Statistics and Data Analysis Practice Questions — GMAT — Page 2

Question 21Question →

In a dataset of $N$ distinct test scores, score $S_1$ is at the 75th percentile and score $S_2$ is at the 40th percentile. A group of 60 new distinct scores is added to the dataset: 12 are strictly less than $S_2$ , 18 are strictly between $S_2$ and $S_1$ , and 30 are strictly greater than $S_1$ . If score $S_1$ is at the 60th percentile of the combined dataset, what is the percentile rank of score $S_2$ in the combined dataset?

16th percentile

25th percentile

28th percentile

40th percentile

46th percentile

Show answer & explanation

Answer: 28th percentile

Answer

28th percentile

In the original dataset of

N

scores,

0.75N

scores are below

S_1

and

0.40N

scores are below

S_2

. Adding 60 scores (12 below

S_2

, 18 between

S_2

and

S_1

, and 30 above

S_1

) increases the total dataset size to

N + 60

. The number of scores below

S_1

becomes

0.75N + 12 + 18 = 0.75N + 30

. Given that

S_1

is at the 60th percentile of the new dataset,

0.75N + 30 = 0.60(N + 60)

, which simplifies to

0.15N = 6

, giving

N = 40

. The original number of scores below

S_2

is

0.40 \times 40 = 16

. Adding the 12 new scores that are below

S_2

gives

16 + 12 = 28

scores below

S_2

in the combined dataset. Out of 100 total scores in the combined dataset, the percentile rank of

S_2

is

\frac{28}{100} \times 100\% = 28\%

, which corresponds to the 28th percentile.

Step-by-Step Solution

1

Set up expressions for the number of scores below

S_1

and

S_2

in the original dataset.

In the original dataset of

N

scores,

0.75N

scores are strictly less than

S_1

, and

0.40N

scores are strictly less than

S_2

.

By definition of percentile rank,

k\%

percentile means

k\%

of the dataset scores fall strictly below that value.

2

Determine the number of scores strictly below

S_1

in the combined dataset.

The total number of new scores added strictly below

S_1

is

12 + 18 = 30

. Thus, the total number of scores below

S_1

in the combined dataset is

0.75N + 30

, while the new total dataset size is

N + 60

.

Scores added below

S_2

and scores added between

S_2

and

S_1

are all strictly less than

S_1

.

3

Solve for

N

using the 60th percentile rank condition for

S_1

in the combined dataset.

0.75N + 30 = 0.60(N + 60) \implies 0.75N + 30 = 0.60N + 36 \implies 0.15N = 6 \implies N = 40

.

Setting the count of scores below

S_1

equal to

60\%

of the new total dataset size

N + 60

forms a single-variable linear equation.

4

Calculate the percentile rank of score

S_2

in the combined dataset.

Original scores below

S_2 = 0.40 \times 40 = 16

. Combined scores below

S_2 = 16 + 12 = 28

. Combined total dataset size =

40 + 60 = 100

. Percentile rank of

S_2 = \frac{28}{100} \times 100\% = 28\%

.

Dividing the total count of scores strictly below

S_2

in the combined set by the total combined dataset size yields the updated percentile rank.

Key Concept

Percentile Rank and Combined Sets

Question 22Question →

A box contains $4$ red balls and $6$ blue balls. If two balls are randomly drawn from the box one after another without replacement, what is the probability that both balls drawn are red?

\frac{4}{25}

\frac{3}{25}

\frac{2}{15}

\frac{1}{3}

\frac{11}{15}

Show answer & explanation

Answer: $\frac{2}{15}$

Answer

The probability that both balls drawn are red is

\frac{2}{15}

.

The probability of selecting a red ball on the first draw is

\frac{4}{10}

. Since the ball is not replaced,

9

total balls remain in the box, of which

3

are red. The probability of selecting a red ball on the second draw given the first was red is

\frac{3}{9}

. By the multiplication rule for dependent events, the probability that both balls drawn are red is

\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}

.

Step-by-Step Solution

1

Find the probability of drawing a red ball on the first pick.

P(\text{First Red}) = \frac{4}{10} = \frac{2}{5}

There are

4

red balls out of a total of

10

balls (

4 + 6 = 10

).

2

Find the conditional probability of drawing a red ball on the second pick after one red ball has been removed.

