Data Analysis Practice Questions — GRE General Test — Page 4

Question 61Question →

The table below shows the frequency distribution of daily rainfall, in millimeters, recorded across 50 weather monitoring stations during a storm event.

Daily Rainfall (mm)	Number of Stations
0 to under 10	10
10 to under 20	15
20 to under 30	17
30 to under 40	5
40 to under 50	3

What percentage of the stations that recorded at least 10 mm of daily rainfall recorded less than 30 mm of daily rainfall?

42.5\%

64\%

75\%

80\%

84\%

Show answer & explanation

Answer: $80\%$

Answer

80\%

To find the desired percentage, first identify the total number of stations that recorded at least 10 mm of rainfall. This includes the intervals 10 to under 20, 20 to under 30, 30 to under 40, and 40 to under 50, giving

15 + 17 + 5 + 3 = 40

stations. Next, find how many of these 40 stations recorded less than 30 mm. These fall into the 10 to under 20 and 20 to under 30 intervals, which total

15 + 17 = 32

stations. Calculating the relative percentage gives

(32 / 40) \times 100\% = 80\%

.

Step-by-Step Solution

1

Determine the conditional base (denominator)

Sum the frequencies for all intervals representing at least 10 mm:

15 + 17 + 5 + 3 = 40

stations.

The question specifies 'stations that recorded at least 10 mm of daily rainfall', restricting our population base to this subset.

2

Determine the target subgroup frequency (numerator)

Sum the frequencies for intervals within the base that are also less than 30 mm (the 10 to under 20 and 20 to under 30 intervals):

15 + 17 = 32

stations.

We need to count the stations that meet both conditions: at least 10 mm and less than 30 mm.

3

Calculate the percentage

\frac{32}{40} \times 100\% = 0.80 \times 100\% = 80\%

.

Divide the target subgroup count by the conditional base count and multiply by 100.

Key Concept

Grouped frequency table interpretation with conditional percentages

Estimated Time:1m 30s

Question 62Question →

Dataset $W$ consists of $101$ distinct real numbers arranged in strictly increasing order: $x_1 < x_2 < \dots < x_{101}$ . The interquartile range of $W$ is $Q$ , its range is $R$ , and its standard deviation is $\sigma$ . A new dataset $W'$ is constructed by adding a positive constant $k$ to each of the highest $25$ observations (that is, $x_{77}, x_{78}, \dots, x_{101}$ become $x_{77}+k, x_{78}+k, \dots, x_{101}+k$ ), while the first $76$ observations remain unchanged. Which of the following statements MUST be true regarding the statistics of $W'$ compared to dataset $W$ ?

The interquartile range of

W'

is equal to

Q + \frac{k}{2}

.

The interquartile range of

W'

is equal to

Q + k

.

The interquartile range of

W'

is equal to

Q

.

The standard deviation of

W'

is equal to

\sigma

.

The median of

W'

is equal to the median of

W

plus

\frac{k}{2}

.

Show answer & explanation

Answer: The interquartile range of $W'$ is equal to $Q + \frac{k}{2}$ .

Answer

The statement specifying that the interquartile range of

W'

is equal to

Q + \frac{k}{2}

must be true.

For an ordered dataset of 101 elements, the first quartile

Q_1

is the median of the lower 50 values (

\frac{x_{25} + x_{26}}{2}

) and the third quartile

Q_3

is the median of the upper 50 values (

\frac{x_{76} + x_{77}}{2}

). Since elements

x_1

through

x_{76}

are unchanged,

Q_1

remains identical. However,

x_{77}

increases by

k

, causing

Q_3

to increase by

\frac{k}{2}

. Consequently, the new interquartile range is

Q' = Q_3' - Q_1' = Q + \frac{k}{2}

.

Step-by-Step Solution

1

Determine the position of quartiles for a 101-element dataset.

For

101

sorted values, the median is

x_{51}

. The lower half consists of

50

values (

x_1

through

x_{50}

), whose median

Q_1

is

\frac{x_{25} + x_{26}}{2}

. The upper half consists of

50

values (

x_{52}

through

x_{101}

), whose median

Q_3

is

\frac{x_{76} + x_{77}}{2}

.

Establishing the exact index formulas for

Q_1

and

Q_3

is required to evaluate changes in IQR.

2

Analyze how

Q_1

changes in dataset

W'

.

x_{25}

and

x_{26}

are among the first

76

observations, so they are unchanged. Thus,

Q_1' = Q_1

.

Since the modified values start at index 77, the first quartile remains completely unaffected.

3

Analyze how

Q_3

changes in dataset

W'

.

In

W'

,

x_{76}' = x_{76}

and

x_{77}' = x_{77} + k

. Therefore,

Q_3' = \frac{x_{76} + (x_{77} + k)}{2} = \frac{x_{76} + x_{77}}{2} + \frac{k}{2} = Q_3 + \frac{k}{2}

.

One of the two values used to calculate

Q_3

is shifted by

k

, so their average increases by

\frac{k}{2}

.

4

Compute the new interquartile range

Q'

.

Q' = Q_3' - Q_1' = \left(Q_3 + \frac{k}{2}\right) - Q_1 = (Q_3 - Q_1) + \frac{k}{2} = Q + \frac{k}{2}

.

Subtracting

Q_1'

from

Q_3'

yields the new interquartile range directly.

Key Concept

Quartile Interpolation and Selective Data Transformations

Question 63Question →

The running times, in minutes, for a fleet of delivery vehicles have a 25th percentile of $134$ minutes and a 75th percentile of $242$ minutes. To analyze the data in hours, a manager converts each vehicle's running time to hours by dividing by $60$ and then adds a fixed delay of $1.25$ hours to every converted value. What is the interquartile range, in hours, of the adjusted running times?

Show answer & explanation

Answer: 1.8

Answer

The interquartile range of the adjusted running times is 1.8 hours.

