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

Question 1Question →

A research study records five numerical observations: $12$ , $16$ , $20$ , $20$ , and $x$ , where $x$ is a positive integer. If the arithmetic mean of these five observations is equal to their median, which of the following values could be $x$ ? Select all such values.

Select all that apply

12

15

17

24

32

Show answer & explanation

Answer: $12$ ; $17$ ; $32$

Answer

The possible values of

x

are

12

,

17

, and

32

.

The mean of the dataset is given by

\frac{68+x}{5}

. Depending on where

x

falls relative to the sorted known values (

12, 16, 20, 20

), the median can be

16

(if

x \le 16

),

x

(if

16 \le x \le 20

), or

20

(if

x \ge 20

). Equating the mean to the median in each case yields three valid solutions:

x = 12

,

x = 17

, and

x = 32

.

Step-by-Step Solution

1

Express the arithmetic mean in terms of

x

.

The sum of the five numbers is

12 + 16 + 20 + 20 + x = 68 + x

. Therefore, the mean is

\frac{68 + x}{5}

.

The mean of

n

numbers is the total sum divided by

n

.

2

Analyze Case 1 where

x \le 16

.

The sorted dataset is

x, 12, 16, 20, 20

(or

12, x, 16, 20, 20

). The median (3rd value) is

16

. Setting mean = median gives

\frac{68 + x}{5} = 16 \implies 68 + x = 80 \implies x = 12

. Since

12 \le 16

,

x = 12

is valid.

The median of a 5-element set is the 3rd element when ordered.

3

Analyze Case 2 where

16 \le x \le 20

.

The sorted dataset is

12, 16, x, 20, 20

. The median is

x

. Setting mean = median gives

\frac{68 + x}{5} = x \implies 68 + x = 5x \implies 4x = 68 \implies x = 17

. Since

16 \le 17 \le 20

,

x = 17

is valid.

When

x

lies between 16 and 20,

x

itself becomes the 3rd element of the ordered set.

4

Analyze Case 3 where

x \ge 20

.

The sorted dataset is

12, 16, 20, 20, x

. The median is

20

. Setting mean = median gives

\frac{68 + x}{5} = 20 \implies 68 + x = 100 \implies x = 32

. Since

32 \ge 20

,

x = 32

is valid.

When

x \ge 20

, 20 is the 3rd element of the ordered set.

Key Concept

Solving for missing observations where mean equals median requires analyzing how the position of the variable affects the sorted order and the resulting median value.

Question 2Question →

A quality control department recorded the processing time, in days, for a batch of requests. The table below displays the frequency distribution of the processing times:

Processing Time (Days)	Number of Requests
1	5
2	8
3	$k$
4	6
5	4

If $k$ is a positive integer and the median processing time for all requests in the batch is equal to 3 days, what is the minimum possible value of $k$ ?

2

3

4

5

6

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. There are

5 + 8 = 13

values less than

3

. For

3

to be the median of the dataset, the median position must be greater than

13

. With

k = 4

, the total number of items is

N = 23 + 4 = 27

. For an odd dataset of size

27

, the median is the

\frac{27 + 1}{2} = 14

th item. Since the first

13

items are less than

3

, the

14

th item is

3

, making

3

the median. Any value of

k \le 3

results in a median less than

3

.

Step-by-Step Solution

1

Calculate the cumulative frequency of values below 3 days.

Number of requests with processing times of 1 or 2 days is

5 + 8 = 13

.

To determine the median position, we first count how many data points lie strictly below 3 days.

2

Express the total number of requests

N

in terms of

k

.

N = 5 + 8 + k + 6 + 4 = 23 + k

.

The total number of requests determines whether

N

is odd or even and where the median position lies.

3

Determine the condition for 3 days to be the median.

For 3 days to be the median, the median position must be greater than 13 so that it falls into the category of 3 days.

Since 13 values are strictly less than 3, the median rank must be at least 14.

4

Test minimum integer values for

k

.

If

k = 3

,

N = 26

. The median is the average of the

13

th value (

2

) and the

14

th value (

3

), which is

2.5

. If

k = 4

,

N = 27

. The median is the

14

th value, which is

3

.

Testing

k = 4

yields

N = 27

where the

14

th value is

3

, satisfying the median requirement.

Key Concept

Median of a Frequency Distribution Table

Estimated Time:1m 30s

Question 3Question →

A dataset $S$ consists of $9$ distinct integers. The median of $S$ is $20$ , and the mean of $S$ is $24$ . The mean of the $4$ smallest integers in $S$ is $10$ . If the largest integer in $S$ is removed to form a new dataset $R$ consisting of $8$ integers, what is the maximum possible median of dataset $R$ ?

18.5

19.0

19.5

20.0

20.5

Show answer & explanation

Answer: $19.5$

Answer

The maximum possible median of dataset

R

is

19.5

.

In the sorted set of

9

distinct integers, the median is the

5

th element, which is

20

. Removing the largest value leaves

8

integers, whose median is the average of the

4

th and

5

th elements. Because all integers are distinct, the

4

th element can be at most

19

. The maximum median is therefore

\frac{19 + 20}{2} = 19.5

.

Step-by-Step Solution

1

Identify the position of the median in the original dataset

S

.

In a dataset of

9

sorted distinct integers

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9

, the median is the

5

th element, so

x_5 = 20

.

For an odd number of ordered elements

n=9

, the median is at index

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

.