P(\text{Second Red} \mid \text{First Red}) = \frac{3}{9} = \frac{1}{3}

Because sampling is done without replacement,

1

red ball and

1

total ball are removed, leaving

3

red balls out of

9

remaining balls.

3

Multiply the probabilities of the dependent events to find the joint probability.

P(\text{Both Red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}

For dependent events,

P(A \text{ and } B) = P(A) \times P(B \mid A)

.

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

Question 23Question →

Set $A$ consists of 5 consecutive even integers, and Set $B$ consists of 5 consecutive even integers. The smallest element of Set $B$ is 6 greater than the largest element of Set $A$ . If $S$ represents the standard deviation of the combined set of 10 integers formed by merging Set $A$ and Set $B$ , what is the value of $S^2$ ?

Show answer & explanation

Answer: 57

Answer

The value of

S^2

is 57.

Each subset of 5 consecutive even integers has an internal variance of 8. Because the mean of Set A is

x

and the mean of Set B is

x+14

, the combined mean is

x+7

. Each group mean is 7 units away from the overall mean. By the parallel axis theorem for variance, the combined variance is the within-group variance plus the squared distance from the combined mean:

8 + 7^2 = 57

.

Step-by-Step Solution

1

Define Set A algebraically and calculate its mean and variance.

Set

A = \{x-4, x-2, x, x+2, x+4\}

. Mean

\mu_A = x

, and variance

\sigma_A^2 = \frac{(-4)^2 + (-2)^2 + 0^2 + 2^2 + 4^2}{5} = \frac{40}{5} = 8

.

Symmetrically defining 5 consecutive even integers simplifies variance calculation.

2

Define Set B relative to Set A.

The largest element of Set

A

is

x+4

. The smallest element of Set

B

is

(x+4) + 6 = x+10

. Thus, Set

B = \{x+10, x+12, x+14, x+16, x+18\}

with mean

\mu_B = x+14

and variance

\sigma_B^2 = 8

.

Translating the word problem into exact algebraic representations for Set B.

3

Calculate the mean of the combined 10-element set.

The combined mean

\mu = \frac{5(x) + 5(x+14)}{10} = x+7

.

The combined mean is essential to evaluate overall variation.

4

Calculate the variance

S^2

of the combined set.

Using the combined variance formula

S^2 = \frac{1}{2}[\sigma_A^2 + (\mu_A - \mu)^2] + \frac{1}{2}[\sigma_B^2 + (\mu_B - \mu)^2] = \frac{1}{2}[8 + (-7)^2] + \frac{1}{2}[8 + 7^2] = 8 + 49 = 57

.

Combines internal set variances with the variance between group means.

Key Concept

Combined Variance and Standard Deviation of Sets

Question 24Question →

A data set consists of $8$ positive integers with an arithmetic mean of $15$ and a unique mode of $12$ . What is the maximum possible value of an integer in this data set?

Show answer & explanation

Answer: 81

Answer

81

The total sum of the

8

integers is

8 \times 15 = 120

. To maximize the single largest integer, the sum of the remaining

7

integers must be minimized. Since

12

is the unique mode,

12

must appear at least twice, and no other value can appear more than once. The smallest positive integers that can occupy the first

5

positions without creating another mode are

1, 2, 3, 4,

and

5

. Placing

12

in the

6\text{th}

and

7\text{th}

positions minimizes the sum of the first

7

elements to

1 + 2 + 3 + 4 + 5 + 12 + 12 = 39

. Thus, the maximum possible value for the largest integer is

120 - 39 = 81

.

Step-by-Step Solution

1

Calculate the total sum of the 8 positive integers.

Total Sum =

8 \times 15 = 120

.

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

2

Minimize the sum of the first 7 integers to maximize the 8th integer.

Minimum sum of the first 7 integers =

1 + 2 + 3 + 4 + 5 + 12 + 12 = 39

.

To maximize the largest integer, the remaining 7 integers must be as small as possible. Since 12 is the unique mode, 12 must appear at least twice, and no other integer can appear more than once. Placing two 12s at the highest available positions among the 7 terms (

a_6 = 12

and

a_7 = 12

) and selecting the smallest distinct positive integers (

1, 2, 3, 4, 5

) for the first 5 terms minimizes their total sum.

3

Subtract the minimum sum of the 7 smallest terms from the total sum.

Maximum integer =

120 - 39 = 81

.