The interquartile range (IQR) measures the spread of the middle 50% of a dataset (

Q_3 - Q_1

). For the original dataset in minutes,

IQR = 242 - 134 = 108

minutes. When data is transformed linearly via

Y = aX + b

, scaling all values by a positive constant

a

scales the IQR by

a

, whereas adding a constant

b

shifts all values equally and does not change the distance between percentiles. Here

a = \frac{1}{60}

and

b = 1.25

. Therefore, the new IQR in hours is

108 \times \frac{1}{60} = 1.8

hours.

Step-by-Step Solution

1

Find the original interquartile range (IQR) in minutes

IQR = 242 - 134 = 108 minutes

The interquartile range is defined as the difference between the 75th percentile (Q3) and the 25th percentile (Q1).

2

Apply the scaling factor for unit conversion

108 / 60 = 1.8 hours

Dividing every value by 60 scales the data by a factor of 1/60, which scales measures of dispersion like IQR by |1/60|.

3

Evaluate the effect of adding the constant shift

Adjusted IQR remains 1.8 hours

Adding a constant shift to every data point shifts the entire distribution without changing the spread or distance between percentiles.

Key Concept

Linear Transformations on Measures of Dispersion

Question 64Question →

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

Select all that apply

The mean of the combined dataset is equal to

86.25

.

The median of the combined dataset is at least

82

.

The median of the combined dataset is at most

86

.

The median of the combined dataset is equal to

84

.

The mode of the combined dataset must lie between

80

and

90

.

Show answer & explanation

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

Answer

The statements asserting that the combined mean is

86.25

, that the combined median is at least

82

, and that the combined median is at most

86

must all be true.

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

86.25

. For the median of

40

combined values (the average of the

20\text{th}

and

21\text{st}

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

21

values are

\ge 82

and at least

21

values are

\le 86

. Therefore, the median must lie within

[82, 86]

.

Step-by-Step Solution

1

Calculate the weighted arithmetic mean for the combined dataset.

Combined Mean =

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

.

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

2

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

At least

21

values in the combined set of

40

items are greater than or equal to

82

.

Since the median of Group A (

15

items) is

82

, at least

8

items are

\ge 82

. Since the median of Group B (

25

items) is

86

, at least

13

items are

\ge 86 \ge 82

. In total,

8 + 13 = 21

items are

\ge 82

. In a sorted list of

40

items, having at least

21

items

\ge 82

implies that the

20\text{th}

and

21\text{st}

items must be

\ge 82

, so the median

\ge 82

.

3

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

At least

21

values in the combined set of

40

items are less than or equal to

86

.

Group A has at least

8

items

\le 82 \le 86

. Group B has at least

13

items

\le 86

. Thus, at least

8 + 13 = 21

items are

\le 86

. In a sorted list of

40

items, having at least

21

items

\le 86

ensures the

20\text{th}

and

21\text{st}

items are

\le 86

, so the median

\le 86

.

4

Evaluate the remaining statements regarding the exact median and mode.

Neither the exact median value of

84

nor the existence/location of a mode is guaranteed.

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

Key Concept

Weighted averages and order statistic bounds for combined datasets

Question 65Question →

A total of 120 software engineers at a firm were surveyed about their attendance at three professional development workshops: Machine Learning ( $M$ ), Cloud Computing ( $C$ ), and Cybersecurity ( $S$ ). The survey results revealed the following:

- 65 engineers attended Machine Learning
- 55 engineers attended Cloud Computing
- 48 engineers attended Cybersecurity
- 25 engineers attended both Machine Learning and Cloud Computing
- 18 engineers attended both Cloud Computing and Cybersecurity
- 20 engineers attended both Machine Learning and Cybersecurity
- 10 engineers attended all three workshops

How many of the surveyed software engineers attended exactly one of the three workshops?

42

48

72

105

115

Show answer & explanation

Answer: 72

Answer

72 engineers attended exactly one of the three workshops.

To find the number of engineers attending exactly one workshop, calculate the exclusive single-set counts for each category using a 3-set Venn diagram model. Subtracting the overlapping regions gives 30 for Machine Learning only, 22 for Cloud Computing only, and 20 for Cybersecurity only. Summing these exclusive regions yields

30 + 22 + 20 = 72

. Alternatively, using the algebraic formula

|\text{Exactly 1}| = (|M| + |C| + |S|) - 2(|M \cap C| + |C \cap S| + |M \cap S|) + 3|M \cap C \cap S| = (65 + 55 + 48) - 2(25 + 18 + 20) + 3(10) = 168 - 126 + 30 = 72

.

Step-by-Step Solution

1

Find the number of engineers in the regions corresponding to exactly two workshops.

Machine Learning and Cloud Computing only =

25 - 10 = 15

; Cloud Computing and Cybersecurity only =

18 - 10 = 8

; Machine Learning and Cybersecurity only =

20 - 10 = 10

.

The given pairwise intersection values include engineers who attended all three workshops.

2

Calculate the number of engineers who attended only one specific workshop.

Machine Learning only =

65 - (15 + 10 + 10) = 30

; Cloud Computing only =

55 - (15 + 8 + 10) = 22

; Cybersecurity only =

48 - (10 + 8 + 10) = 20

.

Subtract all overlapping regions (two-workshop-only and three-workshop) from each set's total.

3

Sum the single-workshop regions to find the total attending exactly one workshop.

30 + 22 + 20 = 72

.

The subsets representing single workshop attendance are disjoint.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Venn Diagram Region Partitioning

Estimated Time:2m 0s

Question 66Question →

A health club surveyed $100$ members regarding their use of cardio machines and weight equipment. Of the members surveyed, $60$ use cardio machines, $50$ use weight equipment, and $15$ use neither cardio machines nor weight equipment. Which of the following statements must be true? Indicate all such statements.

Select all that apply

Exactly

25

members use both cardio machines and weight equipment.

Exactly

35

members use cardio machines but do not use weight equipment.

Exactly

10

members use both cardio machines and weight equipment.

Exactly

110

members use at least one of the two types of equipment.

Exactly

85

members use at least one of the two types of equipment.

Show answer & explanation

Answer: Exactly $25$ members use both cardio machines and weight equipment.; Exactly $35$ members use cardio machines but do not use weight equipment.; Exactly $85$ members use at least one of the two types of equipment.