2

Determine the formula for the median of the modified dataset

R

.

Removing the largest element

x_9

leaves

8

ordered elements

x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8

. The median of

R

is

\frac{x_4 + x_5}{2} = \frac{x_4 + 20}{2}

.

For an even number of ordered elements

n=8

, the median is the average of the two middle terms at indices

4

and

5

.

3

Maximize the value of

x_4

.

Since all integers are distinct and

x_5 = 20

,

x_4

must be an integer strictly less than

20

. Thus, the maximum integer value

x_4

can take is

19

.

Maximizing

x_4

maximizes the average

\frac{x_4 + 20}{2}

.

4

Verify that

x_4 = 19

is achievable under the given mean constraint.

The sum of the smallest

4

integers is

4 \times 10 = 40

. With

x_4 = 19

, we require

x_1 + x_2 + x_3 = 21

. Choosing distinct integers such as

x_1 = 1

,

x_2 = 2

, and

x_3 = 18

satisfies

1 + 2 + 18 = 21

and

x_1 < x_2 < x_3 < x_4

.

Confirming feasibility ensures the upper bound

19

is valid.

5

Calculate the maximum median of dataset

R

.

\frac{19 + 20}{2} = 19.5

.

Substitute the maximum value of

x_4 = 19

into the median expression.

Key Concept

Properties of median for odd vs. even datasets and optimization under distinctness constraints

Question 4Question →

An environmental monitoring station recorded the daily nitrogen dioxide ( $\text{NO}_2$ ) concentrations, in parts per billion (ppb), for 6 consecutive days: $19, 28, 30, 35, 42,$ and $42$ . On the 7th day, a measurement of $x$ ppb was recorded, where $x > 45$ . If the arithmetic mean of the 7 daily concentrations is equal to their median, what is the value of $x$ ?

Show answer & explanation

Answer: 49

Answer

The value of

x

is

49

.

Sorting the first 6 recorded concentrations gives

19, 28, 30, 35, 42, 42

. Because

x > 45

,

x

is strictly greater than all existing values, so the full set in ascending order is

19, 28, 30, 35, 42, 42, x

. The median of a 7-element set is the 4th value, which is

35

. The sum of the 7 concentrations is

19 + 28 + 30 + 35 + 42 + 42 + x = 196 + x

, making the arithmetic mean

\frac{196 + x}{7}

. Setting the mean equal to the median gives

\frac{196 + x}{7} = 35 \implies 196 + x = 245 \implies x = 49

.

Step-by-Step Solution

1

Order the first 6 data points in ascending order

The sorted list is

19, 28, 30, 35, 42, 42

.

Establishing ordered positions is necessary to determine the median.

2

Determine the median of the 7-element dataset

Since

x > 45

, the complete ordered dataset is

19, 28, 30, 35, 42, 42, x

, making the 4th element,

35

, the median.

The median of an odd number of sorted values is the middle element.

3

Express the arithmetic mean in terms of

x

and set it equal to the median

\frac{196 + x}{7} = 35

The problem states that the arithmetic mean equals the median.

4

Solve the linear equation for

x

196 + x = 245 \implies x = 49

Multiplying both sides by 7 and subtracting 196 isolates

x

.

Key Concept

Measures of Central Tendency (Mean and Median)

Question 5Question →

A specialty tea shop sells five distinct sizes of loose-leaf tea canisters. The table below shows the weight per canister and the number of canisters sold during a promotional week:

Canister Size	Weight per Canister (ounces)	Number of Canisters Sold
Sampler	5	16
Small	12	18
Medium	15	$n$
Large	20	6
Executive	25	4

If the mean weight of all canisters sold during the week is $13$ ounces, what is the median weight, in ounces, of the canisters sold?

12

ounces

13

ounces

15

ounces

15.4

ounces

16

ounces

Show answer & explanation

Answer: $15$ ounces

Answer

15 ounces

The correct answer is 15 ounces. Calculating the total weight as a function of

n

gives

516 + 15n

, and the total number of canisters as

44 + n

. Setting the mean

\frac{516 + 15n}{44 + n} = 13

yields

n = 28

. The total number of canisters sold is

44 + 28 = 72

. For 72 items sorted in ascending order, the median is the average of the 36th and 37th values. Cumulative frequencies show that the 1st through 16th items are 5 ounces, the 17th through 34th items are 12 ounces, and the 35th through 62nd items are 15 ounces. Thus, both the 36th and 37th items weigh 15 ounces, making the median 15 ounces.

Step-by-Step Solution

1

Set up an equation for the total number of canisters and total weight.

Total canisters =

16 + 18 + n + 6 + 4 = 44 + n

. Total weight =

5(16) + 12(18) + 15(n) + 20(6) + 25(4) = 516 + 15n

ounces.

The mean of a weighted dataset is the total sum of all values divided by the total number of items.

2

Solve for the missing frequency

n

using the given mean of 13 ounces.

\frac{516 + 15n}{44 + n} = 13 \implies 516 + 15n = 13(44 + n) \implies 516 + 15n = 572 + 13n \implies 2n = 56 \implies n = 28

.

Equating the weighted average expression to 13 allows us to determine the exact number of Medium canisters sold.

3

Determine the total number of items and locate the position of the median.

Total canisters =

44 + 28 = 72

. The median is the average of the

36^{\text{th}}

and

37^{\text{th}}

values when ordered.