Subtracting the smallest possible sum of 7 terms from the fixed sum of 120 yields the maximum possible value for the 8th term.

Key Concept

Maximizing an element in a data set under mean and mode constraints

Question 25Question →

At a semiconductor manufacturing facility, silicon wafers are produced across three consecutive shifts: Shift X, Shift Y, and Shift Z. The ratio of the number of wafers produced in Shift X to Shift Y is $3 : 2$ , and the ratio of the number of wafers produced in Shift Y to Shift Z is $4 : 5$ . The defect rate of wafers produced in Shift X is $2.0\%$ , and the defect rate of wafers produced in Shift Y is $3.5\%$ . If the combined defect rate for all wafers produced across the three shifts is $3.0\%$ , what is the defect rate, expressed as a percentage, of the wafers produced in Shift Z?

Show answer & explanation

Answer: 3.8

Answer

The defect rate of the wafers produced in Shift Z is 3.8%.

To find the defect rate of Shift Z, first determine the combined ratio of wafer production for Shift X, Shift Y, and Shift Z. Given Shift X : Shift Y = 3 : 2 and Shift Y : Shift Z = 4 : 5, scale Shift X : Shift Y to 6 : 4 so that Shift Y has the same ratio value in both expressions. The unified ratio is Shift X : Shift Y : Shift Z = 6 : 4 : 5, representing 15 total parts. Applying the weighted average formula gives (6 * 2.0% + 4 * 3.5% + 5 * r_Z) / 15 = 3.0%. Multiplying both sides by 15 yields 12.0 + 14.0 + 5 * r_Z = 45.0, which simplifies to 5 * r_Z = 19.0. Dividing by 5 gives r_Z = 3.8%.

Step-by-Step Solution

1

Find the combined ratio of production volumes across Shift X, Shift Y, and Shift Z.

Shift X : Shift Y = 3 : 2 = 6 : 4, and Shift Y : Shift Z = 4 : 5. Therefore, Shift X : Shift Y : Shift Z = 6 : 4 : 5.

Combining the separate ratios into a single compound ratio establishes the weighting factor for each shift.

2

Calculate the total parts and weight of each shift.

Total parts = 6 + 4 + 5 = 15 parts.

The weight of each shift in the weighted average corresponds to its share of the total 15 parts.

3

Formulate the weighted average equation for the overall defect rate.

(6 * 2.0 + 4 * 3.5 + 5 * r_Z) / 15 = 3.0

The overall combined defect rate is the weighted sum of individual defect rates divided by the total number of parts.

4

Solve for the unknown defect rate r_Z of Shift Z.

12.0 + 14.0 + 5 * r_Z = 45.0 => 26.0 + 5 * r_Z = 45.0 => 5 * r_Z = 19.0 => r_Z = 3.8

Algebraic simplification yields the exact defect rate for Shift Z.

Key Concept

Weighted Average of Combined Sets

Estimated Time:2m 0s

Question 26Question →

A project manager must schedule 5 distinct client presentations—for clients A, B, C, D, and E—on 5 consecutive days from Monday through Friday, with exactly one presentation per day. If the presentations for client A and client B cannot be scheduled on consecutive days, how many different presentation schedules are possible?

Show answer & explanation

Answer: 72

Answer

The total number of valid presentation schedules is 72.

The correct result is found by subtracting the number of restricted arrangements (where presentation A and presentation B are scheduled on consecutive days) from the total number of unrestricted arrangements of 5 presentations. The total unrestricted arrangements equal

5! = 120

. Treating A and B as a single block leaves 4 items to arrange in

4! = 24

ways, with

2! = 2

internal orderings for A and B, yielding

24 \times 2 = 48

consecutive schedules. Subtracting 48 from 120 results in 72 valid schedules.

Step-by-Step Solution

1

Calculate total unrestricted linear arrangements of the 5 presentations.

5! = 120

Without restrictions, 5 distinct items can be arranged in 5 distinct positions in

5!

ways.

2

Calculate the number of invalid arrangements where presentations A and B are on consecutive days.

2! \times 4! = 48

Grouping A and B into a single unit results in 4 items to arrange (

4! = 24

), and A and B can swap positions inside the block in

2! = 2

ways.

3

Subtract the invalid arrangements from total arrangements.