Answer

The true statements are those stating that exactly 25 members use both types of equipment, exactly 35 members use cardio machines only, and exactly 85 members use at least one type of equipment.

The total surveyed population of

100

consists of members using at least one type of equipment and members using neither (

15

). Thus,

100 - 15 = 85

members use at least one type of equipment. Using the inclusion-exclusion formula

\text{Union} = \text{Cardio} + \text{Weights} - \text{Intersection}

, we have

85 = 60 + 50 - \text{Intersection}

, giving an intersection of

25

members using both types of equipment. Subtracting

25

from the

60

total cardio users shows that

35

members use cardio machines exclusively.

Step-by-Step Solution

1

Calculate the total number of members using at least one type of equipment.

Total using at least one type

= 100 - 15 = 85

.

Members using at least one type of equipment represent the total surveyed population minus those who use neither.

2

Calculate the number of members using both cardio machines and weight equipment using the principle of inclusion-exclusion.

\text{Both} = \text{Cardio} + \text{Weights} - \text{At least one} = 60 + 50 - 85 = 25

.

Adding the individual set counts double-counts the overlapping region, so subtracting the total union gives the intersection size.

3

Calculate the number of members using cardio machines only.

\text{Cardio Only} = \text{Cardio} - \text{Both} = 60 - 25 = 35

.

Subtracting the intersection from the total cardio group yields the exclusive cardio users.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 15s

Question 67Question →

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

20

21

19

25

10

Show answer & explanation

Answer: $20$

Answer

The maximum possible integer value of

x

is

20

.

The sum of the original

11

values is

11 \times 45 = 495

. When the

5

largest integers are each increased by

10

, the sum increases by

+50

. When the

5

smallest integers are each decreased by

x

, the sum decreases by

-5x

. The new sum is

545 - 5x

, making the new mean

\frac{545 - 5x}{11}

. Because

D

has

11

elements, the median is the

6^{\text{th}}

element. Changing the smallest

5

and largest

5

elements does not alter the value of the

6^{\text{th}}

element, so the median remains

40

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

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

. The greatest integer less than

21

is

20

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset

D

.

Since

D

has

11

elements with a mean of

45

, the sum of elements is

11 \times 45 = 495

.

The mean formula is

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

, so

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

.

2

Determine the median and sum of the modified dataset

D'

.

The median remains the

6^{\text{th}}

element, which is

40

. The sum of

D'

is

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

.

Modifying only the

5

smallest and

5

largest elements leaves the

6^{\text{th}}

central element unchanged.

3

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

D'

.

Solve

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

.

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

4

Identify the maximum integer value satisfying the inequality.

The largest integer strictly less than

21

is

20

.

x

must be an integer.

Key Concept

Effect of data modifications on mean and median

Estimated Time:2m 30s

Question 68Question →

Dataset $X$ consists of $20$ distinct real numbers with standard deviation $s_X > 0$ and interquartile range $IQR_X > 0$ . Dataset $Y$ is created by multiplying each number in Dataset $X$ by $-3$ and then adding $7$ . Dataset $Z$ is created by adding a single 21st value, equal to the arithmetic mean of Dataset $X$ , to Dataset $X$ . Which of the following statements must be true regarding the measures of dispersion of these datasets?

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is equal to

s_X

.

The standard deviation of Dataset

Y

is

-3s_X + 7

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

The standard deviation of Dataset

Y

is

3s_X + 7

, and the standard deviation of Dataset

Z

is strictly greater than

s_X

.

The standard deviation of Dataset

Y

is

-3s_X

, and the standard deviation of Dataset

Z

is equal to

s_X

.

Show answer & explanation

Answer: The standard deviation of Dataset $Y$ is $3s_X$ , and the standard deviation of Dataset $Z$ is strictly less than $s_X$ .

Answer

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

Linear transformation scales standard deviation by the absolute value of the multiplier, making

s_Y = |-3|s_X = 3s_X

. Constant additions do not affect spread. Adding a data point equal to the mean does not change the total sum of squared deviations, but increases the sample size

n

, thereby reducing the average squared deviation and yielding

s_Z < s_X

.

Step-by-Step Solution

1

Analyze the linear transformation on Dataset X to form Dataset Y.

For any linear transformation of data

y_i = a x_i + b

, the standard deviation transforms according to

s_Y = |a| s_X

. Here

a = -3

and

b = 7

, so

s_Y = |-3| s_X = 3s_X

.

Adding a constant shift

b

shifts all points equally without altering their spread relative to the mean, whereas multiplying by

a

scales distance by

|a|

.

2

Analyze the effect of inserting the mean into Dataset X to form Dataset Z.

Adding

\mu_X

to Dataset

X

yields a new dataset with mean

\mu_Z = \mu_X

. The sum of squared deviations

\sum (z_i - \mu_Z)^2 = \sum (x_i - \mu_X)^2 + (\mu_X - \mu_X)^2 = \sum (x_i - \mu_X)^2

remains identical.

The extra point contributes

0

squared distance from the mean.

3

Compare the standard deviation

s_Z

to

s_X

.

Because the sum of squared deviations remains unchanged while the total count of elements increases from

20

to

21

, the average squared deviation (variance) decreases. Hence,

s_Z < s_X

.

Dividing the same numerator by a larger denominator (

21

or

20

depending on sample/population formula) results in a smaller variance and smaller standard deviation.

Key Concept

Effect of linear transformations and mean-value insertion on standard deviation

Alternative Method

Consider extreme simple cases: if Dataset X has two values

\{ -1, 1 \}

, mean is 0,

s_X = 1

. Dataset Y becomes

\{ 10, 4 \}

, mean is 7,

s_Y = 3 = 3s_X

. Adding 0 to X gives

\{ -1, 0, 1 \}

, mean remains 0, variance becomes

2/3

, so

s_Z = \sqrt{2/3} < 1

.

