Overlapping Sets, Statistics, and Data Distributions Practice Questions — GMAT — Page 2

Question 21Question →

In a financial firm, a total of $100$ portfolio audits were conducted last year. Each audit was classified as High-Risk, International, or both. Exactly $60$ audits were classified as High-Risk, and exactly $50$ audits were classified as International. What was the average (arithmetic mean) duration, in days, of the $100$ portfolio audits?

(1) The average duration of the High-Risk audits was $14$ days, and the average duration of the International audits was $18$ days.
(2) The average duration of the audits classified as both High-Risk and International was $20$ days.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

The correct answer states that both statements together are sufficient, but neither statement alone is sufficient. By set theory, the number of audits in both categories is

60 + 50 - 100 = 10

. The total sum of days across all

100

audits satisfies

S_{\text{total}} = S_{\text{High-Risk}} + S_{\text{International}} - S_{\text{Both}}

. Statement (1) gives

S_{\text{High-Risk}} = 60 \times 14 = 840

and

S_{\text{International}} = 50 \times 18 = 900

, yielding

S_{\text{total}} = 1740 - S_{\text{Both}}

, which is insufficient alone. Statement (2) gives

S_{\text{Both}} = 10 \times 20 = 200

, which is insufficient alone. Combining both gives

S_{\text{total}} = 1740 - 200 = 1540

days, allowing us to find a unique overall mean of

15.4

days.

Step-by-Step Solution

1

Determine set sizes using Principle of Inclusion-Exclusion.

Total

N = 100

.

N(H) = 60

,

N(I) = 50

. Since every audit is in

H

,

I

, or both,

N(H \cup I) = 100 = 60 + 50 - N(H \cap I) \implies N(H \cap I) = 10

.

Identify the count of audits in the overlapping set (both High-Risk and International).

2

Formulate total duration sum in terms of set sums.

S_{\text{total}} = S_H + S_I - S_{H \cap I}

, where

S_H

is sum of High-Risk durations,

S_I

is sum of International durations, and

S_{H \cap I}

is sum of durations in both categories.

Double counting principle applies to the sum of values across sets.

3

Evaluate Statement (1) alone.

S_H = 60 \times 14 = 840

days,

S_I = 50 \times 18 = 900

days.

S_{\text{total}} = 840 + 900 - S_{H \cap I} = 1740 - S_{H \cap I}

.

Since

S_{H \cap I}

is unknown,

S_{\text{total}}

cannot be calculated. Statement (1) alone is insufficient.

4

Evaluate Statement (2) alone.

S_{H \cap I} = 10 \times 20 = 200

days.

Without

S_H

or

S_I

,

S_{\text{total}}

cannot be calculated. Statement (2) alone is insufficient.

5

Evaluate Statements (1) and (2) together.

S_{\text{total}} = 1740 - 200 = 1540

days. Mean duration =

1540 / 100 = 15.4

days.

A unique value for the overall mean is obtained. Both statements together are sufficient.

Key Concept

Overlapping Sets and Weighted Averages (Sum Double Counting)

Estimated Time:2m 0s

Question 22Question →

A financial advisory firm manages a portfolio of $90$ assets, each of which is classified as a Tech asset, a Biotech asset, or both. Exactly $50$ of the assets are Tech assets, and exactly $60$ of the assets are Biotech assets. Is the average (arithmetic mean) value of the Tech-only assets greater than the average value of the Biotech-only assets?

(1) The average value of all $50$ Tech assets is $\$12$ million, and the average value of all $60$ Biotech assets is $\$12$ million.
(2) The average value of the $20$ assets classified as both Tech and Biotech assets is $\$15$ million.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Combining both statements allows us to solve for the exact average value of the Tech-only assets (

10

million) and the Biotech-only assets (

10.5

million). Because we can answer the stem's question with a definitive 'No' (

10

is not greater than

10.5

), both statements together are sufficient.

Step-by-Step Solution

1

Determine the number of assets in each non-overlapping subgroup using set theory.

Total assets

N = 90

, Tech assets

|T| = 50

, Biotech assets

|B| = 60

. Using

|T \cup B| = |T| + |B| - |T \cap B|

, we get

90 = 50 + 60 - |T \cap B| \implies |T \cap B| = 20

. Therefore, Tech-only assets

= 50 - 20 = 30

, Biotech-only assets

= 60 - 20 = 40

, and Both

= 20

.

Rephrasing the question stem into distinct, non-overlapping subset counts simplifies the algebraic system.

