Data Analysis Practice Questions — GRE General Test

Question 1Question →

A quality control engineer records the thickness, in millimeters, of 9 sample components: $11, 13, 15, 17, 19, 21, 23, 25,$ and $27$ . Each thickness measurement $x$ is then converted to a scaled rating $y$ using the linear formula $y = 1.5x + 4.8$ . What is the interquartile range (IQR) of the transformed dataset of $y$ -values?

Show answer & explanation

Answer: 15

Answer

The interquartile range of the transformed dataset is 15.

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

Q_3 - Q_1

). For the original dataset

11, 13, 15, 17, 19, 21, 23, 25, 27

, the median is

19

.

Q_1

is the median of the lower half

\{11, 13, 15, 17\}

, which is

\frac{13+15}{2} = 14

.

Q_3

is the median of the upper half

\{21, 23, 25, 27\}

, which is

\frac{23+25}{2} = 24

. Thus, the original

\text{IQR} = 24 - 14 = 10

. Under a linear transformation

y = ax + b

, measures of position shift by

a x + b

, so

Q_1(y) = 1.5(14) + 4.8 = 25.8

and

Q_3(y) = 1.5(24) + 4.8 = 40.8

. Subtracting these yields

\text{IQR}(y) = 40.8 - 25.8 = 15

. Notice that this is simply

1.5 \times 10

, as adding a constant shifts the location of the distribution but leaves measures of spread unchanged.

Step-by-Step Solution

1

Find the quartiles of the original 9-element dataset.

Q_1 = 14

and

Q_3 = 24

The median of the dataset is 19 (the 5th value). The lower half of the data consists of

11, 13, 15, 17

, so

Q_1 = \frac{13 + 15}{2} = 14

. The upper half consists of

21, 23, 25, 27

, so

Q_3 = \frac{23 + 25}{2} = 24

.

2

Calculate the interquartile range of the original dataset.

\text{IQR}_x = 10

\text{IQR}_x = Q_3 - Q_1 = 24 - 14 = 10

.

3

Apply the linear transformation rules to find the transformed interquartile range.

\text{IQR}_y = 15

For a linear transformation

y = ax + b

, the interquartile range scales by

|a|

, so

\text{IQR}_y = |a| \cdot \text{IQR}_x = 1.5 \times 10 = 15

. The constant addition of

4.8

shifts all values equally and does not affect the spread/IQR.

Key Concept

Effect of linear transformations on measures of dispersion (IQR, standard deviation, range)

Alternative Method

Transform the individual quartiles directly:

Q_1(y) = 1.5(14) + 4.8 = 25.8

and

Q_3(y) = 1.5(24) + 4.8 = 40.8

. Then calculate the new IQR directly as

40.8 - 25.8 = 15

.

Estimated Time:1m 30s

Question 2Question →

The processing times for a specific type of database query on a server cluster are normally distributed with a mean of $180$ milliseconds and a standard deviation of $25$ milliseconds. Which of the following statements must be true? Indicate all such statements.

Select all that apply

Approximately

68\%

of the query processing times are between

155

milliseconds and

205

milliseconds.

A query processing time of

230

milliseconds corresponds to a

z

-score of

+2.0

.

A query processing time of

205

milliseconds corresponds to approximately the

84\text{th}

percentile rank.

A query processing time of

155

milliseconds corresponds to approximately the

32\text{nd}

percentile rank.

A

z

-score of

+1.0

indicates that the query processing time is

25\%

greater than the mean processing time.

Show answer & explanation

Answer: Approximately $68\%$ of the query processing times are between $155$ milliseconds and $205$ milliseconds.; A query processing time of $230$ milliseconds corresponds to a $z$ -score of $+2.0$ .; A query processing time of $205$ milliseconds corresponds to approximately the $84\text{th}$ percentile rank.

Answer

The correct statements are those asserting that approximately 68% of processing times lie between 155 ms and 205 ms, that a processing time of 230 ms corresponds to a z-score of +2.0, and that 205 ms corresponds to approximately the 84th percentile rank.

