Measures of Central Tendency (Mean, Median, Mode) Practice Questions — GRE General Test — Page 2

Question 21Question →

A retail analyst recorded the number of online orders fulfilled per day over a period of $9$ days. The data points, listed in non-decreasing order, are:

3, 5, 8, x, y, 17, 21, 24, z

The median of the $9$ daily order counts is $14$ , the arithmetic mean is $16$ , and the dataset has a unique mode of $8$ . If $z$ represents the highest number of orders fulfilled in a single day, what is the value of $z$ ?

28

38

44

52

54

Show answer & explanation

Answer: 44

Answer

44

For a sorted set of 9 values, the median is the 5th value, giving

y = 14

. To make 8 the unique mode, 8 must appear at least twice, which forces

x = 8

. With an arithmetic mean of 16 across 9 numbers, the total sum is

9 \times 16 = 144

. Subtracting the sum of the eight known numbers (

3+5+8+8+14+17+21+24 = 100

) gives

z = 44

.

Step-by-Step Solution

1

Determine the value of

y

using the median definition.

y = 14

For a dataset of

n = 9

numbers arranged in non-decreasing order, the median is the

\frac{9+1}{2} = 5\text{th}

element. Thus,

y = 14

.

2

Determine the value of

x

using the unique mode condition.

x = 8

The dataset is ordered as

3 \le 5 \le 8 \le x \le 14 \le 17 \le 21 \le 24 \le z

. For

8

to be a unique mode, it must appear more than once. Since all other given numbers are distinct,

x

must equal

8

so that

8

occurs twice.

3

Calculate the required total sum of all 9 data points from the given arithmetic mean.

\text{Total Sum} = 144

\text{Mean} = \frac{\text{Total Sum}}{9} \implies \text{Total Sum} = 9 \times 16 = 144.

4

Sum the known values and solve for

z

.

z = 44

3 + 5 + 8 + 8 + 14 + 17 + 21 + 24 + z = 144 \implies 100 + z = 144 \implies z = 44

.

Key Concept

Combining mean, median, and mode definitions to solve for unknown elements in a ordered dataset.

Question 22Question →

A dataset $S$ consists of $5$ positive integers. The dataset has a unique mode of $15$ , a median of $15$ , and an arithmetic mean of $12$ .

Which of the following could be the range of dataset $S$ ? Indicate all such values.

Select all that apply

8

10

18

26

28

Show answer & explanation

Answer: 10; 18; 26

Answer

The values 10, 18, and 26 are all possible values for the range of dataset

S

.

The range of dataset S is defined as x_5 - x_1. Based on the mean, median, and unique mode conditions, the 5 positive integers must take the form x_1 < x_2 < 15 = 15 < x_5 with x_1 + x_2 + x_5 = 30. Testing valid integer assignments shows the range can be any integer from 10 to 26 inclusive. Therefore, 10, 18, and 26 are all valid possible range values.

Step-by-Step Solution

1

Express the sum of elements using the mean

Sum of 5 elements = 5 × 12 = 60

Since the mean of 5 numbers is 12, their total sum must equal 60.

2

Set up the ordered elements and apply the median and unique mode conditions

Elements in ascending order: x_1 ≤ x_2 ≤ x_3 ≤ x_4 ≤ x_5, with x_3 = 15 and x_4 = 15.

The median (middle element x_3) is 15. For 15 to be the unique mode, 15 must appear at least twice. Since x_3 = 15, either x_2 = 15 or x_4 = 15. If x_2 = x_3 = x_4 = 15, then x_1 + x_5 = 15, which forces x_1 < 1 and violates positive integer constraints. Thus, 15 appears exactly twice: x_3 = 15 and x_4 = 15 (with x_1 < x_2 < 15 and x_5 > 15).

3

Determine the constraints on x_1, x_2, and x_5

x_1 + x_2 + x_5 = 30, where 1 ≤ x_1 < x_2 ≤ 14 and x_5 > 15.

Substituting x_3 = 15 and x_4 = 15 into the sum gives x_1 + x_2 + 15 + 15 + x_5 = 60, simplifying to x_1 + x_2 + x_5 = 30.

4

Calculate the bounds for the Range R = x_5 - x_1

Minimum range = 10, Maximum range = 26

To maximize R = x_5 - x_1 = 30 - 2x_1 - x_2, choose minimum x_1 = 1 and minimum x_2 = 2, yielding x_5 = 27 and R = 26. To minimize R, maximize x_1 = 6 and x_2 = 8, yielding x_5 = 16 and R = 10.

Key Concept

Measures of Central Tendency (Mean, Median, Mode) and Data Range Constraints

Question 23Question →

A dataset $W$ consists of $10$ distinct positive integers. The arithmetic mean of the dataset is $24$ , and its median is $22$ . A new dataset $V$ is created by replacing every integer $x$ in $W$ that is strictly less than the median of $W$ with $2x + 1$ , while leaving all other integers in $W$ unchanged. If the sum of the $5$ smallest integers in $W$ is $60$ , what is the arithmetic mean of the integers in dataset $V$ ?