For an even number of data points

N = 72

, the median is the arithmetic mean of the

(N/2)^{\text{th}}

and

(N/2 + 1)^{\text{th}}

items.

4

Compute the cumulative frequency distribution to find the values at the

36^{\text{th}}

and

37^{\text{th}}

positions.

5 oz: positions 1 to 16; 12 oz: positions 17 to 34; 15 oz: positions 35 to 62. Both

36^{\text{th}}

and

37^{\text{th}}

positions have a weight of 15 ounces, so the median is 15 ounces.

Cumulative counting reveals that items in positions 35 through 62 all correspond to 15-ounce canisters.

Key Concept

Weighted mean and finding the median of frequency distribution data

Question 6Question →

The frequency distribution table below summarizes the scores of 20 students on a statistics test.

Score	Number of Students
5	3
6	5
7	$x$
8	$y$
9	4

If the arithmetic mean score of the 20 students is $7.15$ , what is the median score of the 20 students?

6.5

7.0

7.15

7.5

8.0

Show answer & explanation

Answer: 7.5

Answer

The median score of the 20 students is 7.5.

To find the median score of 20 students, we first determine the missing student counts

x

and

y

. From the total count of 20 students,

x + y = 20 - (3 + 5 + 4) = 8

. Using the given mean of 7.15, the sum of all scores is

20 \times 7.15 = 143

. The sum expressed in terms of

x

and

y

is

3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Setting

81 + 7x + 8y = 143

gives

7x + 8y = 62

. Solving

x + y = 8

and

7x + 8y = 62

yields

x = 2

and

y = 6

. Listing scores in non-decreasing order, the 10th student scored 7 and the 11th student scored 8. Because the total number of students is even, the median is the average of these two middle values:

(7 + 8) / 2 = 7.5

.

Step-by-Step Solution

1

Set up an equation for the total number of students to find the sum of

x

and

y

.

3 + 5 + x + y + 4 = 20 \implies x + y + 12 = 20 \implies x + y = 8

.

The total frequency of all score categories must equal the total number of students (20).

2

Set up an equation for the sum of all scores using the given arithmetic mean.

Total sum

= 20 \times 7.15 = 143

. Also, total sum

= 3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Therefore,

81 + 7x + 8y = 143 \implies 7x + 8y = 62

.

The sum of all data values equals the mean multiplied by the total number of observations.

3

Solve the system of linear equations for

x

and

y

.

Multiply

x + y = 8

by 7 to get

7x + 7y = 56

. Subtract this from

7x + 8y = 62

to get

y = 6

. Then

x = 8 - 6 = 2

.

Determining the exact frequencies

x=2

and

y=6

is required to locate the middle values.

4

Determine the cumulative frequencies and find the 10th and 11th values.

Score 5: positions 1 to 3 (3 students); Score 6: positions 4 to 8 (5 students); Score 7: positions 9 to 10 (2 students); Score 8: positions 11 to 16 (6 students); Score 9: positions 17 to 20 (4 students). The 10th value is 7 and the 11th value is 8.

For an even number of data points (

N=20

), the median is the arithmetic mean of the

(N/2)

-th and

(N/2 + 1)

-th values.

5

Calculate the median score.

\text{Median} = \frac{7 + 8}{2} = 7.5

.

Average the two middle values.

Key Concept

Calculating the median from a frequency distribution table with missing counts by combining system of linear equations with cumulative frequency analysis.

Question 7Question →

A store manager recorded the number of orders processed online each day during a one-week period: $14$ , $7$ , $25$ , $11$ , $18$ , $2$ , and $16$ . What is the median number of daily online orders processed during this week?

11

14

16

18

93

Show answer & explanation

Answer: $14$

Answer

The median number of daily online orders is

14

.

To find the median, the seven numbers must first be placed in ascending order:

2, 7, 11, 14, 16, 18, 25

. Since there is an odd number of data points (

7

), the median is the single middle value located at the 4th position, which is

14

.

Step-by-Step Solution

1

Arrange the given dataset of seven numbers in ascending order.

The sorted dataset is

2, 7, 11, 14, 16, 18, 25

.

To determine the median of any dataset, the numerical values must first be ordered from smallest to largest.

2

Find the position of the middle element for a dataset with

n = 7

values.

The median position is

\frac{7 + 1}{2} = 4

.

When a dataset contains an odd number of elements

n

, the median is located at the

\frac{n+1}{2}

-th position.

3

Identify the 4th element in the sorted list.

The 4th element is

14

.

In the ordered sequence

2, 7, 11, 14, 16, 18, 25

, the 4th entry is

14

.

Key Concept

The median is the middle value in a set of numerical data that has been arranged in ascending or descending order.

Question 8Question →

A researcher recorded seven daily temperature readings (in degrees Celsius): $14, 5, 22, 9, 25, 2,$ and $17$ . What is the median of these seven temperature readings?

9

14

13.5

17

25

Show answer & explanation

Answer: $14$

Answer

The median of the temperature readings is

14

.

To find the median of a set of numbers, first arrange them in ascending order:

2, 5, 9, 14, 17, 22, 25

. Since there are

7

numbers, the median is the middle value in the

4\text{th}

position, which is

14

.

Step-by-Step Solution

1

Arrange the given dataset in ascending numerical order.

The ordered list of

7

temperatures is:

2, 5, 9, 14, 17, 22, 25

.

To calculate the median of a dataset, the values must first be ordered from smallest to largest.

2

Identify the middle position for an odd number of data values (

n = 7

).

