Data Analysis Practice Questions — GRE General Test — Page 5

Question 81Question →

A technology institute surveyed a cohort of $300$ software engineers regarding their proficiency in three programming paradigms: Functional ( $F$ ), Object-Oriented ( $O$ ), and Concurrent ( $C$ ). Every surveyed engineer is proficient in at least one of these three paradigms. The ratio of the total number of engineers proficient in $F$ , $O$ , and $C$ is $5 : 6 : 4$ , respectively. Furthermore, exactly $20\%$ of the engineers proficient in $F$ are proficient in all three paradigms. If exactly $54$ engineers are proficient in both $F$ and $O$ , $48$ are proficient in both $O$ and $C$ , and $30$ are proficient in both $F$ and $C$ , how many engineers in the cohort are proficient in exactly one programming paradigm?

96

141

222

273

78

Show answer & explanation

Answer: 222

Answer

222 engineers are proficient in exactly one programming paradigm.

Using the Principle of Inclusion-Exclusion for three sets,

|F \cup O \cup C| = |F| + |O| + |C| - (|F \cap O| + |O \cap C| + |F \cap C|) + |F \cap O \cap C|

. Substituting

|F|=5k

,

|O|=6k

,

|C|=4k

,

|F \cap O \cap C|=k

, and the total cohort

300

gives

300 = 16k - 132

, yielding

k = 27

. Consequently, the triple intersection is

27

. Decomposing into disjoint regions:

F

only

= 78

,

O

only

= 87

, and

C

only

= 57

. Summing these gives

78 + 87 + 57 = 222

.

Step-by-Step Solution

1

Set up algebraic representations for the set sizes using the given ratio.

Let

|F| = 5k

,

|O| = 6k

, and

|C| = 4k

for some positive constant

k

.

The total proficiencies follow the ratio

5:6:4

.

2

Express the triple intersection

|F \cap O \cap C|

in terms of

k

.

|F \cap O \cap C| = 0.20 \times |F| = 0.20 \times 5k = k

.

Exactly 20% of engineers proficient in

F

are proficient in all three paradigms.

3

Apply the Principle of Inclusion-Exclusion for three sets to solve for

k

.

|F \cup O \cup C| = |F| + |O| + |C| - (|F \cap O| + |O \cap C| + |F \cap C|) + |F \cap O \cap C| \implies 300 = 5k + 6k + 4k - (54 + 48 + 30) + k \implies 300 = 16k - 132 \implies 16k = 432 \implies k = 27

.

Every engineer is proficient in at least one paradigm, so

|F \cup O \cap C| = 300

.

4

Calculate the total size of each set and each exclusive intersection region.

|F| = 135

,

|O| = 162

,

|C| = 108

, and

|F \cap O \cap C| = 27

.
Exclusively

F \cap O = 54 - 27 = 27

.
Exclusively

O \cap C = 48 - 27 = 21

.
Exclusively

F \cap C = 30 - 27 = 3

.

Subtracting the triple intersection from pairwise intersections yields the two-set-only regions.

5

Determine the number of engineers proficient in exactly one paradigm.

Only

F = 135 - (27 + 3 + 27) = 78

.
Only

O = 162 - (27 + 21 + 27) = 87

.
Only

C = 108 - (3 + 21 + 27) = 57

.
Total exactly one =

78 + 87 + 57 = 222

.

Subtracting all overlapping regions from each total set size gives the single-category populations.

Key Concept

Three-set Principle of Inclusion-Exclusion and Venn Diagram region decomposition.

Question 82Question →

A survey of $300$ urban commuters evaluated their usage of three transit services: the Bus ( $B$ ), the Commuter Rail ( $R$ ), and the Express Ferry ( $F$ ). The survey revealed the following information:

- $160$ commuters use the Bus.
- $140$ commuters use the Commuter Rail.
- $110$ commuters use the Express Ferry.
- $60$ commuters use both the Bus and the Commuter Rail.
- $45$ commuters use both the Commuter Rail and the Express Ferry.
- $50$ commuters use both the Bus and the Express Ferry.
- $25$ commuters use all three transit services.

How many of the surveyed commuters use none of these three transit services?

20

45

70

280

145

Show answer & explanation

Answer: 20

Answer

20 commuters use none of the three transit services.

Using the inclusion-exclusion principle for three overlapping sets, the total number of commuters using at least one of the transit services is

160 + 140 + 110 - (60 + 45 + 50) + 25 = 280

. Subtracting this value from the total surveyed group of

300

commuters gives

300 - 280 = 20

commuters who use none of the three services.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three sets to find the total number of commuters who use at least one transit service,

|B \cup R \cup F|

.

|B \cup R \cup F| = |B| + |R| + |F| - (|B \cap R| + |R \cap F| + |F \cap B|) + |B \cap R \cap F|

Simply adding set sizes double-counts elements in pairwise intersections and triple-counts elements in all three sets.

2

Substitute the given numerical values into the inclusion-exclusion formula.

|B \cup R \cup F| = 160 + 140 + 110 - (60 + 45 + 50) + 25 = 410 - 155 + 25 = 280

Combining the sums and differences yields the exact count of commuters using at least one mode of transit.

3

Subtract the number of commuters using at least one service from the total surveyed population to find those using none.

\text{None} = 300 - 280 = 20

The universe of surveyed commuters consists of those using at least one service plus those using none.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 30s

Question 83Question →

An environmental auditing agency surveyed $250$ manufacturing plants regarding their compliance with three environmental standards: Air Quality ( $A$ ), Water Discharge ( $W$ ), and Waste Management ( $M$ ). The survey yielded the following data:

- $130$ plants meet Air Quality standards ( $A$ ).
- $140$ plants meet Water Discharge standards ( $W$ ).
- $120$ plants meet Waste Management standards ( $M$ ).
- $40$ plants meet all three standards.
- $20$ plants meet none of the three standards.
- The number of plants meeting both Air Quality and Water Discharge standards is equal to the number of plants meeting both Water Discharge and Waste Management standards.
- The number of plants meeting both Air Quality and Waste Management standards is $10$ fewer than the number meeting both Air Quality and Water Discharge standards.