For a normal distribution with mean

\mu = 180

and standard deviation

\sigma = 25

: (1) The interval

\mu \pm 1\sigma = [155, 205]

contains approximately

68\%

of all observations. (2) A score of

230

ms has a

z

-score of

\frac{230-180}{25} = +2.0

. (3) A score of

205

ms corresponds to

z = +1.0

, which accumulates

50\%

(area below mean)

+ 34\%

(area between mean and

+1\sigma

) =

84\%

percentile rank.

Step-by-Step Solution

1

Evaluate the 1 standard deviation interval

\mu \pm 1\sigma = 180 \pm 25 = [155, 205]

ms

The 68-95-99.7 empirical rule states that approximately 68% of observations in a normal distribution fall within 1 standard deviation of the mean.

2

Calculate the z-score for a query processing time of 230 ms

z = \frac{230 - 180}{25} = \frac{50}{25} = +2.0

The standard score formula

z = \frac{x - \mu}{\sigma}

measures how many standard deviations an observation is above or below the mean.

3

Determine the percentile rank for 205 ms and 155 ms

For 205 ms (

z = +1

), percentile =

50\% + 34\% = 84\%

. For 155 ms (

z = -1

), percentile =

50\% - 34\% = 16\%

.

50% of the distribution lies below the mean. For

z = +1

, add the 34% between mean and

z = +1

. For

z = -1

, subtract the 34% from 50%.

4

Evaluate relative percentage change versus standard deviation

A 1 standard deviation increase (

25

ms) above the mean (

180

ms) equals an increase of

\frac{25}{180} \approx 13.89\%

.

Standard deviation measures absolute dispersion in the units of the variable, not a percentage of the mean.

Key Concept

Empirical Rule, z-Score Calculation, and Percentile Ranks in Normal Distributions

Question 3Question →

The table below shows the frequency distribution of daily passenger counts (in hundreds) for a city bus route recorded over a 40-day period.

Daily Passengers (in hundreds)	Frequency (Number of Days)
$10 - 14$	$5$
$15 - 19$	$12$
$20 - 24$	$15$
$25 - 29$	$8$

If the mean of the grouped data is estimated by using the midpoint of each class interval, what is the estimated mean daily passenger count (in hundreds)?

Show answer & explanation

Answer: 20.25

Answer

The estimated mean daily passenger count is 20.25 hundred passengers.

To calculate the estimated mean of grouped data, find the midpoint of each interval, multiply each midpoint by its frequency, sum those products (

810

), and divide by the total number of observations (

40

). This gives

\frac{810}{40} = 20.25

.

Step-by-Step Solution

1

Calculate the midpoints for each of the four class intervals.

The midpoints are

\frac{10+14}{2} = 12

,

\frac{15+19}{2} = 17

,

\frac{20+24}{2} = 22

, and

\frac{25+29}{2} = 27

.

To estimate the mean of grouped frequency data, each interval is represented by its center value (midpoint).

2

Multiply each class midpoint by its frequency and calculate the total sum of these products.

(12 \times 5) + (17 \times 12) + (22 \times 15) + (27 \times 8) = 60 + 204 + 330 + 216 = 810

.

Multiplying each midpoint by its frequency computes the total estimated value contributed by all observations in that class.

3

Divide the total estimated value by the total sample size (total frequency).

\frac{810}{40} = 20.25

.

The weighted average (grouped mean) is the sum of weighted midpoints divided by the total frequency.

Key Concept

Estimated Mean of Grouped Data

Question 4Question →

A dataset of employee salaries has a standard deviation of $6$ . If every salary in the dataset is multiplied by $3$ and then increased by $10$ , what is the standard deviation of the resulting dataset?

6

18

28

16

54

Show answer & explanation

Answer: $18$

Answer

The standard deviation of the resulting dataset is

18

.

Multiplying each data point by

3

increases the spread around the mean by a factor of

3

, changing the standard deviation from

6

to

18

. Adding

10

to each value shifts all values by the same amount, which leaves the distance between data points and the mean unchanged. Thus, the standard deviation remains

18

.

Step-by-Step Solution

1

Evaluate the effect of multiplying each value in the dataset by a positive constant

c = 3

.