Estimated Time:2m 0s

Question 69Question →

A survey of $250$ registered voters was conducted to analyze their primary news sources: Television ( $T$ ), the Internet ( $I$ ), and Print newspapers ( $P$ ). The survey revealed the following results:

- $140$ voters get news from Television.
- $150$ voters get news from the Internet.
- $80$ voters get news from Print newspapers.
- $45$ voters get news from both Television and Print newspapers.
- $60$ voters get news from both the Internet and Print newspapers.
- $30$ voters get news from all three sources.
- $20$ voters do not get news from any of these three sources.

How many of the surveyed voters get news from Television and the Internet, but NOT from Print newspapers?

Show answer & explanation

Answer: 35

Answer

35 voters get news from Television and the Internet, but not from Print newspapers.

Using the principle of inclusion-exclusion for three sets, the total union size is 230 voters (250 total minus 20 who use none). Setting up the formula 230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30 allows us to solve for |T ∩ I| = 65. To find those who use Television and Internet but NOT Print newspapers, we subtract the 30 voters who use all three sources from 65, resulting in 35 voters.

Step-by-Step Solution

1

Find the size of the union of all three sets

|T ∪ I ∪ P| = 250 - 20 = 230

Subtracting the 20 voters who use none of the three news sources from the total sample of 250 gives the total number of voters in at least one category.

2

Set up the Principle of Inclusion-Exclusion for three sets

230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30

The formula sums individual set sizes, subtracts pairwise intersections, and adds back the triple intersection.

3

Solve for the total intersection of Television and Internet

|T ∩ I| = 65

Simplifying the equation gives 230 = 295 - |T ∩ I|, which yields |T ∩ I| = 65.

4

Exclude those who also read Print newspapers

|(T ∩ I) \ P| = 65 - 30 = 35

Subtracting the 30 voters who use all three sources leaves only those who use Television and Internet without Print newspapers.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 70Question →

Dataset $P$ consists of $n$ numerical values with a mean of $40$ and a standard deviation of $6$ . Dataset $Q$ consists of $n$ numerical values with a mean of $60$ and a standard deviation of $6$ . Dataset $R$ is created by combining all $n$ values from Dataset $P$ and all $n$ values from Dataset $Q$ into a single dataset of $2n$ values. If $\sigma_R$ represents the standard deviation of Dataset $R$ , which of the following is the exact value of $\sigma_R$ ?

6

6\sqrt{2}

2\sqrt{34}

12

16

Show answer & explanation

Answer: $2\sqrt{34}$

Answer

The exact value of

\sigma_R

is

2\sqrt{34}

.

The total variance of a combined dataset is given by the law of total variance:

\sigma_R^2 = \text{Mean}(\sigma_P^2, \sigma_Q^2) + \text{Var}(\mu_P, \mu_Q)

. Since

\sigma_P = \sigma_Q = 6

, the average within-group variance is

6^2 = 36

. The combined mean is

50

, and both group means (

40

and

60

) lie

10

units away from

50

, giving a between-group variance of

10^2 = 100

. Combining these yields

\sigma_R^2 = 36 + 100 = 136

, so

\sigma_R = \sqrt{136} = 2\sqrt{34}

.

Step-by-Step Solution

1

Calculate the mean of the combined dataset

R

.

\mu_R = \frac{n(40) + n(60)}{2n} = \frac{100n}{2n} = 50

Since both datasets have equal size

n

, the combined mean is the arithmetic average of the two group means.

2

Express the sum of squared deviations for Dataset

P

around the combined mean

\mu_R = 50

.

\sum_{i=1}^n (p_i - 50)^2 = \sum_{i=1}^n ((p_i - 40) - 10)^2 = \sum_{i=1}^n (p_i - 40)^2 - 20\sum_{i=1}^n (p_i - 40) + 100n = 36n - 0 + 100n = 136n

The variance of

P

gives

\sum (p_i - 40)^2 = 6^2 n = 36n

, and

\sum (p_i - 40) = 0

by the definition of the mean.

3

Express the sum of squared deviations for Dataset

Q

around the combined mean

\mu_R = 50

.

\sum_{j=1}^n (q_j - 50)^2 = \sum_{j=1}^n ((q_j - 60) + 10)^2 = \sum_{j=1}^n (q_j - 60)^2 + 20\sum_{j=1}^n (q_j - 60) + 100n = 36n + 0 + 100n = 136n

The variance of

Q

gives

\sum (q_j - 60)^2 = 6^2 n = 36n

, and

\sum (q_j - 60) = 0

.

4

Compute the combined variance

\sigma_R^2

and take the square root to find

\sigma_R

.

\sigma_R^2 = \frac{136n + 136n}{2n} = \frac{272n}{2n} = 136 \implies \sigma_R = \sqrt{136} = 2\sqrt{34}

The total variance of a combined dataset equals the average within-group variance (

36

) plus the between-group variance around the combined mean (

(40-50)^2 = 100

). Thus

\sigma_R^2 = 36 + 100 = 136

.

Key Concept

Pooled standard deviation for combined datasets with differing means

Estimated Time:2m 0s

Question 71Question →

Dataset $A$ consists of 100 distinct positive numbers. Dataset $B$ is created by replacing every number in Dataset $A$ that is strictly greater than the 75th percentile of Dataset $A$ with the value of the 75th percentile of Dataset $A$ . All numbers less than or equal to the 75th percentile remain unchanged. Which of the following statements comparing Dataset $B$ to Dataset $A$ must be true? Select all such statements.

Select all that apply

The 25th percentile of Dataset

B

is equal to the 25th percentile of Dataset

A

.

The interquartile range of Dataset

B

is equal to the interquartile range of Dataset

A

.

The standard deviation of Dataset

B

is greater than the standard deviation of Dataset

A

.

The range of Dataset

B

is strictly less than the range of Dataset

A

.

The median of Dataset

B

is strictly greater than the median of Dataset

A

.

Show answer & explanation

Answer: The 25th percentile of Dataset $B$ is equal to the 25th percentile of Dataset $A$ .; The interquartile range of Dataset $B$ is equal to the interquartile range of Dataset $A$ .; The range of Dataset $B$ is strictly less than the range of Dataset $A$ .

Answer

The statements asserting that the 25th percentile remains equal, the interquartile range remains equal, and the range is strictly smaller for Dataset B compared to Dataset A are all correct.