2

Evaluate Statement (1) independently.

Let

M_{T}

,

M_{B}

, and

M_{both}

represent the mean values of Tech-only, Biotech-only, and Both subsets respectively. From Statement (1), total value of Tech

= 30 M_T + 20 M_{both} = 50 \times 12 = 600

, and total value of Biotech

= 40 M_B + 20 M_{both} = 60 \times 12 = 720

. Subtracting the equations yields

40 M_B - 30 M_T = 120 \implies 4 M_B - 3 M_T = 12

.

Since

M_T

could be

16

(making

M_B = 15

, so

M_T > M_B

is Yes) or

M_T

could be

8

(making

M_B = 9

, so

M_T > M_B

is No), Statement (1) alone is NOT sufficient.

3

Evaluate Statement (2) independently.

Statement (2) gives

M_{both} = 15

. Without knowing the total asset values or subgroup means, we cannot compare

M_T

and

M_B

.

Statement (2) alone is NOT sufficient.

4

Evaluate Statements (1) and (2) together.

Substituting

M_{both} = 15

into Statement (1) equations:

30 M_T + 20(15) = 600 \implies 30 M_T = 300 \implies M_T = 10

. Similarly,

40 M_B + 20(15) = 720 \implies 40 M_B = 420 \implies M_B = 10.5

. Since

10 > 10.5

is definitively FALSE, we get a clear 'No' response.

In Yes/No Data Sufficiency questions, a definitive 'No' means the combined statements are sufficient.

Key Concept

Data Sufficiency decision logic for overlapping sets combined with weighted averages

Estimated Time:2m 0s

Question 23Question →

In a commercial survey of $120$ retail stores, each store sells products from Brand P, Brand Q, or both. The average (arithmetic mean) monthly revenue of the stores that sell Brand P is $\$40,000$ , and the average monthly revenue of the stores that sell Brand Q is $\$50,000$ . What is the average monthly revenue of all $120$ stores combined?

(1) Exactly $30$ stores sell both Brand P and Brand Q, and their average monthly revenue is $\$60,000$ .
(2) Exactly $40$ stores sell only Brand P.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

The correct response identifies that neither statement alone provides enough information to determine both the exact store counts in each category and the overlap revenue sum, but combining both statements gives a unique system of linear equations that yields an exact combined average revenue of

\frac{\$5,000,000}{120} \approx \$41,666.67

.

Step-by-Step Solution

1

Formulate the total revenue expression in terms of overlapping sets and subgroup means.

Let

n(P)

be the number of stores selling Brand P and

n(Q)

be the number of stores selling Brand Q. Total stores

N = n(P \text{ only}) + n(Q \text{ only}) + n(P \cap Q) = 120

. Total revenue

S_{total} = S(P) + S(Q) - S(P \cap Q) = 40,000 n(P) + 50,000 n(Q) - S(P \cap Q)

.

Stores selling both brands are counted twice when summing

S(P)

and

S(Q)

, so their revenue sum

S(P \cap Q)

must be subtracted once.

2

Evaluate Statement (1) alone.

Statement (1) gives

n(P \cap Q) = 30

and average overlap revenue

= \$60,000 \implies S(P \cap Q) = 30 \times 60,000 = 1,800,000

. Also

n(P) + n(Q) = 120 + 30 = 150

.

Since

n(P)

and

n(Q)

can still vary while their sum is

150

(e.g.,

n(P)=70, n(Q)=80

vs

n(P)=90, n(Q)=60

),

S_{total} = 40,000 n(P) + 50,000 n(Q) - 1,800,000

yields multiple possible total values. Statement (1) alone is INSUFFICIENT.

3

Evaluate Statement (2) alone.

Statement (2) gives

n(P \text{ only}) = 40

. Thus

n(Q) = 120 - 40 = 80

.

No revenue information or overlap count is given. Statement (2) alone is INSUFFICIENT.

4

Evaluate Statements (1) and (2) together.

From Statement (2),

n(P \text{ only}) = 40

. From Statement (1),

n(P \cap Q) = 30

. Thus

n(P) = 40 + 30 = 70

. Since

n(P) + n(Q) = 150

,

n(Q) = 80

.

We now have exact values:

n(P) = 70

,

n(Q) = 80

,

S(P \cap Q) = 1,800,000

.