The position of the median is

\frac{7 + 1}{2} = 4\text{th}

position.

For a dataset with

n

elements where

n

is odd, the median is located at position

\frac{n+1}{2}

.

3

Extract the value located at the

4\text{th}

position in the ordered list.

The

4\text{th}

term is

14

.

The fourth number in

2, 5, 9, 14, 17, 22, 25

is

14

.

Key Concept

Median of a Dataset

Estimated Time:45s

Question 9Question →

A customer service representative resolved the following number of support tickets over five consecutive days: $18$ , $24$ , $15$ , $31$ , and $22$ . What is the arithmetic mean of the number of tickets resolved per day by the representative?

Show answer & explanation

Answer: 22

Answer

The arithmetic mean of the number of tickets resolved per day is

22

.

The mean of a data set is calculated by taking the sum of all values and dividing by the total number of items. The sum of the numbers is

18 + 24 + 15 + 31 + 22 = 110

. Dividing

110

by

5

yields

22

.

Step-by-Step Solution

1

Sum all data values in the set

18 + 24 + 15 + 31 + 22 = 110

To find the mean, the first step is to calculate the total sum of all observations.

2

Divide the total sum by the total number of values

\frac{110}{5} = 22

The arithmetic mean is defined as the sum of the values divided by the count of the values.

Key Concept

Arithmetic Mean

Question 10Question →

A high school basketball team scored the following points in five consecutive games: $4$ , $6$ , $6$ , $7$ , and $12$ . Which of the following statements regarding the measures of central tendency for these scores must be true? Select all such statements.

Select all that apply

The mean score is equal to

7

.

The median score is equal to the mode score.

The median score is equal to

7

.

The mode score is equal to

6

.

The mean score is less than the median score.

Show answer & explanation

Answer: The mean score is equal to $7$ .; The median score is equal to the mode score.; The mode score is equal to $6$ .

Answer

The statements asserting that the mean score is equal to 7, the median score is equal to the mode score, and the mode score is equal to 6 are all true.

For the dataset

4, 6, 6, 7, 12

, the mean is

\frac{35}{5} = 7

, the median (middle score) is

6

, and the mode (most frequent score) is

6

. Consequently, the mean equals

7

, the median and mode are equal (

6 = 6

), and the mode is

6

.

Step-by-Step Solution

1

Calculate the mean of the dataset.

Mean =

\frac{4 + 6 + 6 + 7 + 12}{5} = \frac{35}{5} = 7

.

The mean is calculated by dividing the sum of all data points by the total count.

2

Determine the median of the dataset.

Median =

6

.

The data is already arranged in ascending order:

4, 6, 6, 7, 12

. The 3rd value (middle position) is

6

.

3

Determine the mode of the dataset.

Mode =

6

.

The value

6

occurs twice, whereas all other values occur only once.

4

Evaluate each given statement against the calculated metrics.

Mean =

7

, Median =

6

, Mode =

6

. Thus: Mean (

7

) is true; Median = Mode (

6 = 6

) is true; Mode =

6

is true.

Direct comparison verifies which statements hold true.

Key Concept

Calculating and comparing mean, median, and mode for a discrete set of numerical data.

Estimated Time:1m 0s

Question 11Question →

A company has two sales divisions, Division X and Division Y. Division X has 12 representatives with an average monthly sales volume of $\$14,000$ . Division Y has 18 representatives with an average monthly sales volume of $\$24,000$ . If 2 representatives leave Division Y who together accounted for $\$82,000$ in monthly sales, what is the new combined average monthly sales volume, in dollars, for all remaining representatives in both divisions?

Show answer & explanation

Answer: 18500

Answer

The new combined average monthly sales volume for all remaining representatives is $18,500.

To find the combined mean, calculate the sum of all remaining values across both groups (

518,000

) and divide by the total number of remaining items (

28

), yielding

18,500

.

Step-by-Step Solution

1

Find total monthly sales for Division X

12 \times 14,000 = 168,000

The total value of a group is the product of its count and its arithmetic mean.

2

Find total monthly sales for Division Y before representatives left

18 \times 24,000 = 432,000

Multiply the number of representatives in Division Y by their average sales volume.

3

Determine the remaining total sales in Division Y

432,000 - 82,000 = 350,000

Subtract the sales volume of the 2 departing representatives from Division Y's original total.

4

Calculate the total combined sales for both divisions

168,000 + 350,000 = 518,000

Add the total sales of Division X and the remaining sales of Division Y.

5

Calculate total number of remaining representatives across both divisions

12 + (18 - 2) = 28

Division X retains 12 representatives, while Division Y retains 16 representatives.

6

Compute the combined mean sales volume

\frac{518,000}{28} = 18,500

Divide the total combined sales volume by the total number of remaining representatives.

Key Concept

Weighted Mean and Combining Group Data

Question 12Question →

A dataset consists of $9$ distinct positive integers $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9$ listed in strictly increasing order. The mean of the $9$ integers is $28$ , and the median is $24$ . A $10\text{th}$ positive integer $x$ , where $x > a_9$ , is added to the dataset, causing the new mean of the $10$ integers to become $31$ . If $L$ represents the minimum possible value of $a_9$ , what is the value of $x - L$ ?

2

6

21

37

58

Show answer & explanation

Answer: $21$

Answer

The value of

x - L

is

21

.