The statements confirming that the 25th percentile remains equal, the interquartile range remains equal, and the range decreases are correct. The lower 75% of the ordered dataset is untouched, leaving

P_{25}

and

P_{75}

unchanged, which preserves the IQR. Furthermore, replacing the top 25 distinct values with

P_{75}

lowers the maximum value while keeping the minimum value the same, strictly decreasing the range.

Step-by-Step Solution

1

Analyze how Dataset B is constructed from Dataset A

Since all 100 values in Dataset A are distinct, exactly 25 values are strictly greater than the 75th percentile (

P_{75}

). In Dataset B, these top 25 values are replaced by

P_{75}

, while the bottom 75 values remain unchanged.

Understanding which specific data points change allows us to determine positional and dispersion metrics.

2

Evaluate positional statistics (25th percentile, median, 75th percentile)

The lower 75% of the ordered dataset is identical in Dataset A and Dataset B. Thus,

P_{25}(B) = P_{25}(A)

,

\text{Median}(B) = \text{Median}(A)

, and

P_{75}(B) = P_{75}(A)

.

Percentiles at or below the 75th percentile depend only on the values at or below those percentile ranks.

3

Evaluate Interquartile Range (IQR) and Range

\text{IQR}(B) = P_{75}(B) - P_{25}(B) = P_{75}(A) - P_{25}(A) = \text{IQR}(A)

. For range,

\text{Min}(B) = \text{Min}(A)

, but

\text{Max}(B) = P_{75}(A) < \text{Max}(A)

. Therefore,

\text{Range}(B) < \text{Range}(A)

.

IQR depends on

P_{75}

and

P_{25}

, which are unchanged. Range depends on Max and Min; decreasing the maximum value decreases the range.

4

Evaluate Standard Deviation

All altered values were in the upper tail and were moved closer to the center of the distribution. Reducing extreme values decreases variance and standard deviation.

Standard deviation measures average squared distance from the mean; pulling upper extreme values inward reduces standard deviation.

Key Concept

Effect of upper-tail data transformation on measures of position (percentiles, median) and dispersion (range, IQR, standard deviation).

Question 72Question →

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

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

Select all that apply

The arithmetic mean of the combined dataset of all

10

facilities is greater than the arithmetic mean of Region A.

The median of the combined dataset of all

10

facilities is equal to the median of Region A.

The median of the combined dataset of all

10

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

The mode of the combined dataset of all

10

facilities is greater than the mode of Region A.

The arithmetic mean of the combined dataset of all

10

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

Show answer & explanation

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

Answer

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

The mean of the combined dataset (

18.8

) is greater than the mean of Region A (

18.0

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

16.5

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

Step-by-Step Solution

1

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

Region A sum

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

. Mean

= \frac{108}{6} = 18

. Sorted values are

12, 15, 15, 18, 20, 28

, so Median

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

. Mode

= 15

.

Establishing base central metrics for the first subgroup.

2

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

Region B sum

= 10 + 14 + 22 + 34 = 80

. Mean

= \frac{80}{4} = 20

. Sorted values are

10, 14, 22, 34

, so Median

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

.

Establishing base central metrics for the second subgroup.

3

Combine and order all 10 values to calculate combined metrics.

Combined ordered set:

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

. Total sum

= 108 + 80 = 188

. Combined Mean

= \frac{188}{10} = 18.8

. Combined Median

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

. Combined Mode

= 15

.

Necessary to evaluate combined properties accurately against subgroup properties.

4

Evaluate each candidate statement against calculated values.

Combined mean (

18.8

) > Region A mean (

18

) is TRUE. Combined median (

16.5

) = Region A median (

16.5

) is TRUE. Average of medians (

17.25

) = Combined median (

16.5

) is FALSE. Combined mode (

15

) > Region A mode (

15

) is FALSE. Simple average of means (

19

) = Combined mean (

18.8

) is FALSE.

Determining which statements satisfy the required conditions.

Key Concept

Combining Datasets and Weighted Measures of Central Tendency

Question 73Question →

The table below shows the frequency distribution of order fulfillment times, in minutes, for a sample of $50$ online orders processed at a warehouse facility.

Fulfillment Time $t$ (minutes)	Number of Orders
$10 \le t < 20$	$10$
$20 \le t < 30$	$15$
$30 \le t < 40$	$15$
$40 \le t < 50$	$7$
$50 \le t < 60$	$3$

An order is categorized as "delayed" if its fulfillment time is at least $30$ minutes. What percentage of all orders that took at least $20$ minutes to fulfill were categorized as "delayed"?

30.0%

37.5%

50.0%

62.5%

75.0%

Show answer & explanation

Answer: 62.5%

Answer

62.5%

To find the required percentage, first determine the subset of orders that took at least

20

minutes to fulfill. Summing the frequencies of the last four intervals (

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

) yields

15 + 15 + 7 + 3 = 40

orders. Next, find how many of these orders took at least

30

minutes to fulfill (the delayed orders), which corresponds to the last three intervals:

15 + 7 + 3 = 25

orders. Finally, calculate the percentage:

\frac{25}{40} \times 100\% = 62.5\%

.

Step-by-Step Solution

1

Identify the sub-population that defines the base (denominator) of the percentage.

The target base is all orders with fulfillment times of at least

20

minutes (

t \ge 20

). Summing the frequencies for intervals

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 15 + 7 + 3 = 40

orders.

The question specifically asks 'of all orders that took at least 20 minutes to fulfill', establishing a conditional base.

2

Determine the number of orders meeting the numerator condition.

The delayed orders are those taking at least

30

minutes (

t \ge 30

). Summing frequencies for

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 7 + 3 = 25

orders.

Orders in all intervals starting from

30

minutes onwards qualify as delayed.

3

Compute the required percentage.

\left(\frac{25}{40}\right) \times 100\% = 0.625 \times 100\% = 62.5\%

.

Dividing the target count by the restricted base count gives the conditional proportion.