120 - 48 = 72

Complementary counting gives the total number of arrangements where A and B are not on consecutive days.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)

Question 27Question →

A game features two boxes of tokens. Box X contains $2$ red tokens and $3$ blue tokens. Box Y contains $4$ red tokens and $1$ blue token. A fair coin is flipped to determine which box to draw from: if the coin lands on heads, Box X is chosen; if it lands on tails, Box Y is chosen. Two tokens are then drawn sequentially without replacement from the chosen box. What is the probability that both drawn tokens are red?

Show answer & explanation

Answer: 0.35

Answer

The probability that both drawn tokens are red is 0.35.

The total probability combines the independent choice of box with dependent draws without replacement. Box X yields two red tokens with probability

0.10

, and Box Y yields two red tokens with probability

0.60

. Weighting each by the

0.5

probability of choosing that box gives

(0.5 \times 0.10) + (0.5 \times 0.60) = 0.35

.

Step-by-Step Solution

1

Determine the conditional probability of drawing two red tokens from Box X without replacement.

The probability is

\frac{2}{5} \times \frac{1}{4} = 0.10

.

Because draws are dependent (without replacement), the number of remaining red tokens decreases to 1 and total tokens to 4 after the first red draw.

2

Determine the conditional probability of drawing two red tokens from Box Y without replacement.

The probability is

\frac{4}{5} \times \frac{3}{4} = 0.60

.

Box Y initially contains 4 red out of 5 total tokens; drawing one red leaves 3 red out of 4 total tokens.

3

Combine the independent box selection probabilities with the dependent drawing probabilities.

Overall probability is

(0.5 \times 0.10) + (0.5 \times 0.60) = 0.05 + 0.30 = 0.35

.

The initial coin flip selects Box X or Box Y with equal, independent probability of 0.5.

Key Concept

Combining independent events (environment selection) with dependent events (sampling without replacement) using the Law of Total Probability.

Estimated Time:2m 0s

Question 28Question →

A crate in a warehouse contains $15$ functional electronic components and $10$ defective components. If two components are selected at random from the crate one after another without replacement, what is the probability that both selected components are functional?

Show answer & explanation

Answer: 0.35

Answer

The probability that both selected components are functional is 0.35.

Because selection is made without replacement, the outcome of the second draw depends on the outcome of the first draw. The probability of selecting a functional component first is 15/25 = 3/5. With 14 functional components left among 24 total components, the probability of selecting a second functional component is 14/24 = 7/12. Multiplying these probabilities gives (3/5) * (7/12) = 21/60 = 0.35.

Step-by-Step Solution

1

Calculate the probability of drawing a functional component on the first selection.

P(First Functional) = 15 / 25 = 3/5 = 0.6.

There are 15 functional components in the total pool of 25 components.

2

Calculate the conditional probability of drawing a second functional component given that the first component drawn was functional.

P(Second Functional | First Functional) = 14 / 24 = 7/12.

Because sampling is without replacement, the total count decreases to 24 and the functional count decreases to 14.

3

Multiply the two dependent probabilities to find the overall joint probability.

P(Both Functional) = (3/5) * (7/12) = 21/60 = 0.35.

By the multiplication rule for dependent events, P(A and B) = P(A) * P(B|A).

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

Question 29Question →

During a quality control inspection, five manufactured items were measured and found to have lengths of $44\text{ mm}$ , $48\text{ mm}$ , $50\text{ mm}$ , $52\text{ mm}$ , and $56\text{ mm}$ . What is the standard deviation, in millimeters, of the lengths of these five items?

Show answer & explanation

Answer: 4

Answer

The standard deviation of the lengths of the five items is 4 millimeters.

The arithmetic mean of the five measurements is 50 mm. The sum of the squared deviations from 50 is 36 + 4 + 0 + 4 + 36 = 80. Dividing 80 by 5 yields a variance of 16. Taking the principal square root of 16 gives a standard deviation of 4 mm.

Step-by-Step Solution

1

Calculate the arithmetic mean of the dataset

Mean = 50

The standard deviation measures dispersion relative to the mean.

2

Find the squared difference of each data point from the mean

Squared deviations are 36, 4, 0, 4, and 36

Squaring ensures all deviations are non-negative and penalizes larger deviations.

3

Compute the mean of the squared deviations (variance)

Variance = 80 / 5 = 16

Variance is the average squared distance from the mean.