How many of the surveyed plants meet exactly one of the three environmental standards?

Show answer & explanation

Answer: 110

Answer

110

By setting up the 3-set inclusion-exclusion equation, the unknown pairwise intersections are found to be 70, 70, and 60. Subtracting the 40 plants that meet all three standards gives the exclusive double-overlap regions (30, 30, and 20). Subtracting these along with the central intersection from each single set yields 40 plants meeting only Air Quality, 40 meeting only Water Discharge, and 30 meeting only Waste Management, totaling 110 plants.

Step-by-Step Solution

1

Find total number of plants meeting at least one standard

|A ∪ W ∪ M| = 250 - 20 = 230

Subtracting plants that meet no standards from the total surveyed gives the union of all three sets.

2

Set up algebraic expressions for pairwise intersections

|A ∩ M| = k, |A ∩ W| = k + 10, |W ∩ M| = k + 10

Define the smallest pairwise intersection as k and express the other two based on the given relationships.

3

Apply the Principle of Inclusion-Exclusion (PIE) for three sets to solve for k

230 = 130 + 140 + 120 - (k + 10 + k + 10 + k) + 40 => k = 60

Substitute set sizes and the triple intersection into the 3-set inclusion-exclusion formula.

4

Calculate the number of plants in each exclusive region

Only (A ∩ W) = 30, Only (W ∩ M) = 30, Only (A ∩ M) = 20

Subtract the triple intersection (40) from each pairwise intersection.

5

Calculate plants meeting exactly one standard and sum them

Only A = 40, Only W = 40, Only M = 30; Total = 40 + 40 + 30 = 110

Subtract all double-overlap and triple-overlap regions from each individual set total.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 84Question →

A university surveyed a cohort of $150$ freshmen regarding their membership in three student organizations: the Art Club ( $A$ ), the Music Society ( $M$ ), and the Theater Guild ( $T$ ). The survey revealed the following data:

- $68$ students belong to the Art Club.
- $62$ students belong to the Music Society.
- $54$ students belong to the Theater Guild.
- $22$ students belong to both the Art Club and the Music Society.
- $18$ students belong to both the Music Society and the Theater Guild.
- $15$ students belong to both the Art Club and the Theater Guild.
- $8$ students belong to all three organizations.

How many of the surveyed students belong to exactly one of these three organizations?

Show answer & explanation

Answer: 98

Answer

98 students belong to exactly one of the three organizations.

To find the number of students belonging to exactly one organization, analyze the regions of a 3-set Venn diagram starting from the innermost region (all three clubs =

8

). Subtracting

8

from each pairwise intersection gives the students in exactly two clubs: Art & Music only (

14

), Music & Theater only (

10

), and Art & Theater only (

7

). Next, subtract the overlapping regions from each club total: Art only is

68 - (14 + 7 + 8) = 39

; Music only is

62 - (14 + 10 + 8) = 30

; Theater only is

54 - (7 + 10 + 8) = 29

. Summing these single-club regions gives

39 + 30 + 29 = 98

.

Step-by-Step Solution

1

Find the number of students belonging strictly to each pair of organizations (two-set intersections only).

Art and Music only =

14

; Music and Theater only =

10

; Art and Theater only =

7

.

The given pairwise totals include students who belong to all three organizations (

8

), so subtracting

8

isolates those in exactly two groups.

2

Determine the number of students belonging to each individual organization exclusively.

Art only =

39

; Music only =

30

; Theater only =

29

.

Subtract all shared membership regions (both two-group only and three-group) from each total organization membership.

3

Add the counts of students belonging to exactly one group.

39 + 30 + 29 = 98

.

The question requests the sum of all students in the non-overlapping single-set regions.

Key Concept

Three-Set Venn Diagram Region Partitioning

Question 85Question →

At a technology conference attended by $80$ software engineers, $50$ engineers write code in Python, $40$ write code in Java, and $15$ write code in neither Python nor Java. How many of the software engineers write code in both Python and Java?

10

15

25

30

35

Show answer & explanation

Answer: 25

Answer

The number of software engineers who write code in both Python and Java is 25.

Out of

80

engineers,

15

write neither language, which means

80 - 15 = 65

engineers write Python, Java, or both. Using the inclusion-exclusion principle,

|P \cup J| = |P| + |J| - |P \cap J|

, substituting the values gives

65 = 50 + 40 - |P \cap J|

. Solving for the overlap gives

|P \cap J| = 90 - 65 = 25

. Thus, 25 engineers write code in both languages.

Step-by-Step Solution

1

Find the number of engineers who write code in at least one of the two languages (Python or Java).

|P \cup J| = 80 - 15 = 65

Subtracting the engineers who write neither language from the total gives the union of the two sets.

2

Apply the Principle of Inclusion-Exclusion formula for two sets.

|P \cup J| = |P| + |J| - |P \cap J|

The total number of engineers in the union is equal to the sum of the individual sets minus their intersection.

3

Substitute the known values into the equation to solve for the intersection

|P \cap J|

.