Key Concept

Grouped Frequency Distributions and Conditional Percentages

Question 74Question →

A market research firm surveyed $150$ venture capital investors regarding their investments in three emerging sectors: Artificial Intelligence ( $A$ ), Biotechnology ( $B$ ), and Clean Energy ( $C$ ). The survey revealed the following data:

- $15$ investors did not invest in any of the three sectors.
- $78$ investors invested in Artificial Intelligence.
- $65$ investors invested in Biotechnology.
- $62$ investors invested in Clean Energy.
- $10$ investors invested in all three sectors.

If $50$ investors invested in exactly two of the three sectors, how many investors invested in exactly one of the three sectors?

65

70

75

85

90

Show answer & explanation

Answer: 75

Answer

75 investors invested in exactly one of the three sectors.

The total number of investors who invested in at least one of the three sectors is

150 - 15 = 135

. In a 3-set Venn diagram, the total union is the sum of those who belong to exactly one sector (

E_1

), exactly two sectors (

E_2

), and all three sectors (

E_3

). We are given

E_2 = 50

and

E_3 = 10

. Setting up the region equation

E_1 + E_2 + E_3 = 135

gives

E_1 + 50 + 10 = 135

, which yields

E_1 = 75

.

Step-by-Step Solution

1

Calculate the total number of investors who invested in at least one sector (the union of sets A, B, and C).

|A \cup B \cup C| = 150 - 15 = 135

Subtracting the investors who invested in none of the sectors from the total population yields the size of the union.

2

Express the union as the sum of non-overlapping region counts.

|A \cup B \cup C| = E_1 + E_2 + E_3

, where

E_1

is the number of investors in exactly 1 sector,

E_2

is the number in exactly 2 sectors, and

E_3

is the number in all 3 sectors.

Any element in the union belongs to exactly one, exactly two, or all three sets.

3

Substitute the known values into the partition equation and solve for

E_1

.

135 = E_1 + 50 + 10 \implies E_1 = 135 - 60 = 75

We are given

E_2 = 50

and

E_3 = 10

, leaving

E_1

as the remaining unknown value.

Key Concept

Three-Set Inclusion-Exclusion and Region Partitioning

Estimated Time:2m 0s

Question 75Question →

A specialty coffee roastery surveyed $200$ customers to determine their preferences among three origin beans: Ethiopian ( $E$ ), Colombian ( $C$ ), and Guatemalan ( $G$ ). Every customer surveyed preferred at least one of the three origins. The survey revealed that $105$ customers preferred Ethiopian, $90$ preferred Colombian, and $85$ preferred Guatemalan. Additionally, $35$ customers preferred both Ethiopian and Colombian, $30$ preferred both Colombian and Guatemalan, and $40$ preferred both Ethiopian and Guatemalan. How many customers preferred all three origin beans?

15

20

25

35

50

Show answer & explanation

Answer: $25$

Answer

25 customers preferred all three origin beans.

The correct answer is obtained by setting up the three-set inclusion-exclusion formula

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

. Substituting the given values gives

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

. Solving

200 = 175 + x

yields

x = 25

.

Step-by-Step Solution

1

State the 3-set Inclusion-Exclusion Principle formula.

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

This fundamental formula relates the total union of three overlapping sets to their individual sizes, pairwise intersections, and three-set intersection.

2

Substitute the given values into the formula.

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

, where

x = |E \cap C \cap G|

Since every customer preferred at least one origin, the union of all three sets equals the total number of customers surveyed (

200

).

3

Simplify the sum of individual sets and pairwise intersections.

Sum of individual sets =

105 + 90 + 85 = 280

. Sum of pairwise intersections =

35 + 30 + 40 = 105

.

Grouping terms makes evaluating the equation easier and reduces calculation errors.

4

Solve for the unknown three-set intersection

x

.

200 = 280 - 105 + x \implies 200 = 175 + x \implies x = 25

Subtracting

175

from

200

yields the number of customers who prefer all three origin beans.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 76Question →

An aviation safety committee audited $240$ international flights for operational delays caused by Weather ( $W$ ), Mechanical issues ( $M$ ), and Air Traffic Control ( $T$ ). The audit revealed the following data:

- $112$ flights experienced Weather delays.
- $98$ flights experienced Mechanical delays.
- $104$ flights experienced Air Traffic Control delays.
- $28$ flights experienced both Weather and Mechanical delays.
- $34$ flights experienced both Mechanical and Air Traffic Control delays.
- $42$ flights experienced both Weather and Air Traffic Control delays.
- The number of flights that experienced none of these three types of delays was equal to twice the number of flights that experienced all three types of delays.

How many flights experienced exactly one of these three types of delays?

Show answer & explanation

Answer: 136

Answer

The number of flights that experienced exactly one of the three types of delays is 136.

The total number of audited flights is partitioned into those experiencing at least one delay and those experiencing none. By applying the Inclusion-Exclusion formula

|W \cup M \cup T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

, and setting the total audited flights to

240 = (210 + x) + 2x

, we solve for the triple intersection

x = 10

. Isolating the exclusive single-delay regions yields

52

flights for Weather only,

46

for Mechanical only, and

38

for ATC only, giving a total of

136

flights.

Step-by-Step Solution

1

Define the unknown intersection and external set quantities in terms of a single variable.

Let

x = |W \cap M \cap T|

represent the number of flights experiencing all three types of delays. Consequently, the number of flights experiencing none of the delays is

2x

.

Relating the unknown triple intersection to the 'neither' category allows all regions to be expressed in terms of

x

.

2

Apply the Principle of Inclusion-Exclusion for three overlapping sets to find the total number of flights with at least one delay.

|W \cup M \cup T| = |W| + |M| + |T| - (|W \cap M| + |M \cap T| + |W \cap T|) + |W \cap M \cap T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

.

Subtracting pairwise intersections corrects for double-counting, and adding back the triple intersection corrects for over-subtraction.

3

Formulate and solve the linear equation for the total sample population of 240 flights.

\text{Total Flights} = |W \cup M \cup T| + |\text{Neither}| \implies 240 = (210 + x) + 2x \implies 240 = 210 + 3x \implies 3x = 30 \implies x = 10

.