4

Take the non-negative square root of the variance

Standard deviation = sqrt(16) = 4

Standard deviation converts variance back to the original unit of measurement.

Key Concept

Standard Deviation Calculation for a Data Set

Question 30Question →

A real estate developer purchased two tracts of land. Tract 1 consists of $15$ acres purchased at an average cost of $\$24,000$ per acre. Tract 2 consists of $35$ acres purchased at an average cost of $\$40,000$ per acre. What is the average cost per acre, in thousands of dollars, for the combined land purchase?

Show answer & explanation

Answer: 35.2

Answer

The average cost per acre for the combined land purchase is

35.2

thousand dollars (or

\$35,200

).

The overall weighted average is calculated by dividing the sum of all individual costs by the total number of acres across both tracts. Since

15

acres cost

\$24,000

per acre and

35

acres cost

\$40,000

per acre, the total cost is

15(24,000) + 35(40,000) = 360,000 + 1,400,000 = 1,760,000

dollars. Dividing by the total of

50

acres yields

\$35,200

per acre, or

35.2

thousand dollars.

Step-by-Step Solution

1

Determine the total financial expenditure for each tract

Tract 1 cost =

15 \times 24,000 = \$360,000

; Tract 2 cost =

35 \times 40,000 = \$1,400,000

Total value of a set equals the number of items multiplied by the mean of the items.

2

Calculate total combined cost and total combined acreage

Total cost =

\$360,000 + \$1,400,000 = \$1,760,000

; Total acres =

15 + 35 = 50

Combined sets require summing all total values and all individual counts.

3

Compute the weighted average cost per acre

Weighted Mean =

\frac{1,760,000}{50} = 35,200

dollars, which is

35.2

thousand dollars.

Divide total combined cost by total combined number of units.

Key Concept

Weighted Average of Combined Sets

Question 31Question →

A set of $5$ integers has an arithmetic mean of $12$ . If four of the integers are $8$ , $10$ , $14$ , and $15$ , what is the value of the fifth integer?

Show answer & explanation

Answer: 13

Answer

The value of the fifth integer is

13

.

The total sum of a set of numbers is given by the formula

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

. For

5

numbers with a mean of

12

, the total sum is

5 \times 12 = 60

. The sum of the four provided numbers is

8 + 10 + 14 + 15 = 47

. Subtracting

47

from

60

yields

13

, which is the value of the fifth integer.

Step-by-Step Solution

1

Find the total sum of the 5 integers

The total sum is

60

The sum of a set of numbers equals the arithmetic mean multiplied by the total count of numbers (

12 \times 5 = 60

).

2

Sum the four given integers

The sum of the four integers is

47

Adding the given numbers:

8 + 10 + 14 + 15 = 47

.

3

Subtract the sum of the known integers from the total sum

The fifth integer is

13

Subtracting

47

from

60

gives

60 - 47 = 13

.

Key Concept

Arithmetic Mean

Estimated Time:45s

Question 32Question →

A technology firm consists of two divisions: Division X and Division Y. Division X has $20$ employees with an average monthly salary of $\$5{}000$ , while Division Y has $30$ employees with an average monthly salary of $\$6{}000$ . What is the average monthly salary of all $50$ employees combined?

\$5{}400

\$5{}500

\$5{}600

\$5{}700

\$5{}800

Show answer & explanation

Answer: $\$5{}600$

Answer

The combined average monthly salary for all

50

employees is

\$5{}600

.

The total salary for Division X is

20 \times \$5{}000 = \$100{}000

and for Division Y is

30 \times \$6{}000 = \$180{}000

. Combining both gives a total payout of

\$280{}000

across

50

employees. Dividing

\$280{}000

by

50

yields an exact weighted average of

\$5{}600

.

Step-by-Step Solution

1

Calculate total monthly salary expenditure for Division X

20 \times \$5{}000 = \$100{}000

Total value of a set equals the number of items times the average value per item.

2

Calculate total monthly salary expenditure for Division Y

30 \times \$6{}000 = \$180{}000

Total value of a set equals the number of items times the average value per item.

3

Sum total monthly salary expenditures and total number of employees

Total salary =

\$100{}000 + \$180{}000 = \$280{}000

; Total employees =

20 + 30 = 50

Combined set averages require combined total sum divided by combined total quantity.