The sum of the original

9

elements is

9 \times 28 = 252

, and the sum of the

10

elements is

10 \times 31 = 310

, giving

x = 58

. To minimize

a_9

(

L

), the sum of all other elements must be maximized. The median

a_5 = 24

. The maximum possible values for the first four distinct elements below

24

are

20, 21, 22, 23

(summing to

86

). To minimize

a_9

, the upper four elements must be consecutive integers

(a_9 - 3, a_9 - 2, a_9 - 1, a_9)

. Setting the total sum equation

86 + 24 + (4a_9 - 6) = 252

yields

4a_9 = 148

, so

L = 37

. Consequently,

x - L = 58 - 37 = 21

.

Step-by-Step Solution

1

Calculate the sum of the original 9 integers and determine the value of the 10th integer

x

.

The sum of the original

9

integers is

9 \times 28 = 252

. The sum of the

10

integers after adding

x

is

10 \times 31 = 310

. Therefore,

x = 310 - 252 = 58

.

The sum of a set of numbers equals the number of elements multiplied by the mean.

2

Identify the median of the ordered 9-element set.

In an ordered set of

9

elements, the median is the

5\text{th}

element, so

a_5 = 24

.

For an odd number of ordered elements, the median is the exact middle element.

3

Maximize the sum of the first 4 elements

a_1, a_2, a_3, a_4

to minimize

a_9

.

Since all integers are distinct and strictly increasing,

a_4 < 24

. To maximize

a_1 + a_2 + a_3 + a_4

, choose

a_4 = 23, a_3 = 22, a_2 = 21, a_1 = 20

. Their sum is

20 + 21 + 22 + 23 = 86

.

Maximizing lower elements leaves the smallest possible remainder of the total sum for the upper elements.

4

Express

a_6, a_7, a_8

in terms of

a_9

to minimize

a_9

.

To make

a_9

as small as possible,

a_6, a_7, a_8

should be as large as possible relative to

a_9

, meaning they are consecutive integers:

a_8 = a_9 - 1

,

a_7 = a_9 - 2

,

a_6 = a_9 - 3

.

Making elements above the median consecutive integers directly below

a_9

minimizes

a_9

for a fixed sum.

5

Set up the sum equation for the 9 elements to solve for

L = \text{min}(a_9)

.

(a_1 + a_2 + a_3 + a_4) + a_5 + (a_6 + a_7 + a_8 + a_9) = 252 \implies 86 + 24 + (a_9 - 3 + a_9 - 2 + a_9 - 1 + a_9) = 252 \implies 104 + 4a_9 = 252 \implies 4a_9 = 148 \implies a_9 = 37

. Thus,

L = 37

.

Solving the algebraic equation derived from the total sum yields the minimum integer value for

a_9

while satisfying

a_6 = 34 > 24

.

6

Calculate

x - L

.

x - L = 58 - 37 = 21

.

Subtracting the minimum bound

L

from

x

fulfills the target question requirement.

Key Concept

Optimization of Dataset Values using Central Tendency Constraints

Estimated Time:2m 30s

Question 13Question →

A department consisting of $5$ employees has a mean monthly sales total of $\$12{}000$ . If a new employee with a monthly sales total of $\$18{}000$ joins the department, what is the new mean monthly sales total, in dollars, for the $6$ employees?

Show answer & explanation

Answer: 13000

Answer

13000

To find the new mean, multiply the initial mean by the initial number of employees to get total sales:

5 \times 12{}000 = 60{}000

. Add the new employee's sales:

60{}000 + 18{}000 = 78{}000

. Finally, divide by the new total number of employees (

6

) to get

\frac{78{}000}{6} = 13{}000

.

Step-by-Step Solution

1

Find the total sales of the original 5 employees.

5 \times 12{}000 = 60{}000

The sum of values is equal to the mean multiplied by the number of observations.

2

Calculate the total sales for all 6 employees.

60{}000 + 18{}000 = 78{}000

Add the new employee's sales to the initial total.

3

Calculate the new mean sales per employee.

\frac{78{}000}{6} = 13{}000

Divide the combined total sales by the new total number of employees (6).

Key Concept

Mean of Combined Data Sets

Estimated Time:1m 0s

Question 14Question →

A dataset consists of $25$ distinct integers arranged in increasing order. The mean of all $25$ integers is $52$ . The mean of the smallest $12$ integers is $30$ , and the mean of the largest $12$ integers is $70$ .

If $5$ additional numbers, each equal to the median of the original dataset, are added to the dataset, what is the mean of the new set of $30$ numbers?

Show answer & explanation

Answer: 60

Answer

The mean of the new set of 30 numbers is 60.

The sum of all 25 numbers is

25 \times 52 = 1300

. Since the numbers are ordered, the median is the 13th value. The 12 smallest numbers sum to

12 \times 30 = 360

and the 12 largest sum to

12 \times 70 = 840

. The sum of these 24 numbers is

360 + 840 = 1200

, so the 13th number (the median) must be

1300 - 1200 = 100

. Adding 5 copies of 100 increases the sum to

1300 + 500 = 1800

across

30

numbers. The new mean is

\frac{1800}{30} = 60

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset of 25 numbers.

Sum =

25 \times 52 = 1300

.

The mean multiplied by the number of elements gives the total sum.

2

Calculate the combined sum of the 12 smallest and 12 largest integers.

Sum of 12 smallest =

12 \times 30 = 360

; Sum of 12 largest =

12 \times 70 = 840

; Total =

360 + 840 = 1200

.