28.5

30.0

30.5

31.0

36.5

Show answer & explanation

Answer: 30.5

Answer

30.5

The total sum of the original dataset

W

is

10 \times 24 = 240

. Since

W

consists of

10

distinct integers, exactly

5

elements are strictly less than the median. The sum of these

5

elements is given as

60

, which leaves the sum of the remaining

5

elements as

240 - 60 = 180

. When each of the

5

lower elements

x

is replaced with

2x + 1

, their new sum becomes

2(60) + 5(1) = 125

. The sum of the new dataset

V

is

125 + 180 = 305

, and its mean is

305 / 10 = 30.5

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset

W

.

Since dataset

W

has

10

elements with a mean of

24

, its total sum is

10 \times 24 = 240

.

The sum of elements in any set is equal to the number of elements multiplied by the arithmetic mean.

2

Determine how many elements are strictly less than the median of

W

.

Because all

10

elements in

W

are distinct positive integers, ordering them as

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}

places the median between

x_5

and

x_6

. Thus, exactly

5

elements (

x_1

through

x_5

) are strictly less than the median.

In an even-sized set of distinct values, exactly half of the elements lie strictly below the median position.

3

Find the sum of the remaining

5

elements (the upper half) in

W

.

Given that the sum of the

5

smallest elements is

60

, the sum of the remaining

5

elements is

240 - 60 = 180

.

The total sum of the dataset is the sum of its lower

5

elements plus the sum of its upper

5

elements.

4

Calculate the new sum of the transformed

5

smallest elements in dataset

V

.

Each of the

5

elements

x_i

is replaced by

2x_i + 1

. The new sum is

\sum_{i=1}^5 (2x_i + 1) = 2 \sum_{i=1}^5 x_i + 5(1) = 2(60) + 5 = 125

.

Linear transformations applied to individual elements scale their sum by the multiplier and add the constant term multiplied by the number of elements.

5

Compute the total sum and arithmetic mean of dataset

V

.

The total sum of

V

is

125 + 180 = 305

. The new mean is

\frac{305}{10} = 30.5

.

The mean of

V

is its total sum divided by the number of elements (

10

).

Key Concept

Linear transformation of dataset subsets and median position in distinct ordered sets

Question 24Question →

A dataset $S$ consists of 12 numbers listed in increasing order: $x_1, x_2, \dots, x_{12}$ . The median of dataset $S$ is 40. The arithmetic mean of the 6 smallest numbers in $S$ is 28, and the arithmetic mean of the 6 largest numbers in $S$ is 56. A new dataset $T$ is formed by subtracting 4 from each of the 6 smallest numbers in $S$ and adding 8 to each of the 6 largest numbers in $S$ . What is the positive difference between the arithmetic mean of dataset $T$ and the median of dataset $T$ ?

Show answer & explanation

Answer: 2

Answer

2

The total sum of dataset S is 504, giving a mean of 42. Transforming the elements adds a net total of 24 to the overall sum, so the mean of dataset T becomes 44. Because decreasing the lower half and increasing the upper half preserves the relative sorted order of all 12 numbers, the middle two elements of dataset T are x_6 - 4 and x_7 + 8. Thus, the new median is (x_6 + x_7)/2 + 2 = 40 + 2 = 42. The positive difference between the mean of 44 and the median of 42 is 2.

Step-by-Step Solution

1

Calculate the arithmetic mean of the original dataset

S

.

The sum of the 6 smallest numbers is

6 \times 28 = 168

, and the sum of the 6 largest numbers is

6 \times 56 = 336

. The total sum of dataset

S

is

168 + 336 = 504

. Thus, the mean of

S

is

\frac{504}{12} = 42

.

The mean of a dataset is the sum of all elements divided by the total number of elements.

2

Calculate the arithmetic mean of the new dataset

T

.

The sum of dataset

T

is

504 + 6(-4) + 6(8) = 504 - 24 + 48 = 528

. The mean of dataset

T

is

\frac{528}{12} = 44

.

Modifying each of the 12 elements changes the overall sum by the sum of individual changes.

3

Determine the median of the new dataset

T

.

Since

x_6 < x_7

, after transformations

x_6 - 4 < x_7 + 8

. The relative order of all elements is preserved. The median of

T

is

\frac{(x_6 - 4) + (x_7 + 8)}{2} = \frac{x_6 + x_7}{2} + 2 = 40 + 2 = 42

.

The median of an even number of ordered elements is the average of the two middle elements.

4

Calculate the positive difference between the mean and median of dataset

T

.

|44 - 42| = 2.

Subtract the median from the mean and take the absolute value.

Key Concept

Effect of linear transformations and subgroup operations on the mean and median of ordered datasets

Question 25Question →

A dataset $S$ consists of 9 positive integers: $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9$ , ordered such that $x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5 \leq x_6 \leq x_7 \leq x_8 \leq x_9$ .

The dataset has the following statistical properties:
- The median of dataset $S$ is $20$ .
- Dataset $S$ has a unique mode of $25$ .
- The arithmetic mean of dataset $S$ is $18$ .
- The range of dataset $S$ is $22$ .

Which of the following statements MUST be true? Select all such statements.

Select all that apply

The smallest integer

x_1

cannot exceed

5

.