65 = 50 + 40 - |P \cap J| \implies 65 = 90 - |P \cap J| \implies |P \cap J| = 25

Solving the linear equation yields the number of engineers writing both languages.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 0s

Question 86Question →

An environmental protection agency audited $240$ coastal wetland sites to evaluate contamination by three specific pollutants: Microplastics ( $M$ ), Heavy metals ( $H$ ), and Agricultural runoff ( $A$ ). The audit revealed the following findings:

- Exactly $30$ of the audited wetlands showed no contamination from any of the three pollutants.
- The total number of wetlands containing Microplastics, Heavy metals, and Agricultural runoff were $130$ , $110$ , and $100$ , respectively.
- The number of wetlands containing Microplastics and Heavy metals but NOT Agricultural runoff was $25$ .
- The number of wetlands containing Heavy metals and Agricultural runoff but NOT Microplastics was $35$ .
- The number of wetlands containing Microplastics and Agricultural runoff but NOT Heavy metals was $20$ .

Based on the audit data, which of the following statements must be true? Select all such statements.

Select all that apply

Exactly

25

wetlands contain all three pollutants.

The number of wetlands containing only Microplastics is

60

.

The total number of wetlands containing exactly one pollutant is

85

.

The percentage of all surveyed wetlands that contain at least two pollutants is

50\%

.

The ratio of wetlands containing only Heavy metals to wetlands containing only Agricultural runoff is

5

to

4

.

Show answer & explanation

Answer: Exactly $25$ wetlands contain all three pollutants.; The number of wetlands containing only Microplastics is $60$ .; The ratio of wetlands containing only Heavy metals to wetlands containing only Agricultural runoff is $5$ to $4$ .

Answer

The statements confirming that exactly 25 wetlands contain all three pollutants, that 60 wetlands contain only Microplastics, and that the ratio of wetlands containing only Heavy metals to only Agricultural runoff is 5 to 4 are all correct.

The correct options are those stating that 25 wetlands contain all three pollutants, that 60 wetlands contain only Microplastics, and that the ratio of Heavy metals only to Agricultural runoff only is 5 to 4. Each of these follows directly from setting up the inclusion-exclusion equation for three sets and determining all eight mutually exclusive regions of the Venn diagram.

Step-by-Step Solution

1

Determine the number of wetlands containing at least one pollutant.

Total in union

|M \cup H \cup A| = 240 - 30 = 210

.

The total audited set is 240, and 30 sites have no pollutants.

2

Apply the Principle of Inclusion-Exclusion for three sets.

|M \cup H \cup A| = |M| + |H| + |A| - (|M \cap H| + |H \cap A| + |A \cap M|) + |M \cap H \cap A|

, which simplifies to

210 = 130 + 110 + 100 - S_2 + x \implies S_2 - x = 130

, where

S_2

is the sum of pairwise intersections and

x

is the triple intersection.

This relates the total union to the individual set sizes and intersection regions.

3

Express

S_2

in terms of the given 'exactly two' regions and solve for

x

.

The number of wetlands with exactly two pollutants is

25 + 35 + 20 = 80

. Since

S_2 - 3x = 80

, substituting

S_2 = 130 + x

gives

(130 + x) - 3x = 80 \implies 2x = 50 \implies x = 25

.

The sum of pairwise overlaps counts the triple intersection three times, so subtracting

3x

yields the 'exactly two' region.

4

Calculate single-category region sizes.

Microplastics only =

130 - (25 + 20 + 25) = 60

; Heavy metals only =

110 - (25 + 35 + 25) = 25

; Agricultural runoff only =

100 - (20 + 35 + 25) = 20

. Total single-category =

60 + 25 + 20 = 105

.

Subtracting all dual and triple overlap counts from each set total isolates the exclusive membership.

5

Verify each choice statement against the calculated region counts.

Triple intersection is 25 (True). Microplastics only is 60 (True). Exactly one pollutant total is 105, not 85 (False). Percentage of total sample with at least two pollutants is

105 / 240 = 43.75\%

, not

50\%

(False). Ratio of Heavy metals only to Agricultural runoff only is

25:20 = 5:4

(True).

Validates exact values against option assertions.

Key Concept

Three-set Principle of Inclusion-Exclusion and Venn diagram region decomposition

Question 87Question →

Dataset

S

consists of

4n

distinct real numbers arranged in ascending order, where

n \ge 5

. The interquartile range of Dataset

S

is

IQR_S

, and its standard deviation is

\sigma_S

. A new dataset,

T

, is formed by transforming each value

x

in Dataset

S

into a corresponding value

y

as follows:

y = \begin{cases} x + k & \text{if } x \le Q_3 \\ x + 2k & \text{if } x > Q_3 \end{cases}

where

Q_3

is the third quartile (75th percentile) of Dataset

S

, and

k

is a positive constant. Which of the following statements must be true regarding the interquartile range

IQR_T

and standard deviation

\sigma_T

of Dataset

T

relative to Dataset

S

?

IQR_T = IQR_S

and

\sigma_T > \sigma_S

IQR_T > IQR_S

and

\sigma_T > \sigma_S

IQR_T = IQR_S

and

\sigma_T = \sigma_S

IQR_T > IQR_S

and

\sigma_T = \sigma_S

IQR_T < IQR_S

and

\sigma_T > \sigma_S

Show answer & explanation

Answer: $IQR_T = IQR_S$ and $\sigma_T > \sigma_S$

Answer

The interquartile range remains unchanged (

IQR_T = IQR_S

) while the standard deviation strictly increases (

\sigma_T > \sigma_S

).

The statement asserting that

IQR_T = IQR_S

and

\sigma_T > \sigma_S

is correct. Both

Q_1

and

Q_3

belong to the condition

x \le Q_3

, meaning both quartile values increase by exactly

k

. Thus,

IQR_T = (Q_3 + k) - (Q_1 + k) = Q_3 - Q_1 = IQR_S

. Meanwhile, the highest 25% of data values are shifted by an additional distance of

k

, increasing the overall spread of values around the mean, which strictly increases the standard deviation.

