set:

$\left { 12, 15, 17, 19, 19, 19, 22, 23, 23 \right }$

Which is true of the arithmetic mean $\mu$ , the median , and the midrange ?

Answer

The mean of a data set is equal to the sum of the entries divided by the number of entries. There are nine entries in the set, so

$\mu =( 12+15+ 17+ 19+ 19+ 19+ 22+ 23+ 23 ) \div 9 = 169 \div 9 \approx 18.8$

The median of a data set with an odd number of values is the value of the entry in the middle when the values are arranged in ascending order:

$\left { 12, 15, 17, 19, \textbf{19}, 19, 22, 23, 23 \right }$

This value is .

The midrange of a data set is the arithmetic mean of the least and greatest elements in the set. This element is

$m = (12+23)\div 2 = 17.5$

The correct response is that

$m < \mu < M$ .

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Question

Consider the following data set:

$\left { 4, 5, 5, 6, 8, 9, 10, 10, x, x \right }$

where is an integer from 1 to 10 inclusive.

How many possible values of make the median of the set?

Answer

The median of a set of ten data values - and even number - is the arithmetic mean of the two values that appear in the middle when the values are ranked. For to be the median, it would have to hold that either the two middle values are both . The data set, in ascending order, would be as follows:

$\left { 4, 5, 5, 6, x, x, 8, 9, 10, 10 \right }$ .

must be between 6 and 8 inclusive. Since it is given that is an integer, there are three possibilities: 6, 7, and 8.

Three is the correct response.

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