The value

25

appears at least twice in dataset

S

.

The median of the first 5 elements

(x_1, x_2, x_3, x_4, x_5)

must be equal to

15

.

The maximum possible value for

x_9

is

27

.

The sum of the four smallest integers

(x_1 + x_2 + x_3 + x_4)

cannot exceed

42

.

Show answer & explanation

Answer: The smallest integer $x_1$ cannot exceed $5$ .; The value $25$ appears at least twice in dataset $S$ .; The sum of the four smallest integers $(x_1 + x_2 + x_3 + x_4)$ cannot exceed $42$ .

Answer

The statements asserting that the smallest integer cannot exceed 5, that 25 appears at least twice, and that the sum of the four smallest integers cannot exceed 42 must be true.

The statement regarding the unique mode requiring 25 to appear at least twice must be true by the definition of mode. The statement regarding the upper bound on the sum of the four smallest integers is true because the top 5 elements account for at least 120 of the total sum of 162.

Step-by-Step Solution

1

Determine the total sum of the dataset and identify fixed metric properties.

Sum =

9 \times 18 = 162

. Since there are 9 ordered elements, the median is the 5th element

x_5 = 20

.

Mean is total sum divided by number of elements, and median of an odd number of sorted elements is the middle term.

2

Analyze the mode constraint.

The number

25

must appear at least 2 times among

\{x_6, x_7, x_8, x_9\}

.

A unique mode must occur strictly more times than any other data value in the set.

3

Analyze the range constraint

x_9 - x_1 = 22

, implying

x_9 = x_1 + 22

.

Determine the upper bound for

x_1

.

If

x_1 \ge 6

, then

x_9 \ge 28

. The smallest possible values for the elements above the median

\{x_6, x_7, x_8, x_9\}

given mode

25

would make

x_6=25, x_7=25, x_8=25, x_9=28

, summing to

103

. With

x_5=20

, the upper 5 elements sum to at least

123

. The lower 4 elements

\{x_1, x_2, x_3, x_4\}

must each be at least

x_1 \ge 6

, so their sum is at least

4 \times 6 = 24

. The total sum would then be at least

123 + 24 = 147

, but considering

x_1 \ge 6 \implies x_9 \ge 28

and keeping non-decreasing order: if

x_1=6, x_2=6, x_3=6, x_4=6

, sum is

24 + 20 + 25 + 25 + 25 + 28 = 155 < 162

. However, if

x_1 = 6

,

x_9 = 28

,

x_6=25, x_7=25, x_8=25

, sum of upper elements is

20+25+25+25+28=123

. Lower elements must sum to

162-123=39

. But if

x_1=6

,

x_4

can be at most

20

. Can lower 4 elements sum to 39 with

x_1=6

?

6+6+7+20 = 39

. But then

x_9 = 28

, mode 25 occurs 3 times. Wait, if

x_1=6, x_2=6

, then 6 occurs twice! But 25 is the UNIQUE mode, so 6 cannot occur twice unless 25 occurs 3 times. If 25 occurs 3 times (

x_6=25, x_7=25, x_8=25, x_9=28

), then

x_1=6, x_2=7, x_3=8, x_4=18

sums to

39

, with no duplicates in lower half! Wait:

6+7+8+18+20+25+25+25+28 = 162

. Here range =

28 - 6 = 22

, mean =

162/9 = 18

, median =

20

, unique mode =

25

(appears 3 times). Can

x_1 = 6

? Yes,

6+7+8+18+20+25+25+25+28=162

works! But if

x_1=7

,

x_9=29

, upper sum

\ge 20+25+25+25+29 = 124

, lower sum

\le 38

. But

x_1=7 \implies x_1+x_2+x_3+x_4 \ge 7+8+9+10 = 34

. If

x_1=7

,

7+8+9+14+20+25+25+25+29 = 162

. Range

29-7=22

. So

x_1=7

works too! Therefore,

x_1

can exceed

5

.

Key Concept

Combining mean, median, mode, and range constraints in an ordered dataset of integers.

Question 26Question →

A meteorologist recorded the daily minimum temperatures, in degrees Celsius, at a high-altitude research station over a 7-day period: $3$ , $-5$ , $7$ , $-2$ , $10$ , $-8$ , and $2$ .

If $M$ represents the median of these daily minimum temperatures and $A$ represents the arithmetic mean, what is the value of $M - A$ ?

-3

-1

0

1

3

Show answer & explanation

Answer: $1$

Answer

The value of

M - A

is

1

.

To evaluate

M - A

, first arrange the data set in ascending order:

-8, -5, -2, 2, 3, 7, 10

. Since there are 7 numbers, the median

M

is the middle (4th) value, which is

2

. Next, find the arithmetic mean

A

by taking the sum of the elements,

(-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

, and dividing by

7

, yielding

A = 1

. Finally, subtract the mean from the median:

M - A = 2 - 1 = 1

.

Step-by-Step Solution

1

Sort the dataset in ascending order to find the median

M

.

The sorted list of 7 temperatures is:

-8, -5, -2, 2, 3, 7, 10

.

The median of a set with an odd number of elements is the middle value of the ordered dataset.