Step-by-Step Solution

1

Analyze the impact of the transformation on the first quartile (

Q_1

) and third quartile (

Q_3

).

Since

Q_1 < Q_3

, the value corresponding to

Q_1

is less than or equal to

Q_3

, so it is shifted to

Q_1 + k

. The value corresponding to

Q_3

satisfies

x \le Q_3

, so it is also shifted to

Q_3 + k

.

The definition of the piecewise rule adds

k

to all values less than or equal to

Q_3

.

2

Calculate the new interquartile range

IQR_T

.

IQR_T = (Q_3 + k) - (Q_1 + k) = Q_3 - Q_1 = IQR_S

.

The constant shift

k

cancels out when taking the difference between the upper and lower quartiles.

3

Analyze the impact on standard deviation

\sigma_T

.

The lower 75% of elements are shifted by

+k

, while the upper 25% of elements are shifted further by

+2k

. This increases the relative distance between upper-tail data points and the rest of the dataset.

A non-uniform shift that spreads the upper tail farther from the rest of the distribution increases total variation around the mean, resulting in

\sigma_T > \sigma_S

.

Key Concept

Effect of non-linear piecewise transformations on dispersion measures (IQR invariance under equal quartile shifts vs. standard deviation sensitivity to upper-tail displacement)

Question 88Question →

A survey of $500$ university researchers evaluated their usage of three high-performance computing resources: Cloud Containers ( $C$ ), GPU Accelerators ( $G$ ), and Distributed Storage ( $D$ ). Exactly $50$ researchers use none of these three resources. The survey revealed that equal numbers of researchers use Cloud Containers and GPU Accelerators ( $|C| = |G| = 240$ ), while $210$ researchers use Distributed Storage ( $|D| = 210$ ). Exactly $40$ researchers use all three resources. Furthermore, the number of researchers who use both $C$ and $G$ but not $D$ is equal to the number who use both $G$ and $D$ but not $C$ , and this quantity is exactly twice the number of researchers who use both $C$ and $D$ but not $G$ .

How many researchers use GPU Accelerators ( $G$ ) ONLY?

32

72

104

112

144

Show answer & explanation

Answer: 72

Answer

72 researchers use GPU Accelerators (

G

) only.

The total number of researchers using at least one resource is

500 - 50 = 450

. Assigning

x

to the region using Cloud Containers and Distributed Storage only, the regions for C-and-G-only and G-and-D-only are each

2x

. Applying the three-set inclusion-exclusion principle gives

450 = 240 + 240 + 210 - (5x + 120) + 40

, which simplifies to

5x = 160

, so

x = 32

. The exclusive overlapping regions containing GPU Accelerators are

2(32) = 64

and

2(32) = 64

. Subtracting these overlaps along with the triple intersection (

40

) from

|G| = 240

yields

240 - (64 + 64 + 40) = 72

.

Step-by-Step Solution

1

Determine the total number of researchers using at least one resource.

|C \cup G \cup D| = 500 - 50 = 450

.

Subtract researchers who use none of the resources from the total surveyed.

2

Define variables for the two-set exclusive intersection regions.

Let

|C \cap D \cap G^c| = x

. Then

|C \cap G \cap D^c| = 2x

and

|G \cap D \cap C^c| = 2x

.

The problem states that the C-and-D-only region is half of the other two exclusive two-set intersection regions.

3

Express the full pairwise intersections including the triple intersection (

|C \cap G \cap D| = 40

).

|C \cap G| = 2x + 40

,

|G \cap D| = 2x + 40

, and

|C \cap D| = x + 40

.

Each pairwise intersection is the sum of its exclusive two-set intersection and the three-set intersection.

4

Apply the 3-Set Principle of Inclusion-Exclusion.

450 = 240 + 240 + 210 - [(2x + 40) + (2x + 40) + (x + 40)] + 40 \implies 450 = 690 - (5x + 120) + 40 \implies 450 = 610 - 5x \implies 5x = 160 \implies x = 32

.

Inclusion-exclusion formula:

|C \cup G \cup D| = |C| + |G| + |D| - (|C \cap G| + |G \cap D| + |C \cap D|) + |C \cap G \cap D|

.

5

Calculate the number of researchers using GPU Accelerators (

G

) only.

Exclusive

G

-only

= |G| - (|C \cap G \cap D^c| + |G \cap D \cap C^c| + |C \cap G \cap D|) = 240 - (64 + 64 + 40) = 240 - 168 = 72

.

Subtract all overlapping regions within set

G

from the total size of set

G

.

Key Concept

3-Set Inclusion-Exclusion Principle and Venn Diagram Region Partitioning

Question 89Question →

A statistics instructor compiles the exam scores of 50 students in Dataset $S$ . The dataset has a range $R$ , an interquartile range $IQR$ , and a standard deviation $s$ . A revised dataset $T$ is created by multiplying each score in Dataset $S$ by $-1.5$ and then adding $25$ to each result. Which of the following expressions correctly states the measures of dispersion for Dataset $T$ in terms of the measures of dispersion for Dataset $S$ ?

\text{Range}_T = -1.5R + 25

,

IQR_T = -1.5 IQR + 25

, and

s_T = -1.5s + 25

\text{Range}_T = 1.5R + 25

,

IQR_T = 1.5 IQR + 25

, and

s_T = 1.5s + 25

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 1.5s

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 2.25s

\text{Range}_T = 1.5R

,

IQR_T = IQR

, and

s_T = 1.5s

Show answer & explanation

Answer: $\text{Range}_T = 1.5R$ , $IQR_T = 1.5 IQR$ , and $s_T = 1.5s$

Answer

The measures of dispersion for Dataset T are given by Range_T = 1.5R, IQR_T = 1.5 IQR, and s_T = 1.5s.