The 25 numbers consist of the 12 smallest, the 1 median (13th element), and the 12 largest.

3

Determine the value of the median.

Median =

1300 - 1200 = 100

.

Subtracting the sum of the 24 non-median values from the total sum yields the 13th element, which is the median.

4

Find the sum and count of the modified dataset.

New Sum =

1300 + 5(100) = 1800

; New Count =

25 + 5 = 30

.

Adding 5 numbers each equal to 100 increases the sum by 500 and the count by 5.

5

Calculate the mean of the new dataset.

New Mean =

\frac{1800}{30} = 60

.

Divide the new total sum by the new total count of numbers.

Key Concept

Relationship between Mean, Median, and Total Sum in Partitioned Datasets

Estimated Time:2m 30s

Question 15Question →

A software engineer logged the response time, in milliseconds, for seven independent server requests: $8$ , $16$ , $3$ , $21$ , $14$ , $5$ , and $10$ . What is the median response time, in milliseconds, for these seven requests?

8

10

11

14

21

Show answer & explanation

Answer: $10$

Answer

The median response time is

10

milliseconds.

To find the median, first order the seven data points in ascending order:

3, 5, 8, 10, 14, 16, 21

. Since there are

7

values, the median is the middle value at position

\frac{7 + 1}{2} = 4

. The fourth value in this ordered list is

10

.

Step-by-Step Solution

1

Arrange the given dataset in ascending order.

The sorted dataset is

3, 5, 8, 10, 14, 16, 21

.

Finding the median of a numerical dataset requires ordering the values from smallest to largest first.

2

Determine the position of the median element.

For

n = 7

items, the position is

\frac{7 + 1}{2} = 4

th item.

When the number of observations

n

is odd, the median is the exact middle value located at position

\frac{n+1}{2}

.

3

Identify the value at the 4th position.

The 4th value in the sorted list is

10

.

The 4th element in

3, 5, 8, 10, 14, 16, 21

is

10

.

Key Concept

Median of a finite numerical dataset with an odd count

Estimated Time:45s

Question 16Question →

Group A consists of $15$ test scores with an arithmetic mean of $80$ and a median of $82$ . Group B consists of $25$ test scores with an arithmetic mean of $90$ and a median of $86$ . The two groups are combined into a single dataset of $40$ test scores. Which of the following statements MUST be true about the combined dataset? Indicate all such statements.

Select all that apply

The mean of the combined dataset is equal to

86.25

.

The median of the combined dataset is at least

82

.

The median of the combined dataset is at most

86

.

The median of the combined dataset is equal to

84

.

The mode of the combined dataset must lie between

80

and

90

.

Show answer & explanation

Answer: The mean of the combined dataset is equal to $86.25$ .; The median of the combined dataset is at least $82$ .; The median of the combined dataset is at most $86$ .

Answer

The statements asserting that the combined mean is

86.25

, that the combined median is at least

82

, and that the combined median is at most

86

must all be true.

The combined mean is calculated directly from the total sum divided by total elements, giving exactly

86.25

. For the median of

40

combined values (the average of the

20\text{th}

and

21\text{st}

values in sorted order), counting how many values are guaranteed to be above or below specific thresholds shows that at least

21

values are

\ge 82

and at least

21

values are

\le 86

. Therefore, the median must lie within

[82, 86]

.

Step-by-Step Solution

1

Calculate the weighted arithmetic mean for the combined dataset.

Combined Mean =

\frac{15 \times 80 + 25 \times 90}{15 + 25} = \frac{1200 + 2250}{40} = \frac{3450}{40} = 86.25

.

The mean of combined sets is the total sum of all values divided by the total number of items.

2

Analyze the lower bound for the median of the combined set.

At least

21

values in the combined set of

40

items are greater than or equal to

82

.

Since the median of Group A (

15

items) is

82

, at least

8

items are

\ge 82

. Since the median of Group B (

25

items) is

86

, at least

13

items are

\ge 86 \ge 82

. In total,

8 + 13 = 21

items are

\ge 82

. In a sorted list of

40

items, having at least

21

items

\ge 82

implies that the

20\text{th}

and

21\text{st}

items must be

\ge 82

, so the median

\ge 82

.

3

Analyze the upper bound for the median of the combined set.

At least

21

values in the combined set of

40

items are less than or equal to

86

.

Group A has at least

8

items

\le 82 \le 86

. Group B has at least

13

items

\le 86

. Thus, at least

8 + 13 = 21

items are

\le 86

. In a sorted list of

40

items, having at least

21

items

\le 86

ensures the

20\text{th}

and

21\text{st}

items are

\le 86

, so the median

\le 86

.

4

Evaluate the remaining statements regarding the exact median and mode.

Neither the exact median value of

84

nor the existence/location of a mode is guaranteed.

Medians do not combine linearly like means, and mode depends entirely on specific frequency counts which are not given.

Key Concept

Weighted averages and order statistic bounds for combined datasets

Question 17Question →

A dataset $D$ consists of $11$ distinct positive integers. The median of $D$ is $40$ , and the arithmetic mean of $D$ is $45$ . A new dataset $D'$ is created by increasing each of the $5$ largest integers in $D$ by $10$ and decreasing each of the $5$ smallest integers in $D$ by a positive integer $x$ . If the median of $D'$ is strictly less than the arithmetic mean of $D'$ , what is the maximum possible integer value of $x$ ?