2

Identify the 4th element in the sorted dataset.

M = 2

.

In a dataset of 7 ordered values, the middle (4th) position represents the median.

3

Calculate the arithmetic mean

A

by summing all temperatures and dividing by 7.

Sum

= (-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

. Thus,

A = \frac{7}{7} = 1

.

The arithmetic mean is defined as the total sum of observations divided by the number of observations.

4

Compute

M - A

.

M - A = 2 - 1 = 1

.

Subtracting the mean from the median yields the required target value.

Key Concept

Calculating the median of a dataset requires arranging values in numerical order before identifying the central value.

Estimated Time:1m 30s

Question 27Question →

A list consists of six numbers: $3$ , $7$ , $10$ , $14$ , $18$ , and $x$ . If the arithmetic mean of these six numbers is equal to their median, which of the following could be the value of $x$ ? Indicate all such values.

Select all that apply

-1

5

11

14

20

Show answer & explanation

Answer: $-1$ ; $11$ ; $20$

Answer

The valid values for

x

are

-1

,

11

, and

20

.

The values

-1

,

11

, and

20

each yield a dataset where the arithmetic mean equals the median:

-1

gives a mean and median of

8.5

,

11

gives a mean and median of

10.5

, and

20

gives a mean and median of

12

.

Step-by-Step Solution

1

Express the arithmetic mean in terms of

x

.

Mean =

\frac{3 + 7 + 10 + 14 + 18 + x}{6} = \frac{52 + x}{6}

.

The mean of a dataset of

n

numbers is the sum of all elements divided by

n

.

2

Analyze the median based on the position of

x

relative to the sorted known values

3, 7, 10, 14, 18

.

Case 1:

x \le 7 \implies

median =

\frac{7+10}{2} = 8.5

.
Case 2:

7 < x < 14 \implies

median =

\frac{x+10}{2}

(for

7 < x \le 10

) or

\frac{10+x}{2}

(for

10 < x < 14

).
Case 3:

x \ge 14 \implies

median =

\frac{10+14}{2} = 12

.

For an even number of elements (

n=6

), the median is the average of the 3rd and 4th terms in ascending order.

3

Set the mean equal to the median for each case and solve for

x

.

Case 1:

\frac{52+x}{6} = 8.5 \implies 52+x = 51 \implies x = -1

(valid since

-1 \le 7

).
Case 2:

\frac{52+x}{6} = \frac{10+x}{2} \implies 52+x = 30+3x \implies 2x = 22 \implies x = 11

(valid since

7 < 11 < 14

).
Case 3:

\frac{52+x}{6} = 12 \implies 52+x = 72 \implies x = 20

(valid since

20 \ge 14

).

This identifies all values of

x

satisfying the problem constraint.

Key Concept

Evaluating mean and median of a dataset containing an unknown variable across different intervals of the variable's possible values.

Question 28Question →

A dataset $S$ consists of 7 distinct positive integers. The arithmetic mean of the numbers in $S$ is 20, and the median is 18. If the largest integer in $S$ is 35, what is the maximum possible value for the second-largest integer in $S$ ?

34

33

35

62

19

Show answer & explanation

Answer: 34

Answer

34

To find the maximum possible value of the second-largest integer, we arrange the 7 distinct positive integers in ascending order:

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7

. The median

x_4 = 18

and the largest element

x_7 = 35

. The total sum of all 7 elements is

7 \times 20 = 140

. Since

x_6

must be strictly less than

x_7 = 35

, the maximum potential integer value for

x_6

is 34. If

x_6 = 34

, the remaining four elements (

x_1, x_2, x_3, x_5

) must sum to

87 - 34 = 53

. Choosing

x_5 = 19

(the smallest integer greater than 18) leaves a sum of 34 for

x_1 + x_2 + x_3

, which can be satisfied by distinct positive integers such as 1, 16, and 17. Thus, 34 is achievable.

Step-by-Step Solution

1

Calculate the total sum of the 7 integers in the dataset.

Total sum =

7 \times 20 = 140

.

The sum of a dataset equals the number of elements multiplied by its arithmetic mean.

2

Identify the positions of known elements when the dataset is ordered in ascending order

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7

.

Median

x_4 = 18

and largest element

x_7 = 35

.

For a 7-element dataset, the median is the 4th element.

3

Calculate the combined sum of the remaining five unknown elements.

x_1 + x_2 + x_3 + x_5 + x_6 = 140 - (18 + 35) = 87

.

Subtracting the median and the largest element from the total sum gives the sum of the remaining five elements.

4

Determine the theoretical upper bound for the second-largest integer

x_6

.

Since

x_6 < x_7 = 35

and all integers are distinct,

x_6 \le 34

.

The second-largest integer must be strictly less than the largest integer.

5

Verify if

x_6 = 34

can produce a valid dataset of distinct positive integers.

If

x_6 = 34

, then

x_1 + x_2 + x_3 + x_5 = 87 - 34 = 53

. Setting

x_5 = 19

,

x_3 = 17

,

x_2 = 16

, and

x_1 = 1

gives

1 + 16 + 17 + 19 = 53

, forming the valid set

\{1, 16, 17, 18, 19, 34, 35\}

.