When a linear transformation

y = ax + b

is applied to a dataset, any measure of dispersion

D

(such as range, interquartile range, or standard deviation) transforms according to

D_y = |a| D_x

. The additive constant

b = 25

shifts all values equally and does not change the distances between data points, so it has no effect on dispersion. The multiplicative factor

a = -1.5

scales all distances by

|-1.5| = 1.5

. Therefore,

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 1.5s

.

Step-by-Step Solution

1

Analyze the general effect of a linear transformation y = ax + b on measures of dispersion.

Measures of dispersion (range, interquartile range, standard deviation) quantify spread and distance between data points. Adding a constant b shifts the whole distribution without altering distance between points, so b has zero impact on spread.

Additive constants shift position, not dispersion.

2

Determine the effect of multiplying by scale factor a = -1.5 on range, IQR, and standard deviation.

Multiplying data by a constant scale factor a scales all distance-based measures by |a|. Since dispersion measures must be non-negative, the scale factor used is |-1.5| = 1.5.

Distances between values scale by the absolute value of the multiplicative factor.

3

Combine the results to state Range_T, IQR_T, and s_T in terms of R, IQR, and s.

Range_T = 1.5R, IQR_T = 1.5 IQR, and s_T = 1.5s.

All three metrics scale by 1.5 and are unaffected by the +25 term.

Key Concept

Linear Transformations of Dispersion Metrics

Estimated Time:2m 0s

Question 90Question →

A market research firm surveyed a group of $200$ consumers regarding their active subscriptions to three digital services: FilmStream ( $F$ ), AudioVibe ( $A$ ), and PrintPlus ( $P$ ). Every surveyed consumer subscribed to at least one of the three services. The survey gathered the following information:
- $120$ consumers subscribed to FilmStream
- $100$ consumers subscribed to AudioVibe
- $75$ consumers subscribed to PrintPlus
- $45$ consumers subscribed to both FilmStream and AudioVibe
- $35$ consumers subscribed to both AudioVibe and PrintPlus
- $30$ consumers subscribed to both FilmStream and PrintPlus

How many consumers subscribed to all three digital services?

15

25

35

55

95

Show answer & explanation

Answer: $15$

Answer

15 consumers subscribed to all three digital services.

According to the 3-set inclusion-exclusion principle,

|F \cup A \cup P| = |F| + |A| + |P| - (|F \cap A| + |A \cap P| + |F \cap P|) + |F \cap A \cap P|

. Substituting the given values gives

200 = 120 + 100 + 75 - (45 + 35 + 30) + |F \cap A \cap P|

, which simplifies to

200 = 185 + |F \cap A \cap P|

. Subtracting

185

from

200

yields

15

.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three sets.

|F \cup A \cup P| = |F| + |A| + |P| - (|F \cap A| + |A \cap P| + |F \cap P|) + |F \cap A \cap P|

This formula accounts for elements counted multiple times across overlapping sets.

2

Substitute the known values into the equation.

200 = 120 + 100 + 75 - (45 + 35 + 30) + |F \cap A \cap P|

Every consumer subscribes to at least one service, so

|F \cup A \cup P| = 200

.

3

Simplify the numerical terms on the right side of the equation.

200 = 295 - 110 + |F \cap A \cap P| \implies 200 = 185 + |F \cap A \cap P|

Sum of individual sets is

295

, and sum of pairwise intersections is

110

.

4

Solve for the target three-set intersection

|F \cap A \cap P|

.

|F \cap A \cap P| = 200 - 185 = 15

Isolating the variable gives the number of consumers in all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 30s

Question 91Question →

A biotechnology consortium surveyed $180$ research laboratories regarding their implementation of three diagnostic platforms: Platform $A$ , Platform $B$ , and Platform $C$ .

- $95$ laboratories use Platform $A$ .
- $85$ laboratories use Platform $B$ .
- $80$ laboratories use Platform $C$ .
- $15$ laboratories use all three platforms.
- $20$ laboratories use none of the three platforms.
- The number of laboratories that use Platform $A$ and Platform $B$ but NOT Platform $C$ is equal to the number of laboratories that use Platform $B$ and Platform $C$ but NOT Platform $A$ .
- $30$ laboratories use Platform $A$ and Platform $C$ but NOT Platform $B$ .

How many laboratories use Platform $A$ ONLY?

15

20

30

45

50

Show answer & explanation

Answer: 30

Answer

30 laboratories use Platform A only.

Subtracting the 20 laboratories that use none of the platforms from the total population of 180 gives a union size of 160. Applying the inclusion-exclusion formula yields

160 = (95 + 85 + 80) - (x + 15 + x + 15 + 30 + 15) + 15

, which resolves to

x = 20

laboratories that use Platforms A and B only. Subtracting all overlap regions from Platform A's total gives

95 - 20 - 30 - 15 = 30

laboratories that use Platform A only.

Step-by-Step Solution

1

Calculate the total number of laboratories that use at least one platform.

|A \cup B \cup C| = 180 - 20 = 160

Laboratories using at least one platform equal the total surveyed minus those using none.

2

Set up the Principle of Inclusion-Exclusion for three sets.

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

Standard formula relating total union size to individual set sizes and intersections.

3

Express pairwise intersections in terms of non-overlapping regions and solve for the unknown region

x

.

Let

x

be the number of labs using Platform

A

and

B

only. Given

A \cap C \text{ only} = 30

and

A \cap B \cap C = 15

, we have:

160 = (95 + 85 + 80) - [(x + 15) + (x + 15) + (30 + 15)] + 15

160 = 260 - (2x + 75) + 15 = 200 - 2x \implies 2x = 40 \implies x = 20

Substituting the given equality of regions allows finding

x = 20

.