20

21

19

25

10

Show answer & explanation

Answer: $20$

Answer

The maximum possible integer value of

x

is

20

.

The sum of the original

11

values is

11 \times 45 = 495

. When the

5

largest integers are each increased by

10

, the sum increases by

+50

. When the

5

smallest integers are each decreased by

x

, the sum decreases by

-5x

. The new sum is

545 - 5x

, making the new mean

\frac{545 - 5x}{11}

. Because

D

has

11

elements, the median is the

6^{\text{th}}

element. Changing the smallest

5

and largest

5

elements does not alter the value of the

6^{\text{th}}

element, so the median remains

40

. Requiring the median to be strictly less than the new mean gives

40 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21

. The greatest integer less than

21

is

20

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset

D

.

Since

D

has

11

elements with a mean of

45

, the sum of elements is

11 \times 45 = 495

.

The mean formula is

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

, so

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

.

2

Determine the median and sum of the modified dataset

D'

.

The median remains the

6^{\text{th}}

element, which is

40

. The sum of

D'

is

495 + 5(10) - 5(x) = 545 - 5x

.

Modifying only the

5

smallest and

5

largest elements leaves the

6^{\text{th}}

central element unchanged.

3

Set up and solve the inequality comparing the median to the mean of

D'

.

Solve

40 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21

.

The problem specifies that the median must be strictly less than the mean.

4

Identify the maximum integer value satisfying the inequality.

The largest integer strictly less than

21

is

20

.

x

must be an integer.

Key Concept

Effect of data modifications on mean and median

Estimated Time:2m 30s

Question 18Question →

A beverage manufacturing company tracks the monthly production counts (in thousands of units) for its facilities located in two separate regions. Region A consists of $6$ facilities with monthly production counts of $12$ , $15$ , $15$ , $18$ , $20$ , and $28$ . Region B consists of $4$ facilities with monthly production counts of $10$ , $14$ , $22$ , and $34$ .

Which of the following statements must be true regarding the individual regional datasets and the combined dataset of all $10$ facilities? Select all such statements.

Select all that apply

The arithmetic mean of the combined dataset of all

10

facilities is greater than the arithmetic mean of Region A.

The median of the combined dataset of all

10

facilities is equal to the median of Region A.

The median of the combined dataset of all

10

facilities is equal to the arithmetic mean of the median of Region A and the median of Region B.

The mode of the combined dataset of all

10

facilities is greater than the mode of Region A.

The arithmetic mean of the combined dataset of all

10

facilities is equal to the simple average of the mean of Region A and the mean of Region B.

Show answer & explanation

Answer: The arithmetic mean of the combined dataset of all $10$ facilities is greater than the arithmetic mean of Region A.; The median of the combined dataset of all $10$ facilities is equal to the median of Region A.

Answer

The correct statements are the one asserting that the combined arithmetic mean is greater than Region A's mean, and the one asserting that the combined median is equal to Region A's median.

The mean of the combined dataset (

18.8

) is greater than the mean of Region A (

18.0

), making the statement comparing the combined mean to Region A's mean correct. Additionally, both Region A's median and the combined dataset's median evaluate to

16.5

, making the statement asserting equality between these two medians correct.

Step-by-Step Solution

1

Calculate the sum, mean, and median for Region A.

Region A sum

= 12 + 15 + 15 + 18 + 20 + 28 = 108

. Mean

= \frac{108}{6} = 18

. Sorted values are

12, 15, 15, 18, 20, 28

, so Median

= \frac{15 + 18}{2} = 16.5

. Mode

= 15

.

Establishing base central metrics for the first subgroup.

2

Calculate the sum, mean, and median for Region B.

Region B sum

= 10 + 14 + 22 + 34 = 80

. Mean

= \frac{80}{4} = 20

. Sorted values are

10, 14, 22, 34

, so Median

= \frac{14 + 22}{2} = 18

.

Establishing base central metrics for the second subgroup.

3

Combine and order all 10 values to calculate combined metrics.

Combined ordered set:

10, 12, 14, 15, 15, 18, 20, 22, 28, 34

. Total sum

= 108 + 80 = 188

. Combined Mean

= \frac{188}{10} = 18.8

. Combined Median

= \frac{15 + 18}{2} = 16.5

. Combined Mode

= 15

.

Necessary to evaluate combined properties accurately against subgroup properties.

4

Evaluate each candidate statement against calculated values.

Combined mean (

18.8

) > Region A mean (

18

) is TRUE. Combined median (

16.5

) = Region A median (

16.5

) is TRUE. Average of medians (

17.25

) = Combined median (

16.5

) is FALSE. Combined mode (

15

) > Region A mode (

15

) is FALSE. Simple average of means (

19

) = Combined mean (

18.8

) is FALSE.

Determining which statements satisfy the required conditions.

Key Concept

Combining Datasets and Weighted Measures of Central Tendency

Question 19Question →

A dataset $S$ consists of $7$ distinct positive integers arranged in ascending order: $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ . The arithmetic mean of the entire dataset is $28$ , and the median is $25$ . The arithmetic mean of the $3$ smallest integers in $S$ is $12$ . If $M$ is the maximum possible value of $x_7$ and $m$ is the minimum possible value of $x_7$ , what is the value of $M - m$ ?

32

36

37

40

52

Show answer & explanation

Answer: 36

Answer

The value of

M - m

is

36

.