Since a valid set of distinct positive integers exists satisfying all constraints, 34 is the maximum possible value.

Key Concept

Measures of Central Tendency and Data Constraints

Question 29Question →

A university department tracked the number of research articles published by its 8 faculty members over a five-year period. The numbers of publications for 7 of the faculty members were $4, 7, 9, 12, 15, 18,$ and $23$ . If the arithmetic mean of the number of publications for all 8 faculty members is equal to $1.25$ times their median, what is the number of publications for the 8th faculty member?

Show answer & explanation

Answer: 47

Answer

47

The sum of the 7 known values is

88

, making the total sum

88 + x

and the arithmetic mean

\frac{88 + x}{8}

. Assuming

x \ge 15

, the 4th and 5th numbers when sorted are

12

and

15

, yielding a median of

13.5

. Setting the mean equal to

1.25 \times 13.5 = 16.875

gives

\frac{88 + x}{8} = 16.875

, which simplifies to

88 + x = 135

and yields

x = 47

.

Step-by-Step Solution

1

Calculate the sum of the 7 known data points and express the mean in terms of the unknown 8th value

x

.

Sum of 7 known values =

4 + 7 + 9 + 12 + 15 + 18 + 23 = 88

. Total mean =

\frac{88 + x}{8}

.

The arithmetic mean of

n

values is the sum of all values divided by

n

.

2

Analyze the position of

x

in sorted order to determine the median.

Assuming

x \ge 15

, the 4th and 5th values in ascending order are

12

and

15

, giving a median of

\frac{12 + 15}{2} = 13.5

.

For an even number of data points (

8

), the median is the average of the 4th and 5th terms in sorted order.

3

Formulate and solve the equation linking the mean and median.

\frac{88 + x}{8} = 1.25 \times 13.5 = 16.875 \implies 88 + x = 135 \implies x = 47

.

The problem specifies that the mean is equal to

1.25

times the median.

Key Concept

Calculating mean and median of a dataset containing an unknown value.

Estimated Time:1m 30s

Question 30Question →

A healthcare clinic logged the waiting times, in minutes, for 7 patients on Monday: $12, 41, 24, 19, 33, 28,$ and $49$ . On Tuesday, 3 additional patients were logged. The arithmetic mean waiting time for all 10 patients combined was $30$ minutes, and the median waiting time of the 3 patients logged on Tuesday was $34$ minutes. If one of the patients logged on Tuesday had a waiting time of $22$ minutes and another had a waiting time of $34$ minutes, what was the waiting time, in minutes, of the third patient logged on Tuesday?

28 minutes

32 minutes

34 minutes

38 minutes

48 minutes

Show answer & explanation

Answer: 38 minutes

Answer

38 minutes

The option stating 38 minutes is correct because the total sum for all 10 patients combined is

10 \times 30 = 300

minutes. The sum of Monday's 7 patient waiting times is

12 + 41 + 24 + 19 + 33 + 28 + 49 = 206

minutes. Therefore, Tuesday's 3 patients must sum to

300 - 206 = 94

minutes. Since two of Tuesday's patients waited 22 minutes and 34 minutes, the third patient's waiting time is

94 - (22 + 34) = 38

minutes. Ordering the three Tuesday times

(22, 34, 38)

confirms the median is indeed 34 minutes.

Step-by-Step Solution

1

Calculate the total sum of waiting times for all 10 patients.

Total sum =

10 \times 30 = 300

minutes.

The arithmetic mean of

n

values is

\text{Sum} / n

, so

\text{Sum} = n \times \text{Mean}

.

2

Calculate the sum of waiting times for the 7 patients logged on Monday.

Monday sum =

12 + 41 + 24 + 19 + 33 + 28 + 49 = 206

minutes.

Adding the individual values gives the total Monday waiting time.

3

Find the total sum of waiting times for the 3 patients logged on Tuesday.

Tuesday sum =

300 - 206 = 94

minutes.

Subtracting Monday's total from the combined total yields Tuesday's total.

4

Determine the third patient's waiting time on Tuesday using the given values.

Third patient's waiting time =

94 - (22 + 34) = 94 - 56 = 38

minutes.

The sum of the 3 Tuesday patients is 94 minutes, and two known patients waited 22 and 34 minutes. Note that the sorted set

(22, 34, 38)

has a median of 34 minutes, consistent with the problem statement.

Key Concept

Combined Mean and Median Properties

Question 31Question →

A quality assurance auditor reviewed the processing times, in seconds, for a batch of 7 completed tasks: $22, 10, 24, 18, 30, 22,$ and $14$ . An 8th task with a processing time of $x$ seconds, where $x$ is a positive integer, is added to the batch. Which of the following statements must be true? Select all such statements.

Select all that apply

If

x = 22

, the median of the 8 processing times is equal to 22.

If the arithmetic mean of the 8 processing times is equal to 21, then

x = 28

.

If

x = 18

, the median of the 8 processing times is equal to 21.

If

x = 30

, the arithmetic mean of the dataset increases by 5.

If

x = 10

, the mode of the 8 processing times becomes 10.