4

Calculate the number of laboratories that use Platform

A

only.

\text{Platform } A \text{ only} = |A| - (A \cap B \text{ only}) - (A \cap C \text{ only}) - (A \cap B \cap C) = 95 - 20 - 30 - 15 = 30

Subtracting all overlap regions containing Platform A from the total Platform A count gives the single-region count.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Partitioning

Question 92Question →

A committee is to be selected from a group of $5$ distinct people: $P$ , $Q$ , $R$ , $S$ , and $T$ . Which of the following statements regarding the possible selections or arrangements of people from this group are true? Select all that apply.

Select all that apply

The number of different

2

-person committees that can be formed from the group is

10

.

The number of different

3

-person committees that can be formed from the group is

10

.

The number of different ways to assign

2

distinct officer roles (President and Vice President) from the group is

10

.

The total number of different ways to arrange all

5

people in a single line is

25

.

The number of different ways to select and arrange

3

of the

5

people in a line is

60

.

Show answer & explanation

Answer: The number of different $2$ -person committees that can be formed from the group is $10$ .; The number of different $3$ -person committees that can be formed from the group is $10$ .; The number of different ways to select and arrange $3$ of the $5$ people in a line is $60$ .

Answer

The statements confirming that

10

different

2

-person committees can be formed,

10

different

3

-person committees can be formed, and

60

different

3

-person linear arrangements can be formed are all correct.

Selecting committees without specific roles requires combinations

\binom{n}{k}

, giving

\binom{5}{2} = 10

and

\binom{5}{3} = 10

. Arranging 3 people in ordered positions requires permutations

P(5,3) = 5 \times 4 \times 3 = 60

. Thus, all three corresponding statements are correct.

Step-by-Step Solution

1

Evaluate the 2-person committee selection statement.

\binom{5}{2} = \frac{5 \times 4}{2} = 10

Selection of a committee without specific roles is an unordered combination.

2

Evaluate the 3-person committee selection statement.

\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

Choosing 3 items out of 5 yields the same number of outcomes as choosing 2 items out of 5.

3

Evaluate the 2-person officer assignment statement.

P(5,2) = 5 \times 4 = 20

Assigning distinct officer positions means order matters, requiring permutations rather than combinations.

4

Evaluate the 5-person line arrangement statement.

5! = 120

The total number of linear arrangements of 5 distinct objects is given by 5 factorial.

5

Evaluate the 3-person line arrangement statement.

P(5,3) = 5 \times 4 \times 3 = 60

Ordering 3 out of 5 people in a line uses the fundamental counting principle with decreasing choices per slot.

Key Concept

Distinguishing between combinations (where selection order does not matter) and permutations (where selection or position order does matter).

Question 93Question →

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 94Question →

A security code consists of three distinct digits chosen from the non-zero digits $1$ through $9$ . If the first digit must be odd and the third digit must be even, how many such three-digit security codes can be formed?

16

70

140

160

180

Show answer & explanation

Answer: 140

Answer

140

To find the number of three-digit codes with distinct digits from

1

through

9

satisfying the constraints, count the options for each slot: the first position has

5

odd options (

1, 3, 5, 7, 9

), the third position has

4

even options (

2, 4, 6, 8

), and the middle position has

9 - 2 = 7

remaining options. By the Fundamental Counting Principle, multiplying these options gives

5 \times 7 \times 4 = 140

.

Step-by-Step Solution

1

Determine the number of possibilities for the first digit.

5 choices (the odd digits:

1, 3, 5, 7, 9

).

The question specifies that the first digit must be odd.

2

Determine the number of possibilities for the third digit.

4 choices (the even digits:

2, 4, 6, 8

).

The question specifies that the third digit must be even.

3

Determine the number of possibilities for the middle (second) digit.

7 choices.

There are

9

total non-zero digits (

1

through

9

). Since

2

distinct digits have already been used for the first and third positions,

9 - 2 = 7

digits remain available for the middle position.

4

Apply the Fundamental Counting Principle to find the total number of codes.

5 \times 7 \times 4 = 140

.

The total number of sequential independent choices is found by multiplying the number of options for each position.

Key Concept

Fundamental Counting Principle with Restricted Positions and Distinct Elements

Estimated Time:45s

Question 95Question →

A 6-digit security code is to be formed using distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ . The code must satisfy the following conditions:
1. The code must be an even number (its final digit must be $2$ , $4$ , $6$ , or $8$ ).
2. Both digits $1$ and $2$ must be included in the 6-digit code.
3. Digits $1$ and $2$ cannot occupy adjacent positions in the code.

Which of the following statements regarding the number of possible 6-digit security codes must be true? Select all such statements.

Select all that apply

The total number of valid 6-digit security codes that satisfy all conditions is

3,600

.

The number of valid security codes in which the final digit is

2

is

1,440

.

The number of valid security codes in which the final digit is an even digit other than

2

is

2,160

.

If the non-adjacency constraint between

1

and

2

were removed, the total number of valid security codes would be

4,320

.

There are

20

valid ways to place digits

1

and

2

non-adjacently in the first 5 positions of the code.

Show answer & explanation

Answer: The total number of valid 6-digit security codes that satisfy all conditions is $3,600$ .; The number of valid security codes in which the final digit is $2$ is $1,440$ .; The number of valid security codes in which the final digit is an even digit other than $2$ is $2,160$ .

Answer

The statements confirming that the total number of codes is 3,600, that 1,440 codes end in 2, and that 2,160 codes end in an even digit other than 2 are all correct.