4

Compute the combined weighted average salary

\frac{\$280{}000}{50} = \$5{}600

Weighted average formula:

\text{Weighted Average} = \frac{\text{Total Sum}}{\text{Total Count}}

.

Key Concept

Weighted Average of Combined Sets

Estimated Time:1m 0s

Question 33Question →

A data set $S$ consists of $9$ positive integers arranged in non-decreasing order: $x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8 \le x_9$ . The arithmetic mean of set $S$ is $22$ , the median is $20$ , and the set has a unique mode of $15$ . If the range of set $S$ is $30$ , what is the maximum possible value of the largest element, $x_9$

31

40

45

50

52

Show answer & explanation

Answer: 45

Answer

The maximum possible value of the largest element is 45.

To maximize the largest value in a set with a fixed range of 30, we must maximize the smallest value because the maximum value equals the minimum value plus 30. Since 15 is the unique mode of the ordered set, 15 must be an element of the set, which restricts the minimum value to at most 15. Setting the smallest element to 15 allows the largest element to reach 15 + 30 = 45, which can be verified to satisfy all mean, median, and mode constraints.

Step-by-Step Solution

1

Relate total sum, range, and median to the elements of the set.

The sum of all 9 elements is

9 \times 22 = 198

. For a 9-element set ordered as

x_1 \le x_2 \le \dots \le x_9

, the median is the 5th element

x_5 = 20

. The range is

x_9 - x_1 = 30

, so

x_9 = x_1 + 30

.

Establishing the basic relationships among the statistical measures given in the problem.

2

Determine the upper bound for the smallest element

x_1

.

Because

15

is the unique mode of the set,

15

must be present in the set. Since the set is in non-decreasing order,

x_1 \le 15

.

If

x_1 > 15

, then no element in the set could equal

15

, violating the condition that

15

is the mode.

3

Maximize

x_9

using the relationship

x_9 = x_1 + 30

.

To maximize

x_9

,

x_1

must be as large as possible. The maximum possible value for

x_1

is

15

, which gives

x_9 = 15 + 30 = 45

.

Connecting the range equation to the upper bound on

x_1

.

4

Verify that a valid set exists for

x_1 = 15

and

x_9 = 45

.

Consider the set

\{15, 15, 15, 15, 20, 20, 25, 28, 45\}

. Sum

= 198

, Mean

= 22

, Median

= 20

(5th element), Mode

= 15

(frequency 4), Range

= 45 - 15 = 30

. All conditions are satisfied.

Ensuring the upper bound is achievable under all given statistical constraints.

Key Concept

Interplay between statistical measures (mean, median, mode, range) and set boundaries

Question 34Question →

The scores of a student on five quizzes are $72, 85, 90, 78,$ and $95$ . What is the positive difference between the median and the arithmetic mean of these five quiz scores?

Show answer & explanation

Answer: 1

Answer

The positive difference between the median and the arithmetic mean of the five quiz scores is

1

.

To find the arithmetic mean, sum all five scores (

72 + 85 + 90 + 78 + 95 = 420

) and divide by

5

, yielding

84

. To find the median, list the scores in ascending order (

72, 78, 85, 90, 95

); the 3rd score is

85

. The positive difference between the median (

85

) and the arithmetic mean (

84

) is

1

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the given set of scores.

The sum of the scores is

72 + 78 + 85 + 90 + 95 = 420

. Dividing by 5 gives an arithmetic mean of

84

.

The mean is defined as the total sum of all values divided by the number of values.

2

Order the scores in ascending order to identify the median.

The ordered list is

72, 78, 85, 90, 95

. The middle term (the 3rd value) is

85

.

The median of a set with an odd number of elements is the middle value when arranged in numerical order.

3

Subtract the arithmetic mean from the median to find the positive difference.

85 - 84 = 1

.

The question asks for the positive difference between the median and the arithmetic mean.

Key Concept

Calculating and comparing the arithmetic mean and median of a finite numerical data set.

Estimated Time:1m 0s

Question 35Question →

In a certain training program, Class A has $15$ participants with an average score of $80$ on a final exam, and Class B has $25$ participants with an average score of $88$ on the same exam. What is the average score for all $40$ participants combined?

Show answer & explanation

Answer: 85

Answer

The combined average score for all 40 participants is 85.

To find the combined average of two groups of different sizes, sum the total values of both groups (

15 \times 80 + 25 \times 88 = 3,400

) and divide by the total number of participants (

15 + 25 = 40

). This gives a weighted average of

85

.