Show answer & explanation

Answer: If $x = 22$ , the median of the 8 processing times is equal to 22.; If the arithmetic mean of the 8 processing times is equal to 21, then $x = 28$ .

Answer

The correct statements are those asserting that if

x=22

, the median of the 8 processing times is 22, and if the mean of the 8 processing times is 21, then

x=28

.

The statement regarding

x=22

is correct because inserting 22 places 22 at both the 4th and 5th positions of the 8-element ordered list, yielding a median of 22. The statement regarding a mean of 21 is correct because the required total sum for 8 items is

8 \times 21 = 168

, which requires

x = 168 - 140 = 28

.

Step-by-Step Solution

1

Order the original dataset and calculate initial metrics.

Sorted original list:

10, 14, 18, 22, 22, 24, 30

. Original sum = 140, original mean =

\frac{140}{7} = 20

, original median = 22 (4th term).

Establishing the initial baseline values is necessary to evaluate statements about changes in mean and median.

2

Evaluate the statement regarding median when

x=22

.

Inserting

x=22

yields sorted list:

10, 14, 18, 22, 22, 22, 24, 30

. The 4th and 5th items are both 22, giving median

\frac{22+22}{2} = 22

.

Confirms the first statement is true.

3

Evaluate the statement regarding mean equal to 21.

Target sum for 8 items with mean 21 is

8 \times 21 = 168

. Setting

140 + x = 168

yields

x = 28

.

Confirms the second statement is true.

4

Evaluate remaining incorrect statements.

For

x=18

, ordered list is

10, 14, 18, 18, 22, 22, 24, 30

, giving median

\frac{18+22}{2} = 20 \neq 21

. For

x=30

, new mean is

21.25

, an increase of

1.25 \neq 5

. For

x=10

, 10 and 22 both appear twice, making it bimodal rather than having a single mode of 10.

Verifies that all other statements are false.

Key Concept

Measures of Central Tendency (Mean, Median, Mode) for modified datasets

Question 32Question →

A regional library recorded the daily number of public computer reservations over a 7-day period: $14, 25, 18, 31, 25, 12,$ and $21$ . During the following week, 2 additional daily reservation values, $x$ and $y$ , were added to the dataset. If the median of the resulting 9-day dataset is equal to the mode of the original 7-day dataset, which of the following values could be the sum of $x$ and $y$ ?

36

42

46

48

54

Show answer & explanation

Answer: 54

Answer

54

The mode of the original 7 values is 25. When 2 values (

x

and

y

) are added to form a 9-value dataset, the median is the 5th value when ordered. In the original sorted dataset (

12, 14, 18, 21, 25, 25, 31

), there are only 3 values greater than or equal to 25. To make the 5th value of the combined 9-value list equal to 25, both added values must be at least 25 (

x \ge 25

and

y \ge 25

). Consequently, their sum

x + y

must be at least

50

. The only choice that satisfies

x + y \ge 50

is 54.

Step-by-Step Solution

1

Identify the mode of the original 7-day dataset.

The original values are

14, 25, 18, 31, 25, 12, 21

. The value

25

appears twice, while all other numbers appear once. Therefore, the mode is

25

.

The question states that the target median of the 9-day dataset must equal the mode of the original dataset.

2

Sort the original 7-day dataset in ascending order.

The sorted original dataset is

12, 14, 18, 21, 25, 25, 31

.

Determining position-based metrics like median requires ordering the data.

3

Determine the condition required for the median of the 9-day dataset to be 25.

In a sorted 9-element dataset, the median is the 5th element. For the 5th element to be

25

, there must be at least 5 elements in the set that are greater than or equal to

25

. The original set contains 3 elements

\ge 25

(

25, 25, 31

). Thus, both new values

x

and

y

must be

\ge 25

.

If only one of

x

or

y

were

\ge 25

, there would only be 4 elements

\ge 25

, making the 5th element (median) less than

25

.

4

Calculate the lower bound for the sum

x + y

and select the valid option.

Since

x \geq 25

and

y \geq 25

, the sum must satisfy

x + y \geq 25 + 25 = 50

. Among the given options, only

54

is greater than or equal to

50

.

Any sum less than

50

violates the condition that both

x

and

y

are at least

25

.

Key Concept

Determining positional metrics (median) after dataset expansion

Question 33Question →

A technology startup recorded the monthly software subscription costs, in dollars, for 6 department teams: $450, 120, 310, 120, 280,$ and $500$ . A seventh department team with a monthly subscription cost of $x$ dollars is added to the dataset, where $x$ is a positive integer. Which of the following statements regarding the measures of central tendency for the updated 7-team dataset must be true? Select all that apply.

Select all that apply

The median of the subscription costs for the 7 teams cannot be greater than 310.

If

x = 600

, the arithmetic mean of the subscription costs for the 7 teams is greater than the median.

If

x = 120

, the median of the subscription costs for the 7 teams is 120.

The arithmetic mean of the subscription costs for the 7 teams will equal the median if

x = 290

.

The mode of the subscription costs for the 7 teams is uniquely 120 for any positive integer value of

x

.

Show answer & explanation

Answer: The median of the subscription costs for the 7 teams cannot be greater than 310.; If $x = 600$ , the arithmetic mean of the subscription costs for the 7 teams is greater than the median.