The solution requires partitioning into two mutually exclusive scenarios depending on whether digit 2 occupies the final position. When digit 2 is at the end, digit 1 can occupy any of the first 4 positions (excluding position 5 to avoid adjacency), yielding

4 \times P(6,4) = 1,440

codes. When the final position is occupied by 4, 6, or 8 (3 choices), digits 1 and 2 have

P(5,2) - 8 = 12

valid non-adjacent placements across the first 5 positions, and the remaining 3 positions can be filled in

P(5,3) = 60

ways, yielding

3 \times 12 \times 60 = 2,160

codes. The sum of these two cases gives

3,600

total valid codes. Thus, the three statements asserting total codes of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are all correct.

Step-by-Step Solution

1

Analyze the conditions and split into two disjoint cases based on the last digit.

Case 1: The last digit (6th position) is

2

. Case 2: The last digit (6th position) is

4

,

6

, or

8

.

Digit

2

plays a dual role: it satisfies the even-ending condition and is one of the restricted digits.

2

Calculate Case 1 (last digit is 2).

Position 6 is fixed as

2

(

1

choice). Digit

1

must be in the first 5 positions but cannot be adjacent to position 6 (so position 5 is excluded). Thus digit

1

has

4

choices (positions 1, 2, 3, 4). The remaining

4

positions are filled from the remaining

6

available digits

\{3, 4, 5, 6, 7, 8\}

in

P(6, 4) = 6 \times 5 \times 4 \times 3 = 360

ways. Total for Case 1:

1 \times 4 \times 360 = 1,440

.

This determines the valid arrangements when 2 is forced to the end of the code.

3

Calculate Case 2 (last digit is 4, 6, or 8).

There are

3

choices for the 6th position. Digits

1

and

2

must be placed among the first 5 positions non-adjacently. Total ordered placements of

1

and

2

in 5 positions is

P(5, 2) = 20

. The number of adjacent position pairs is

4

(positions (1,2), (2,3), (3,4), (4,5)), with

2! = 2

orderings per pair, giving

4 \times 2 = 8

adjacent placements. Non-adjacent placements =

20 - 8 = 12

. The remaining 3 open positions are filled from the remaining 5 available digits in

P(5, 3) = 60

ways. Total for Case 2:

3 \times 12 \times 60 = 2,160

.

This accounts for codes ending in 4, 6, or 8 while respecting the non-adjacency of 1 and 2.

4

Combine the cases and verify statements.

Total valid codes =

1,440 + 2,160 = 3,600

. The statements asserting totals of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are true.

Summing mutually exclusive cases gives the overall number of valid outcomes.

Key Concept

Fundamental Counting Principle with Permutations under Restricted Adjacency

Question 96Question →

The frequency distribution table below summarizes the monthly water consumption, $c$ (in cubic meters, $\text{m}^3$ ), recorded for a sample of $150$ municipal water accounts.

Monthly Water Consumption ( $\text{m}^3$ )	Number of Accounts
$0 \le c < 10$	$25$
$10 \le c < 20$	$45$
$20 \le c < 30$	$50$
$30 \le c < 40$	$20$
$40 \le c < 50$	$10$

If one water account is selected at random from among all accounts with a monthly water consumption of at least $10\text{ m}^3$ , what is the probability that the selected account has a monthly water consumption of less than $30\text{ m}^3$ ? (Give your answer as a decimal rounded to two decimal places.)

Show answer & explanation

Answer: 0.76

Answer

0.76

To calculate the required probability, first restrict the sample space to accounts with a monthly consumption of at least

10\text{ m}^3

. Summing the frequencies for the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

gives

45 + 50 + 20 + 10 = 125

accounts. Among these

125

accounts, those with a consumption of less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

, giving a count of

45 + 50 = 95

accounts. Dividing the favorable outcomes by the total outcomes in the restricted sample space yields

\frac{95}{125} = 0.76

.

Step-by-Step Solution

1

Determine the total number of accounts meeting the condition of having consumption of at least

10\text{ m}^3

.

Total eligible accounts =

45 + 50 + 20 + 10 = 125

.

Accounts with consumption of at least

10\text{ m}^3

fall into the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

.

2

Determine the number of accounts among the eligible set with consumption less than

30\text{ m}^3

.

Number of favorable accounts =

45 + 50 = 95

.

Within the eligible set, accounts with consumption less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

.

3

Calculate the conditional probability as a decimal.

\frac{95}{125} = 0.76

Dividing the favorable outcomes (

95

) by the total possible outcomes in the restricted sample space (

125

) yields

0.76

.

Key Concept

Conditional probability and sample space restriction in grouped frequency tables

Question 97Question →

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

Question 98Question →

A bookshelf holds $4$ distinct fiction novels and $3$ distinct non-fiction books. If a reader chooses exactly $1$ fiction novel and $1$ non-fiction book to take on a trip, how many different pairs of books can the reader select?

Show answer & explanation

Answer: 12

Answer

The total number of different pairs of books that can be selected is

12

.

According to the Fundamental Counting Principle, if one task can be performed in

m

ways and a second task can be performed in

n

ways, the two tasks together can be performed in

m \times n

ways. Choosing a fiction novel (

4

options) and a non-fiction book (

3

options) results in

4 \times 3 = 12

unique pairs.

Step-by-Step Solution

1

Determine the number of ways to choose one fiction novel

There are

4

possible choices.

The shelf contains

4

distinct fiction novels.

2

Determine the number of ways to choose one non-fiction book

There are

3

possible choices.

The shelf contains

3

distinct non-fiction books.

3

Calculate total pairs using the Fundamental Counting Principle

4 \times 3 = 12

The selection of a fiction novel and a non-fiction book are independent decisions, so the number of outcomes is the product of the number of choices for each decision.