S_{total} = 40,000(70) + 50,000(80) - 1,800,000 = 2,800,000 + 4,000,000 - 1,800,000 = 5,000,000

. Combined mean revenue

= \frac{5,000,000}{120} = \$41,666.67

. Both statements together are SUFFICIENT.

Key Concept

Weighted averages in overlapping sets requiring exact subgroup counts and sum reconciliations.

Question 24Question →

In a laboratory, $80$ experimental trials were conducted. Each trial was monitored by Sensor P, Sensor Q, or both. The average (arithmetic mean) duration of the trials monitored by Sensor P only was $45$ minutes, and the average duration of the trials monitored by Sensor Q only was $60$ minutes. What was the average duration of all $80$ trials?

(1) Exactly $20$ trials were monitored by Sensor P only, and $30$ trials were monitored by Sensor Q only.
(2) The average duration of all trials monitored by Sensor Q was $55$ minutes, and the average duration of the trials monitored by both sensors was $50$ minutes.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to answer the question, but neither statement alone is sufficient.

Evaluating Statement (1) gives the counts for all three disjoint regions (

20

P-only,

30

Q-only, and

30

Both), but does not provide the mean duration of the

30

trials in the 'Both' category. Evaluating Statement (2) gives the mean duration of the 'Both' category (

50

minutes) and establishes that

n_{B} = n_{Q}

, but does not give absolute counts. Combining both statements provides all subgroup sizes and their respective means, allowing us to calculate the exact weighted average of

52.5

minutes.

Step-by-Step Solution

1

Formulate the algebraic expressions for total trials and total duration.

Let

n_{P}

,

n_{Q}

, and

n_{B}

be the counts for Sensor P only, Sensor Q only, and Both sensors respectively.

n_{P} + n_{Q} + n_{B} = 80

. The total duration is

45 n_{P} + 60 n_{Q} + M_{B} n_{B}

, where

M_{B}

is the mean duration of the trials monitored by both sensors.

Deconstruct the overlapping sets into three mutually exclusive subgroups: P only, Q only, and Both.

2

Evaluate Statement (1) alone.

n_{P} = 20

and

n_{Q} = 30 \implies n_{B} = 80 - 20 - 30 = 30

. Total duration

= 45(20) + 60(30) + M_{B}(30) = 2700 + 30 M_{B}

.

Since the mean duration

M_{B}

for trials monitored by both sensors is unknown, Statement (1) is insufficient.

3

Evaluate Statement (2) alone.

The total group Q consists of Q only (

n_{Q}

, mean

60

) and Both (

n_{B}

, mean

50

). Average for Q is

55 \implies \frac{60 n_{Q} + 50 n_{B}}{n_{Q} + n_{B}} = 55 \implies 5 n_{B} = 5 n_{Q} \implies n_{B} = n_{Q}

.

Statement (2) proves

n_{B} = n_{Q}

, which gives

n_{P} + 2 n_{Q} = 80

, but without specific numbers for

n_{Q}

or

n_{P}

, the overall average cannot be uniquely determined. Thus Statement (2) is insufficient.

4

Evaluate Statements (1) and (2) together.

From (1),

n_{P} = 20

and

n_{Q} = 30

, so

n_{B} = 30

. From (2),

M_{B} = 50

. Total duration

= 45(20) + 60(30) + 50(30) = 900 + 1800 + 1500 = 4200

. Overall average

= \frac{4200}{80} = 52.5

minutes.

Combining both statements yields a unique numeric value for the overall average duration.

Key Concept

Weighted Averages and Overlapping Sets Partitioning

Question 25Question →

A summer academic camp has a total of $120$ enrolled students. Every student participates in at least one of two workshops: Data Analysis or Public Speaking. Exactly $70$ students participate in Data Analysis and exactly $80$ students participate in Public Speaking. What is the average (arithmetic mean) test score of the students who participate ONLY in Public Speaking?

Statement (1): The average test score of all $120$ students in the camp is $84$ .
Statement (2): The average test score of the students who participate in BOTH workshops is $90$ , and the average test score of the students who participate ONLY in Data Analysis is $78$ .

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to uniquely determine the average score of the students participating only in Public Speaking, but neither statement alone is sufficient.

The correct response identifies that both statements together provide enough information to solve for the target average, whereas neither statement alone is sufficient. By deconstructing the overlapping sets into three distinct groups (Data Analysis only = 40, Both = 30, Public Speaking only = 50), the weighted average equation connects the overall average to the three subgroup averages. Statement (1) supplies only the overall average, leaving two unknown subgroup averages. Statement (2) supplies two subgroup averages, leaving the overall average unknown. Combining both statements yields a single equation with only one unknown (

A_{P\text{-only}} = 85.2

), producing a unique numerical solution.