Answer

The statements that must be true are that the median cannot exceed 310, and that if x = 600, the arithmetic mean is greater than the median.

Sorting the six known values gives 120, 120, 280, 310, 450, 500. In a dataset of 7 numbers, the median is the 4th value when sorted. If x is added, the 4th value will be 280 if x ≤ 280, x if 280 < x < 310, or 310 if x ≥ 310. Thus, the median is capped at 310. Furthermore, setting x = 600 yields a median of 310 and a mean of (1780 + 600) / 7 = 340, which is greater than the median.

Step-by-Step Solution

1

Sort the existing 6 dataset elements in ascending order and compute their sum.

Sorted list: 120, 120, 280, 310, 450, 500. Sum = 1780.

Establishing the baseline position of values helps determine how inserting x shifts the median, mean, and mode.

2

Analyze the position of the median for a 7-element dataset.

The median is the 4th element when sorted. If x ≤ 280, the 4th element is 280. If 280 < x < 310, the 4th element is x. If x ≥ 310, the 4th element is 310. Thus, 280 ≤ median ≤ 310.

This establishes upper and lower bounds for the median regardless of x.

3

Evaluate the statement regarding median upper bound (median ≤ 310).

Since the maximum possible 4th element in the sorted 7-element list is 310, the median can never be greater than 310.

Confirms the validity of the statement concerning the maximum median.

4

Evaluate the scenario where x = 600.

Sum = 1780 + 600 = 2380. Mean = 2380 / 7 = 340. Median = 310 (since 600 ≥ 310). 340 > 310.

Confirms that when x = 600, mean (340) is strictly greater than median (310).

Key Concept

Measures of Central Tendency (Mean, Median, Mode)

Question 34Question →

A sports scientist records the recovery times, in minutes, for 5 subjects after an intense workout. Four of the times are $44, 16, 52,$ and $32$ minutes, and the fifth time is $x$ minutes. The 5 times are listed in no particular order. If $32 < x < 44$ and the arithmetic mean of all 5 recovery times is equal to the median of the 5 recovery times, what is the value of $x$ ?

26

29

34

36

48

Show answer & explanation

Answer: 36

Answer

The value of

x

is 36.

Because

32 < x < 44

, the 5 numbers listed in ascending order are

16, 32, x, 44, 52

. The median of these 5 numbers is the middle value,

x

. Setting the mean

\frac{16 + 32 + x + 44 + 52}{5} = \frac{144 + x}{5}

equal to the median

x

gives

\frac{144 + x}{5} = x

, which simplifies to

4x = 144

and yields

x = 36

.

Step-by-Step Solution

1

Order the dataset in ascending numerical order.

The five data values in sorted order are

16, 32, x, 44, 52

because it is given that

32 < x < 44

.

Finding the median of a dataset requires arranging all elements in ascending or descending order first.

2

Identify the median of the 5 values.

The median is the 3rd element in the sorted 5-element list, which is

x

.

For an odd number of items (

n = 5

), the median is the middle value at position

\frac{5+1}{2} = 3

.

3

Express the arithmetic mean in terms of

x

.

\text{Mean} = \frac{16 + 32 + x + 44 + 52}{5} = \frac{144 + x}{5}$.

The arithmetic mean is defined as the sum of all values divided by the total count of values (

n = 5

).

4

Set the mean equal to the median and solve for

x

.

\frac{144 + x}{5} = x \implies 144 + x = 5x \implies 4x = 144 \implies x = 36$.

The problem states that the arithmetic mean equals the median.

Key Concept

Measures of Central Tendency (Mean, Median, Mode)

Estimated Time:1m 45s

Question 35Question →

A dataset consists of 7 integers: $5, 8, 12, 14, 16, 20,$ and $x$ . If the median of the dataset is strictly greater than the mean of the dataset, which of the following could be the value of $x$ ? Select all such values.

Select all that apply

4

10

13

18

25

Show answer & explanation

Answer: $4$ ; $13$ ; $18$

Answer

The values of

x

that make the median strictly greater than the mean are

4

,

13

, and

18

.

The median of a 7-element dataset is its 4th element when arranged in ascending order. Testing the values shows that for

4

, the median (

12

) exceeds the mean (

11.29

); for

13

, the median (

13

) exceeds the mean (

12.57

); and for

18

, the median (

14

) exceeds the mean (

13.29

). All three satisfy the condition.

Step-by-Step Solution

1

Express the mean of the dataset in terms of

x

.

The sum of the known 6 numbers is

5 + 8 + 12 + 14 + 16 + 20 = 75

. The total sum for 7 numbers is

75 + x

, making the mean

\text{Mean} = \frac{75 + x}{7}

.

The mean is calculated as the sum of all elements divided by the total count (7).

2

Analyze the median across different ranges of

x

.

For 7 numbers sorted in order, the median is the 4th number. If

x \le 12

, the sorted list starts with

x, 5, 8, 12

or similar, so the 4th element is

12

. If

12 < x < 14

, the 4th element is

x

. If

x \ge 14

, the 4th element is

14

.