The sum of flights in the union of the sets and flights outside all three sets must equal the total number of flights audited.

4

Determine the number of flights in the two-set intersection regions excluding the triple intersection.

Weather and Mechanical only:

28 - 10 = 18

; Mechanical and ATC only:

34 - 10 = 24

; Weather and ATC only:

42 - 10 = 32

.

Each pairwise intersection given in the stem includes the triple intersection of 10 flights.

5

Compute the number of flights belonging exclusively to single categories and sum them.

Weather only:

112 - (18 + 32 + 10) = 52

; Mechanical only:

98 - (18 + 24 + 10) = 46

; ATC only:

104 - (32 + 24 + 10) = 38

. Sum of single categories:

52 + 46 + 38 = 136

.

Subtracting all overlapping components from each set total isolates the exclusive single-category elements.

Key Concept

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

Estimated Time:2m 30s

Question 77Question →

Data set $A$ consists of 7 numbers that form an arithmetic progression with a non-zero common difference. Data set $B$ is formed by removing the smallest number and the largest number from Data set $A$ . If $\sigma_A^2$ represents the population variance of Data set $A$ and $\sigma_B^2$ represents the population variance of Data set $B$ , what is the value of $\frac{\sigma_A^2}{\sigma_B^2}$ ?

Show answer & explanation

Answer: 2

Answer

2

By writing the 7 terms of Data set A symmetrically about their mean

\mu

with common difference

d

, the sum of squared deviations is

28d^2

, giving a population variance of

\sigma_A^2 = \frac{28d^2}{7} = 4d^2

. Removing the minimum and maximum values leaves 5 terms centered at

\mu

with a sum of squared deviations of

10d^2

, resulting in a population variance of

\sigma_B^2 = \frac{10d^2}{5} = 2d^2

. The ratio of the variance of Data set A to the variance of Data set B is therefore

\frac{4d^2}{2d^2} = 2

.

Step-by-Step Solution

1

Represent the elements of Data set A using its mean

\mu

and common difference

d

.

Let the 7 elements of Data set A be

\mu - 3d, \mu - 2d, \mu - d, \mu, \mu + d, \mu + 2d, \mu + 3d

, where

d > 0

.

Symmetrical representation around the mean simplifies variance computation for arithmetic sequences.

2

Calculate the population variance

\sigma_A^2

of Data set A.

\sigma_A^2 = \frac{1}{7} \sum_{k=-3}^{3} (kd)^2 = \frac{d^2}{7}(9 + 4 + 1 + 0 + 1 + 4 + 9) = \frac{28d^2}{7} = 4d^2.

Variance is defined as the mean of the squared deviations from the dataset mean.

3

Determine the elements and population variance

\sigma_B^2

of Data set B.

Data set B contains the 5 remaining central elements, having mean

\mu

. Thus, $\sigma_B^2 = \frac{1}{5} \sum_{k=-2}^{2} (kd)^2 = \frac{d^2}{5}(4 + 1 + 0 + 1 + 4) = \frac{10d^2}{5} = 2d^2.

Removing the minimum and maximum terms leaves a symmetric 5-element dataset with unchanged mean

\mu

.

4

Compute the ratio

\frac{\sigma_A^2}{\sigma_B^2}

.

\frac{\sigma_A^2}{\sigma_B^2} = \frac{4d^2}{2d^2} = 2.

The parameter

d^2

cancels out completely, yielding a constant numerical ratio.

Key Concept

Effect of trimming extreme symmetric values on the population variance of an arithmetic sequence

Question 78Question →

A dataset $W$ consists of 25 distinct positive numbers with mean $M$ , standard deviation $s > 0$ , and interquartile range $IQR > 0$ . Dataset $V$ is constructed by adding $2M$ to every number in $W$ that is strictly greater than the median of $W$ , and subtracting $2M$ from every number in $W$ that is strictly less than the median of $W$ . The value equal to the median of $W$ itself remains unchanged. Which of the following statements MUST be true regarding dataset $V$ compared to dataset $W$ ?

Both the standard deviation and the interquartile range of dataset

V

are greater than those of dataset

W

.

The standard deviation of dataset

V

is greater than

s

, but the interquartile range of dataset

V

is equal to

IQR

.

The standard deviation of dataset

V

is equal to

s

, but the interquartile range of dataset

V

is greater than

IQR

.

Both the standard deviation and the interquartile range of dataset

V

are equal to those of dataset

W

.

The standard deviation of dataset

V

is less than

s

, while the interquartile range of dataset

V

is greater than

IQR

.

Show answer & explanation

Answer: Both the standard deviation and the interquartile range of dataset $V$ are greater than those of dataset $W$ .

Answer

Both the standard deviation and the interquartile range of dataset

V

are greater than those of dataset

W

.

The correct response identifies that both standard deviation and interquartile range increase. In dataset

W

, the 12 elements strictly below the median are reduced by

2M

, and the 12 elements strictly above the median are increased by

2M

. This leaves the mean

M

unchanged but increases the distance of every non-median element from

M

, thereby strictly increasing the standard deviation. Furthermore, the first quartile

Q_1

shifts down by

2M

and the third quartile

Q_3

shifts up by

2M

, expanding the interquartile range from

IQR

to

IQR + 4M

.

Step-by-Step Solution

1

Analyze the impact of the transformation on the first and third quartiles (

Q_1

and

Q_3

).

In an ordered dataset of 25 distinct values, the median is the 13th element. The first quartile

Q_1

is in the lower half (below the median) and the third quartile

Q_3

is in the upper half (above the median).

Understanding where

Q_1

and

Q_3

fall relative to the median determines how their values change.

2

Calculate the new Interquartile Range (

IQR_V

).

Since

Q_1

is in the lower half, its new value is

Q_1 - 2M

. Since

Q_3

is in the upper half, its new value is

Q_3 + 2M

. Thus,

IQR_V = (Q_3 + 2M) - (Q_1 - 2M) = (Q_3 - Q_1) + 4M = IQR + 4M > IQR

because

M > 0

.

Evaluating the difference between the transformed 75th and 25th percentiles.