Key Concept

Fundamental Counting Principle

Question 99Question →

The frequency distribution table below summarizes the calibration offset errors, $x$ (in microvolts, $\mu\text{V}$ ), measured for a sample of $200$ precision voltage sensors in a robotics laboratory.

Offset Error Interval ( $\mu\text{V}$ )	Frequency
$0 \le x < 10$	$32$
$10 \le x < 20$	$f_1$
$20 \le x < 30$	$68$
$30 \le x < 40$	$f_2$
$40 \le x < 50$	$24$

The estimated mean offset error calculated using the midpoints of the five class intervals is equal to $24.8\ \mu\text{V}$ . If a sensor is selected at random from among those with an offset error of at least $20\ \mu\text{V}$ , what is the probability that its offset error is less than $40\ \mu\text{V}$ ?

\frac{14}{17}

\frac{14}{25}

\frac{25}{31}

\frac{11}{34}

\frac{17}{25}

Show answer & explanation

Answer: $\frac{14}{17}$

Answer

The correct answer is

\frac{14}{17}

, which represents the conditional probability that a sensor's offset error is less than

40\ \mu\text{V}

given that it is at least

20\ \mu\text{V}

.

The correct answer is

\frac{14}{17}

. Solving the system of equations formed by the total sample size (

f_1 + f_2 = 76

) and the estimated midpoint mean (

3f_1 + 7f_2 = 404

) yields

f_1 = 32

and

f_2 = 44

. The number of sensors with offset

\ge 20\ \mu\text{V}

is

68 + 44 + 24 = 136

. Among these, the number of sensors with offset

< 40\ \mu\text{V}

is

68 + 44 = 112

. Thus, the conditional probability is

\frac{112}{136} = \frac{14}{17}

.

Step-by-Step Solution

1

Set up an equation for the total frequency of the sample.

32 + f_1 + 68 + f_2 + 24 = 200 \implies f_1 + f_2 = 76

The sum of all class frequencies must equal the given total sample size of 200.

2

Set up an equation for the estimated mean using interval midpoints.

Midpoints are

5, 15, 25, 35, 45

. Total weighted sum

= 32(5) + f_1(15) + 68(25) + f_2(35) + 24(45) = 160 + 15f_1 + 1700 + 35f_2 + 1080 = 2940 + 15f_1 + 35f_2

. Mean

= \frac{2940 + 15f_1 + 35f_2}{200} = 24.8 \implies 15f_1 + 35f_2 = 2020 \implies 3f_1 + 7f_2 = 404

.

The estimated mean of grouped data is the sum of products of interval midpoints and frequencies divided by total sample size.

3

Solve the linear system of equations for

f_1

and

f_2

.

Multiply

f_1 + f_2 = 76

by

3

to get

3f_1 + 3f_2 = 228

. Subtracting from

3f_1 + 7f_2 = 404

gives

4f_2 = 176 \implies f_2 = 44

. Then

f_1 = 76 - 44 = 32

.

Elimination yields exact unknown frequencies for the remaining intervals.

4

Determine the conditional sample space and target frequency.

Condition (offset

\ge 20\ \mu\text{V}

): Intervals

[20, 30), [30, 40), [40, 50)

with total frequency

68 + 44 + 24 = 136

. Target condition (offset

< 40\ \mu\text{V}

within condition): Intervals

[20, 30)

and

[30, 40)

with frequency

68 + 44 = 112

.

Conditional probability restricts the denominator to sensors meeting the given condition.

5

Calculate the final probability fraction.

Probability

= \frac{112}{136} = \frac{14}{17}

.

Dividing target count by conditional total count yields the simplified fraction.

Key Concept

Grouped Data Mean Estimation & Conditional Probability from Frequency Distributions

Estimated Time:2m 30s

Question 100Question →

A environmental monitoring group collected $200$ soil samples from a nature reserve and recorded their pH levels in the frequency table below:

pH Level Range	Number of Samples
$5.0 – 5.9$	$40$
$6.0 – 6.9$	$70$
$7.0 – 7.9$	$50$
$8.0 – 8.9$	$30$
$9.0 – 9.9$	$10$

What percent of the soil samples with a pH level of at least $6.0$ have a pH level in the range $6.0$ to $7.9$ ?

43.75\%

60.00\%

75.00\%

80.00\%

68.75\%

Show answer & explanation

Answer: $75.00\%$

Answer

75.00\%

To find the desired percentage, first restrict the sample space to all soil samples having a pH level of at least

6.0

. Adding the frequencies for

6.0–6.9

(

70

),

7.0–7.9

(

50

),

8.0–8.9

(

30

), and

9.0–9.9

(

10

) gives a base total of

160

samples. Next, find the number of samples within that group that fall in the range

6.0

to

7.9

, which is

70 + 50 = 120

. The percentage is calculated as

(120 / 160) \times 100\% = 75.00\%

.

Step-by-Step Solution

1

Determine the conditional base population (samples with pH

\ge 6.0

).

Sum of frequencies for ranges

6.0-6.9

,

7.0-7.9

,

8.0-8.9

, and

9.0-9.9

:

70 + 50 + 30 + 10 = 160

.

The question asks 'of the soil samples with a pH level of at least 6.0', which restricts the total base to these four intervals.

2

Determine the target frequency (samples with pH in range

6.0

to

7.9

).

Sum of frequencies for ranges

6.0-6.9

and

7.0-7.9

:

70 + 50 = 120

.

This captures all qualifying samples within the specified target interval.

3

Calculate the conditional percentage.

\frac{120}{160} \times 100\% = 0.75 \times 100\% = 75.00\%

.

Divide the target sub-group count by the conditional base count and convert to a percentage.

Key Concept

Conditional Relative Frequency in Grouped Data Tables

Estimated Time:1m 30s

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