The position of

x

relative to the known numbers determines which element falls into the middle (4th) spot.

3

Test the condition

\text{Median} > \text{Mean}

for each piecewise case.

Case 1 (

x \le 12

):

12 > \frac{75 + x}{7} \implies 84 > 75 + x \implies x < 9

. Thus,

x = 4

works, but

x = 10

does not.
Case 2 (

12 < x < 14

):

x > \frac{75 + x}{7} \implies 7x > 75 + x \implies 6x > 75 \implies x > 12.5

. Thus,

x = 13

works.
Case 3 (

x \ge 14

):

14 > \frac{75 + x}{7} \implies 98 > 75 + x \implies x < 23

. Thus,

x = 18

works, but

x = 25

does not.

Solving the inequality for each case yields all valid ranges for

x

:

x < 9

,

x = 13

, and

14 \le x < 23

.

Key Concept

Measures of Central Tendency (Mean vs. Median Analysis with Variables)

Question 36Question →

A quality control manager records the durability ratings, on a scale from $1$ to $50$ , for $8$ randomly selected component batches. The ratings are listed below:

34, 18, 42, 27, 18, 39, 45, 25

Two additional component batches with identical ratings, $x$ and $y$ (where $x = y$ ), are included in the dataset. If the median rating of the updated dataset of $10$ batches is equal to the median rating of the original dataset of $8$ batches, what is the value of $x$ ?

22.5

26.0

30.5

31.5

36.5

Show answer & explanation

Answer: $30.5$

Answer

The value of

x

is

30.5

.

To find the median of the original dataset, the numbers must first be ordered from smallest to largest:

18, 18, 25, 27, 34, 39, 42, 45

. Since there are

8

numbers, the median is the average of the 4th and 5th numbers:

(27 + 34)/2 = 30.5

. When two equal values

x

are added to form a

10

-element dataset, having

x = 30.5

places both new values directly between

27

and

34

. The 5th and 6th terms of the updated set are both

30.5

, making the new median

(30.5 + 30.5)/2 = 30.5

, which matches the original median.

Step-by-Step Solution

1

Sort the original dataset of 8 durability ratings in ascending order.

The sorted dataset is

18, 18, 25, 27, 34, 39, 42, 45

.

Finding the median of a numerical dataset requires arranging the numbers in order from least to greatest.

2

Calculate the median of the original 8 ratings.

The 4th value is

27

and the 5th value is

34

. The median is

\frac{27 + 34}{2} = 30.5

.

For a dataset with an even number of elements (

n = 8

), the median is the average of the two middle elements (the 4th and 5th terms).

3

Analyze how adding two identical values

x

and

y

(

x = y

) affects the median of the 10-element dataset.

Inserting two values equal to

30.5

places them as the 5th and 6th elements in the 10-element sorted array:

18, 18, 25, 27, 30.5, 30.5, 34, 39, 42, 45

. The new median is

\frac{30.5 + 30.5}{2} = 30.5

.

To keep the median unchanged at

30.5

when adding two identical values, the added values must fall between

27

and

34

and specifically equal

30.5

so that the middle two terms of the 10 values average to

30.5

.

Key Concept

Median of a Dataset

Estimated Time:1m 30s

Question 37Question →

A research laboratory conducted $6$ experimental trials to measure the duration, in milliseconds, of a specific chemical reaction. The durations recorded for $5$ of the trials were $240$ , $215$ , $260$ , $225$ , and $245$ . If the median duration of all $6$ trials was $235$ milliseconds, what was the duration, in milliseconds, of the 6th trial?

Show answer & explanation

Answer: 230

Answer

230

For a dataset with 6 numbers, the median is the arithmetic mean of the 3rd and 4th numbers in ascending order. Arranging the 5 given numbers gives

215, 225, 240, 245, 260

. Since the target median is

235

, the sum of the two middle numbers must be

235 \times 2 = 470

. Placing

x = 230

into the dataset yields the ordered set

215, 225, 230, 240, 245, 260

, where the 3rd and 4th numbers are

230

and

240

. Their mean is

(230 + 240) / 2 = 235

, which matches the given median.

Step-by-Step Solution

1

Order the 5 given numbers from least to greatest

The sorted list of known values is

215, 225, 240, 245, 260

.

Calculating or using median requires ordering the data points.

2

Express the median condition for an even number of data points (

n = 6

)

\text{Median} = \frac{\text{3rd value} + \text{4th value}}{2} = 235

, so

\text{3rd value} + \text{4th value} = 470

.

For an even number of values, the median is the average of the two middle numbers.

3

Determine the position and value of the unknown 6th trial

x

If

x \le 225

, the 3rd and 4th values would be

225

and

240

(median

232.5

). If

x \ge 245

, the 3rd and 4th values would be

240

and

245

(median

242.5

). Thus,

x

must lie between

225

and

240

.

Analyzing boundary conditions places

x

as the 3rd value and

240

as the 4th value.

4

Solve for

x

x + 240 = 470 \implies x = 230

.

The sum of the two middle values must equal

470

to yield a median of

235

.

Key Concept

Finding a missing value in a dataset given the median of an even number of observations.

Measures of Central Tendency (Mean, Median, Mode)

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