The sum of all

7

distinct positive integers is

7 \times 28 = 196

. Since the dataset is ordered and has

7

elements, the median is

x_4 = 25

. The sum of the smallest

3

integers is

3 \times 12 = 36

. Therefore, the sum of the remaining three integers

x_5 + x_6 + x_7 = 196 - 36 - 25 = 135

.

To maximize

x_7

,

x_5

and

x_6

must be as small as possible. Since all elements are distinct integers greater than

x_4 = 25

, the smallest possible values are

x_5 = 26

and

x_6 = 27

. Thus,

M = 135 - 26 - 27 = 82

.

To minimize

x_7

,

x_5

,

x_6

, and

x_7

must be as close together as possible while preserving

25 < x_5 < x_6 < x_7

. Dividing

135

by

3

gives

45

. The consecutive integers centered around

45

are

44, 45, 46

, which sum to

135

and satisfy all inequalities. Thus,

m = 46

.

The difference

M - m = 82 - 46 = 36

, which corresponds to the value

36

.

Step-by-Step Solution

1

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

Total sum =

7 \times 28 = 196

.

The arithmetic mean of

n

numbers is the total sum divided by

n

.

2

Identify the median value and the sum of the smallest 3 integers.

Median

x_4 = 25

, and

x_1 + x_2 + x_3 = 3 \times 12 = 36

.

For an odd number of ordered elements (

7

), the middle term

x_4

is the median. The mean of the first 3 terms gives their sum.

3

Determine the sum of the top 3 integers

(x_5 + x_6 + x_7)

.

x_5 + x_6 + x_7 = 196 - 36 - 25 = 135

.

Subtracting

x_1 + x_2 + x_3

and

x_4

from the total sum leaves the sum of the remaining three elements.

4

Calculate the maximum possible value

M

of

x_7

.

M = 82

.

To maximize

x_7

, minimize

x_5

and

x_6

. Since elements are distinct integers and

x_4 = 25

, the minimum values are

x_5 = 26

and

x_6 = 27

. Thus

x_7 = 135 - 26 - 27 = 82

.

5

Calculate the minimum possible value

m

of

x_7

.

m = 46

.

To minimize

x_7

, maximize

x_5

and

x_6

such that

25 < x_5 < x_6 < x_7

and

x_5 + x_6 + x_7 = 135

. Setting

x_5 = 44, x_6 = 45, x_7 = 46

gives

44 + 45 + 46 = 135

, maintaining strict inequalities.

6

Compute

M - m

.

M - m = 82 - 46 = 36

.

Subtract the minimum possible value of

x_7

from its maximum possible value.

Key Concept

Measures of Central Tendency with Extreme Value Optimization

Estimated Time:2m 30s

Question 20Question →

A company has 8 departments. The dataset of the number of employees in these 8 departments has a median of 42, a range of 25, and a unique mode of 38, which appears exactly 3 times. If no department has more than 55 employees, what is the maximum possible arithmetic mean of the number of employees across all 8 departments?

Show answer & explanation

Answer: 44.25

Answer

The maximum possible arithmetic mean of the number of employees across all 8 departments is 44.25.

To maximize the mean, the sum of the 8 department sizes must be maximized under all given constraints. Ordering the dataset as

x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8

, the median requirement gives

x_4 + x_5 = 84

. Since no element exceeds 55 and the range is 25,

x_1

cannot be 38 because

38 + 25 = 63 > 55

. Hence, the three 38s must be

x_2 = x_3 = x_4 = 38

, which forces

x_5 = 46

. To maximize the sum,

x_8

is set to its maximum limit of 55, forcing

x_1 = 55 - 25 = 30

. Next,

x_7

is set to 55, and

x_6

is set to 54 so that 55 appears only twice and 38 remains the unique mode. The maximum sum is

30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354

, giving a maximum mean of

354 / 8 = 44.25

.

Step-by-Step Solution

1

Order the dataset variables and define constraints.

Let the department sizes be ordered as

x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8

.

Arranging values in ascending order allows direct analysis of median, range, and mode bounds.

2

Use the median to form an equation for the middle two elements.

\frac{x_4 + x_5}{2} = 42 \implies x_4 + x_5 = 84

.

For

n=8

elements, the median is the arithmetic mean of the 4th and 5th terms.

3

Determine the exact position of the three occurrences of 38.

x_2 = x_3 = x_4 = 38

, forcing

x_5 = 84 - 38 = 46

.

If

x_1 = 38

, then

x_8 = 38 + 25 = 63

, exceeding the upper bound of 55. Thus 38 cannot start at

x_1

, so it must occupy

x_2, x_3, x_4

.

4

Maximize the remaining elements

x_1, x_6, x_7, x_8

.

x_8 = 55

,

x_1 = 30

,

x_7 = 55

, and

x_6 = 54

.

To maximize the sum, set

x_8 = 55

, which fixes

x_1 = 55 - 25 = 30

. Set

x_7 = 55

.

x_6

can be at most 54 because setting

x_6 = 55

would give 55 a frequency of 3, violating the unique mode requirement.

5

Calculate the maximum sum and arithmetic mean.

Sum

= 30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354

; Mean

= 354 / 8 = 44.25

.

Dividing the maximum total sum of 354 by 8 gives the maximum possible arithmetic mean.

Key Concept

Optimization of Means Subject to Central Tendency and Range Constraints

Estimated Time:2m 30s

Measures of Central Tendency (Mean, Median, Mode)

Answer

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