Step-by-Step Solution

1

Rephrase the question stem by calculating the exact number of students in each mutually exclusive subset.

Using the overlapping sets formula

N(D \cup P) = N(D) + N(P) - N(D \cap P)

, we get

120 = 70 + 80 - N(D \cap P)

, so

N(D \cap P) = 30

students in both workshops. Therefore, students in Data Analysis only =

70 - 30 = 40

, students in Public Speaking only =

80 - 30 = 50

, and students in both =

30

.

Deconstructing the overlapping sets into three distinct, non-overlapping groups (Data Analysis only, Both, Public Speaking only) is essential for formulating a weighted average equation.

2

Formulate the weighted average equation relating all subset averages to the total average.

Total Score Sum =

120 \cdot A_{total} = 40 \cdot A_{D\text{-only}} + 30 \cdot A_{both} + 50 \cdot A_{P\text{-only}}

, where

A_{P\text{-only}}

is the target value.

The total sum of all scores is the weighted sum of scores from the three mutually exclusive subsets.

3

Evaluate Statement (1) independently.

Statement (1) gives

A_{total} = 84

. Substituting gives

120(84) = 10080 = 40 \cdot A_{D\text{-only}} + 30 \cdot A_{both} + 50 \cdot A_{P\text{-only}}

. Since

A_{D\text{-only}}

and

A_{both}

remain unknown,

A_{P\text{-only}}

cannot be uniquely solved.

One linear equation with three unknown variables does not yield a unique solution. Statement (1) alone is NOT sufficient.

4

Evaluate Statement (2) independently.

Statement (2) gives

A_{both} = 90

and

A_{D\text{-only}} = 78

. Substituting gives

120 \cdot A_{total} = 40(78) + 30(90) + 50 \cdot A_{P\text{-only}} = 5820 + 50 \cdot A_{P\text{-only}}

. Since

A_{total}

is unknown,

A_{P\text{-only}}

cannot be uniquely solved.

One linear equation with two unknown variables does not yield a unique solution. Statement (2) alone is NOT sufficient.

5

Evaluate Statement (1) and Statement (2) combined.

Combining both statements provides

A_{total} = 84

,

A_{both} = 90

, and

A_{D\text{-only}} = 78

. The equation becomes

120(84) = 40(78) + 30(90) + 50 \cdot A_{P\text{-only}}

, which simplifies to

10080 = 5820 + 50 \cdot A_{P\text{-only}} \implies 4260 = 50 \cdot A_{P\text{-only}} \implies A_{P\text{-only}} = 85.2

.

We have one linear equation with only one unknown variable, giving a single unique numerical value. Both statements together are SUFFICIENT.

Key Concept

Combining overlapping set cardinalities with group weighted averages in Data Sufficiency

Estimated Time:2m 0s

Question 26Question →

A medical research trial evaluated $100$ patients, each of whom experienced side effect A, side effect B, or both. Exactly $65$ patients experienced side effect A, and exactly $55$ patients experienced side effect B. What was the average (arithmetic mean) duration of side effect A, in days, among all patients who experienced side effect A?

(1) The average duration of side effect A for patients who experienced ONLY side effect A was $8$ days.
(2) The average duration of side effect A for patients who experienced BOTH side effects was $12$ days.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

The correct choice indicates that both statements combined provide sufficient data while neither is sufficient on its own. Using the standard inclusion-exclusion principle for two sets,

100 = 65 + 55 - N(\text{Both})

, giving exactly

20

patients with both side effects and

45

patients with only side effect A. Statement (1) provides the sum of durations for the

45

patients (

45 \times 8 = 360

), and Statement (2) provides the sum of durations for the

20

patients (

20 \times 12 = 240

). Neither statement alone gives both sums, but together they yield a total sum of

600

days for

65

patients, resulting in a single definitive mean of

\frac{600}{65}

days.

Step-by-Step Solution

1

Rephrase the question stem using overlapping sets principles.

Let

N(A) = 65

,

N(B) = 55

, and total patients

N(A \cup B) = 100

. Using the set formula

N(A \cup B) = N(A) + N(B) - N(A \cap B)

, we get

100 = 65 + 55 - N(A \cap B)

, so

N(A \cap B) = 20

.