3

Analyze the impact on the mean and standard deviation.

Equal numbers of values (12 values) are shifted down by

2M

and shifted up by

2M

, so the mean of dataset

V

remains equal to

M

. For every shifted point

x_i

, its squared distance from the mean

(x_i - M)^2

strictly increases. Therefore, the variance and standard deviation strictly increase (

s_V > s

).

Standard deviation measures dispersion relative to the mean; pushing points further from the mean increases standard deviation.

Key Concept

Effect of non-uniform linear shifts on measures of dispersion (Standard Deviation and IQR)

Estimated Time:2m 0s

Question 79Question →

An urban forestry department measured the trunk diameters, in inches, of a sample of $60$ oak trees in a municipal park. The table below summarizes the frequency distribution of the measurements.

Trunk Diameter (inches)	Frequency
$10$ to $14$	$8$
$15$ to $19$	$15$
$20$ to $24$	$22$
$25$ to $29$	$11$
$30$ to $34$	$4$

Based on the table, what fraction of the trees with a trunk diameter of at least $15$ inches have a trunk diameter of less than $25$ inches?

\frac{37}{60}

\frac{37}{52}

\frac{45}{52}

\frac{11}{26}

\frac{37}{15}

Show answer & explanation

Answer: $\frac{37}{52}$

Answer

The fraction of trees with a trunk diameter of at least 15 inches that have a trunk diameter of less than 25 inches is 37/52.

To calculate the required fraction, first identify the subset of trees that have a trunk diameter of at least 15 inches. Summing the frequencies of the intervals 15–19, 20–24, 25–29, and 30–34 gives 15 + 22 + 11 + 4 = 52 trees. Next, identify how many of these 52 trees have a trunk diameter of less than 25 inches. These are the trees in the intervals 15–19 and 20–24, giving 15 + 22 = 37 trees. Therefore, the fraction of trees meeting the condition is 37/52.

Step-by-Step Solution

1

Determine the total number of trees satisfying the condition 'at least 15 inches'.

Sum the frequencies for intervals 15–19, 20–24, 25–29, and 30–34:

15 + 22 + 11 + 4 = 52

trees.

This establishes the restricted denominator (base group) specified by the conditional statement.

2

Determine the number of trees within this group that satisfy the condition 'less than 25 inches'.

Sum the frequencies for intervals 15–19 and 20–24:

15 + 22 = 37

trees.

These are the target intervals that are both at least 15 inches and strictly less than 25 inches.

3

Calculate the required fraction.

\frac{37}{52}

.

Divide the target count by the conditional total count to form the part-to-whole ratio.

Key Concept

Conditional relative frequencies and part-to-whole calculations from grouped data tables.

Estimated Time:1m 30s

Question 80Question →

A technology firm surveyed $100$ software developers regarding their proficiency in three programming languages: Python ( $P$ ), Java ( $J$ ), and C++ ( $C$ ). The survey revealed the following results:
- $55$ developers are proficient in Python.
- $50$ developers are proficient in Java.
- $35$ developers are proficient in C++.
- $25$ developers are proficient in both Python and Java.
- $20$ developers are proficient in both Java and C++.
- $15$ developers are proficient in both Python and C++.
- $10$ developers are proficient in none of these three languages.

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

Select all that apply

Exactly

10

developers are proficient in all three programming languages.

Exactly

50

developers are proficient in at least two of these programming languages.

The number of developers proficient in Python only is

25

.

The total number of developers proficient in exactly one of these languages is

50

.

The percentage of Python-proficient developers who are proficient in Python only is

25\%

.

Show answer & explanation

Answer: Exactly $10$ developers are proficient in all three programming languages.; The number of developers proficient in Python only is $25$ .; The total number of developers proficient in exactly one of these languages is $50$ .

Answer

The correct statements are those asserting that exactly 10 developers know all three languages, that 25 developers know Python only, and that 50 developers know exactly one of these languages.

The statements confirming that 10 developers know all three languages, 25 know Python only, and 50 know exactly one language are correct based on standard 3-set Venn diagram calculations. Using Inclusion-Exclusion,

|P \cup J \cup C| = 90

, which gives

|P \cap J \cap C| = 10

. Subtracting overlapping regions yields

25

for Python only,

15

for Java only,

10

for C++ only, summing to

50

for exactly one language.

Step-by-Step Solution

1

Calculate the total number of developers proficient in at least one language.

N_{\text{at least one}} = 100 - 10 = 90

.

Subtracting the developers proficient in none of the languages from the total survey population yields the union

|P \cup J \cup C|

.

2

Apply the Principle of Inclusion-Exclusion for 3 sets to find the triple intersection

|P \cap J \cap C|

.

90 = 55 + 50 + 35 - (25 + 20 + 15) + |P \cap J \cap C| \implies 90 = 80 + |P \cap J \cap C| \implies |P \cap J \cap C| = 10

.

The formula

|P \cup J \cup C| = |P| + |J| + |C| - (|P \cap J| + |J \cap C| + |P \cap C|) + |P \cap J \cap C|

links all given set quantities.

3

Calculate the count of developers proficient in exactly two languages for each pair.

Python and Java only =

25 - 10 = 15

; Java and C++ only =

20 - 10 = 10

; Python and C++ only =

15 - 10 = 5

.

Subtracting the triple intersection count (

10

) from each pairwise intersection yields the exclusive two-set regions.

4

Calculate the single-language proficiency counts (exactly one language).

Python only =

55 - (15 + 5 + 10) = 25

; Java only =

50 - (15 + 10 + 10) = 15

; C++ only =

35 - (5 + 10 + 10) = 10

. Total exactly one =

25 + 15 + 10 = 50

.

Subtracting all overlapping regions containing each language from its total count yields the single-language region size.

5

Evaluate the statement options based on computed regional values.

Statements stating 10 all three, 25 Python only, and 50 exactly one language are true.

Comparing calculated values (

10

for all three,

25

for Python only,

50

for exactly one language) against each option validates the true choices.

Key Concept

Three-set inclusion-exclusion principle and Venn diagram region decomposition

Data Analysis

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