Step-by-Step Solution

1

Find total score of Class A

1,200 points

Multiply the number of participants in Class A by their average score.

2

Find total score of Class B

2,200 points

Multiply the number of participants in Class B by their average score.

3

Find the total combined score

3,400 points

Add the total points from Class A and Class B together.

4

Calculate the weighted average score

85

Divide the total combined score by the total number of participants (15 + 25 = 40).

Key Concept

Weighted Average of Combined Sets

Estimated Time:1m 0s

Question 36Question →

In a class of $40$ students, each student received a distinct score on a final examination. If a particular student's score is at the $85\text{th}$ percentile of the class scores, how many students scored lower than this student?

34

85

6

35

33

Show answer & explanation

Answer: $34$

Answer

34 students scored lower than this student.

The 85th percentile of a group of 40 students means that 85% of the class scored below this student. Calculating

85\%

of

40

yields

0.85 \times 40 = 34

. Therefore, exactly 34 students scored lower.

Step-by-Step Solution

1

Understand the definition of percentile rank.

Being at the

85\text{th}

percentile means that the student's score is strictly higher than

85\%

of the other scores in the dataset.

By definition, the

p\text{th}

percentile of a set of distinct values separates the lower

p\%

of the data from the rest.

2

Calculate

85\%

of the total number of students (

40

).

0.85 \times 40 = 34

.

Multiplying the percentage by the sample size gives the exact count of students below the percentile threshold.

Key Concept

Percentile Rank

Question 37Question →

A data set $S$ consists of $11$ positive integers. The set has an arithmetic mean of $15$ , a median of $14$ , and a unique mode of $18$ . If $M$ is the maximum possible value of an element in $S$ and $m$ is the minimum possible value of an element in $S$ , what is the maximum possible value of the range $M - m$ ?

68

71

72

73

74

Show answer & explanation

Answer: $73$

Answer

The maximum possible value of the range

M - m

is

73

.

To maximize the range

M - m = x_{11} - x_1

, we minimize

x_1

and maximize

x_{11}

. Since elements are positive integers,

m = x_1 = 1

. The sum of all

11

elements is

11 \times 15 = 165

. With median

x_6 = 14

, the unique mode

18

must lie above the median. Setting the frequency of

18

to

3

allows other numbers to appear up to

2

times. Minimizing

x_1 \dots x_5

gives

1, 1, 2, 2, 3

(sum

9

). Minimizing

x_7 \dots x_{10}

by setting

x_7 = 14

and

x_8 = x_9 = x_{10} = 18

gives sum

68

. The sum of the first

10

terms is

9 + 14 + 68 = 91

, leaving

M = x_{11} = 165 - 91 = 74

. Thus, the maximum range is

74 - 1 = 73

.

Step-by-Step Solution

1

Calculate the sum of all elements in the set.

Sum

= 11 \times 15 = 165

.

The arithmetic mean of

11

elements is

15

.

2

Identify the position of the median and order the elements.

Let elements be

x_1 \le x_2 \le \dots \le x_6 \le \dots \le x_{11}

, where

x_6 = 14

.

For an odd number of elements (

11

), the median is the 6th element.

3

Determine the mode frequency constraint to maximize

x_{11} - x_1

.

The unique mode is

18

. If

18

appears

3

times, the maximum frequency of any other value is

2

.

To maximize

x_{11}

, we minimize

x_1 \dots x_{10}

. Allowing the mode to appear

3

times permits other elements to appear up to

2

times.

4

Minimize the sum of the lower five elements

x_1, x_2, x_3, x_4, x_5

.

x_1 = 1, x_2 = 1, x_3 = 2, x_4 = 2, x_5 = 3

, giving sum

= 1 + 1 + 2 + 2 + 3 = 9

.

The smallest positive integers with maximum frequency

2

are

1, 1, 2, 2, 3

.

5

Minimize elements

x_7, x_8, x_9, x_{10}

.

x_7 = 14

(frequency

2

for

14

), and

x_8 = 18, x_9 = 18, x_{10} = 18

. Sum

= 14 + 18 + 18 + 18 = 68

.

Since

x_6 = 14

, setting

x_7 = 14

minimizes

x_7

while keeping the frequency of

14

at

2

, which is less than the mode frequency of

3

.