Determining the exact count of patients in the overlap region (

A \cap B

) and the exclusive region (

A \text{ only}

) is necessary to set up the weighted average formula.

2

Calculate the subgroup sizes for patients experiencing side effect A.

Patients with ONLY side effect A =

N(A) - N(A \cap B) = 65 - 20 = 45

. Patients with BOTH side effects =

20

. Total patients with side effect A =

65

.

The total duration of side effect A across all

65

patients is the sum of durations from the

45

'only A' patients and the

20

'both' patients.

3

Evaluate Statement (1) independently.

Statement (1) states that the mean for the

45

'only A' patients is

8

days. The total duration for this group is

45 \times 8 = 360

days. However, the duration for the

20

'both' patients is unknown.

Without the total duration for the overlap group, the overall mean cannot be determined. Thus, Statement (1) alone is NOT sufficient.

4

Evaluate Statement (2) independently.

Statement (2) states that the mean for the

20

'both' patients is

12

days. The total duration for this group is

20 \times 12 = 240

days. However, the duration for the

45

'only A' patients is unknown.

Without the total duration for the exclusive A group, the overall mean cannot be determined. Thus, Statement (2) alone is NOT sufficient.

5

Evaluate Statement (1) and Statement (2) together.

Combining both statements, the total duration for all

65

patients with side effect A is

360 + 240 = 600

days. The overall average duration is

\frac{600}{65} = \frac{120}{13}

days.

A unique numerical value is obtained for the target question, so both statements together are sufficient.

Key Concept

Combining overlapping set subgroup sizes with weighted arithmetic mean formulas to evaluate Data Sufficiency statements.

Estimated Time:2m 0s

Question 27Question →

A dataset consists of $15$ distinct positive integers. Is the median of the dataset greater than its arithmetic mean?

(1) The sum of the $7$ smallest integers in the dataset is $42$ , and the sum of the $7$ largest integers in the dataset is $140$ .
(2) The dataset forms an arithmetic progression.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Answer

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

The option stating that Statement (2) alone is sufficient while Statement (1) alone is not sufficient is correct because Statement (2) establishes that the dataset is an arithmetic progression, where median always equals mean. This provides a definitive 'No' to the question of whether the median is strictly greater than the mean, satisfying sufficiency. Statement (1) allows values of the median both above and below

13

, yielding both 'Yes' and 'No' outcomes.

Step-by-Step Solution

1

Rephrase the question stem mathematically.

Let the 15 distinct positive integers be ordered as

x_1 < x_2 < \dots < x_7 < x_8 < x_9 < \dots < x_{15}

. The median is

x_8

, and the arithmetic mean is

\mu = \frac{\sum_{i=1}^{15} x_i}{15}

. We need to determine whether

x_8 > \mu

.

Simplifying the question target clarifies what information is necessary for sufficiency.

2

Evaluate Statement (1) independently.

Statement (1) gives

\sum_{i=1}^7 x_i = 42

and

\sum_{i=9}^{15} x_i = 140

. The total sum is

42 + x_8 + 140 = 182 + x_8

, so the mean is

\mu = \frac{182 + x_8}{15}

. The inequality

x_8 > \frac{182 + x_8}{15}

simplifies to

14 x_8 > 182

, or

x_8 > 13

. If

x_8 = 10

(with

x_1 \dots x_7 = \{3,4,5,6,7,8,9\}

and

x_9 \dots x_{15} = \{11,12,13,14,15,16,59\}

), then

x_8 \le 13

yields 'No'. If

x_8 = 15

(with

x_1 \dots x_7 = \{1,2,3,4,5,13,14\}

and

x_9 \dots x_{15} = \{16,17,18,19,20,21,29\}

), then

x_8 > 13

yields 'Yes'.

Since Statement (1) allows for both 'Yes' and 'No' answers, it is NOT sufficient.

3

Evaluate Statement (2) independently.

Statement (2) states that the elements form an arithmetic progression. In any set of numbers that forms an arithmetic progression, the distribution is perfectly symmetric, so the arithmetic mean is exactly equal to the median (

x_8 = \mu

). Therefore,

x_8 > \mu

is false, giving a definitive answer of 'No'.

In Data Sufficiency Yes/No questions, a definitive 'No' answer establishes that the statement IS sufficient.

Key Concept

Data Sufficiency evaluation of data distributions, comparing median and arithmetic mean in symmetric versus asymmetric datasets.

Overlapping Sets, Statistics, and Data Distributions

Answer

Step-by-Step Solution

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