6

Calculate the maximum value

M = x_{11}

and the range

M - m

.

Sum of first 10 elements

= 9 + 14 + 68 = 91

. Thus

x_{11} = 165 - 91 = 74

. Range

= 74 - 1 = 73

.

Subtracting the sum of the first 10 elements from the total sum gives

M

, and subtracting

m = 1

yields the range.

Key Concept

Range, Mean, Median, and Mode Constraints in Data Sets

Question 38Question →

A dataset consists of five identical integers: $12, 12, 12, 12,$ and $12$ . What is the standard deviation of this dataset?

0

12

-12

144

2\sqrt{3}

Show answer & explanation

Answer: $0$

Answer

0

The correct value is

0

. Standard deviation measures how spread out the numbers in a dataset are around the mean. Since every value in the set is

12

, the mean is

12

and every value lies exactly on the mean (distance of

0

). Thus, the standard deviation is

0

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the dataset

Mean

\mu = \frac{12 + 12 + 12 + 12 + 12}{5} = 12

The mean is needed to measure individual deviations.

2

Calculate the deviation of each data point from the mean

Deviations are

12 - 12 = 0

for all five elements

Standard deviation measures the average distance of data points from the mean.

3

Compute the standard deviation

Standard Deviation

SD = \sqrt{\frac{0^2 + 0^2 + 0^2 + 0^2 + 0^2}{5}} = 0

When all data points are identical, the spread around the mean is zero.

Key Concept

Standard deviation measures the dispersion or spread of a set of numbers around its mean. If all elements in a set are identical, there is no variation, so both the range and the standard deviation are equal to 0.

Question 39Question →

A board of directors consists of $7$ members. In how many different ways can a subcommittee of $3$ members be chosen from the board?

Show answer & explanation

Answer: 35

Answer

35 different subcommittees can be chosen.

The total number of ways to choose a subcommittee of 3 members from a group of 7 members when order does not matter is given by the combination formula

\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Step-by-Step Solution

1

Determine whether the selection requires permutations or combinations.

Since selecting members A, B, and C forms the exact same committee as selecting B, C, and A, the order of selection does not matter. Therefore, this is a combination problem.

Committees are unordered groups.

2

Calculate the number of combinations of 7 items taken 3 at a time using

\binom{7}{3} = \frac{7!}{3!(7-3)!}

.

\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Simplify the factorial expression by canceling out common terms.

Key Concept

Combinations (Selection Without Regard to Order)

Question 40Question →

A dataset consists of 20 distinct numerical values arranged in increasing order. The lower quartile ( $Q_1$ ) of the dataset is equal to the average of the 5th and 6th values, and the upper quartile ( $Q_3$ ) is equal to the average of the 15th and 16th values. Given that the lower quartile $Q_1 = 42$ , the interquartile range ( $\text{IQR} = Q_3 - Q_1$ ) is 38, and the 15th value in the dataset is 74, what is the value of the 16th term?

Show answer & explanation

Answer: 86

Answer

86

To find the 16th value, first determine the upper quartile

Q_3

using the interquartile range:

Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80

. Since

Q_3

is the average of the 15th and 16th terms, set up the equation

\frac{74 + x_{16}}{2} = 80

. Multiplying by 2 gives

74 + x_{16} = 160

, so

x_{16} = 86

.

Step-by-Step Solution

1

Calculate the upper quartile (

Q_3

) using the given lower quartile (

Q_1

) and interquartile range (IQR).

Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80

By definition, the interquartile range is the difference between the upper and lower quartiles (

\text{IQR} = Q_3 - Q_1

).

2

Relate the upper quartile (

Q_3

) to the 15th and 16th values of the ordered dataset.

Q_3 = \frac{x_{15} + x_{16}}{2} = 80

For an ordered dataset of 20 elements,

Q_3

is the median of the upper half of the data (the 11th through 20th terms), which equals the arithmetic mean of the 15th and 16th terms.

3

Substitute the known value of the 15th term (

x_{15} = 74

) into the equation and solve for the 16th term (

x_{16}

).

\frac{74 + x_{16}}{2} = 80 \implies 74 + x_{16} = 160 \implies x_{16} = 86

Multiplying both sides by 2 gives 160, and subtracting 74 yields 86.

Key Concept

Interquartile Range and Quartile Calculations

Statistics and Data Analysis

Answer

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