The standard deviation is scaled by

3

, giving a new standard deviation of

6 \times 3 = 18

.

Scaling every data value by a constant factor

c

multiplies the measures of dispersion (such as standard deviation, range, and IQR) by

|c|

.

2

Evaluate the effect of adding a constant

k = 10

to each value in the dataset.

The standard deviation remains

18

.

Adding a constant to every data value shifts the entire distribution along the number line without changing the relative distances between values or the overall spread.

Key Concept

Linear Transformations on Standard Deviation

Estimated Time:45s

Question 5Question →

A museum surveyed $150$ visitors regarding their attendance at two special exhibitions: a Fine Art exhibit and a Natural History exhibit. Among the visitors surveyed, $85$ attended the Fine Art exhibit, $70$ attended the Natural History exhibit, and $20$ attended neither exhibit. How many visitors attended both the Fine Art exhibit and the Natural History exhibit?

5

25

45

65

155

Show answer & explanation

Answer: $25$

Answer

The correct answer is

25

visitors.

Using the inclusion-exclusion formula

\text{Total} = |A| + |B| - |A \cap B| + \text{Neither}

, we substitute the given values:

150 = 85 + 70 - |A \cap B| + 20

. Simplifying gives

150 = 175 - |A \cap B|

, which yields

|A \cap B| = 25

. Thus,

25

visitors attended both exhibits.

Step-by-Step Solution

1

Identify the given set values and formula.

Total visitors

= 150

, Fine Art attendees

|A| = 85

, Natural History attendees

|H| = 70

, Neither

= 20

.

The principle of inclusion-exclusion for two sets states that

\text{Total} = |A| + |H| - |A \cap H| + \text{Neither}

.

2

Calculate the number of visitors who attended at least one exhibit.

|A \cup H| = 150 - 20 = 130

.

Subtracting those who attended neither exhibit from the total population yields the total number of unique visitors who attended at least one of the two exhibits.

3

Solve for the intersection

|A \cap H|

.

130 = 85 + 70 - |A \cap H| \implies 130 = 155 - |A \cap H| \implies |A \cap H| = 25

.

Subtracting the union

|A \cup H|

from the sum of the individual sets

|A| + |H|

eliminates the double-counted intersection.

Key Concept

Two-Set Principle of Inclusion-Exclusion

Estimated Time:1m 0s

Question 6Question →

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

Select all that apply

12

15

17

24

32

Show answer & explanation

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

Answer

The possible values of

x

are

12

,

17

, and

32

.

The mean of the dataset is given by

\frac{68+x}{5}

. Depending on where

x

falls relative to the sorted known values (

12, 16, 20, 20

), the median can be

16

(if

x \le 16

),

x

(if

16 \le x \le 20

), or

20

(if

x \ge 20

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

x = 12

,

x = 17

, and

x = 32

.

Step-by-Step Solution

1

Express the arithmetic mean in terms of

x

.

The sum of the five numbers is

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

. Therefore, the mean is

\frac{68 + x}{5}

.

The mean of

n

numbers is the total sum divided by

n

.

2

Analyze Case 1 where

x \le 16

.

The sorted dataset is

x, 12, 16, 20, 20

(or

12, x, 16, 20, 20

). The median (3rd value) is

16

. Setting mean = median gives

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

. Since

12 \le 16

,

x = 12

is valid.

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

3

Analyze Case 2 where

16 \le x \le 20

.

The sorted dataset is

12, 16, x, 20, 20

. The median is

x

. Setting mean = median gives

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

. Since

16 \le 17 \le 20

,

x = 17

is valid.

When

x

lies between 16 and 20,

x

itself becomes the 3rd element of the ordered set.

4

Analyze Case 3 where

x \ge 20

.

The sorted dataset is

12, 16, 20, 20, x

. The median is

20

. Setting mean = median gives

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

. Since

32 \ge 20

,

x = 32

is valid.

When

x \ge 20

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

Key Concept

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

Question 7Question →

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

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

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

2

3

4

5

6

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. There are

5 + 8 = 13

values less than

3

. For

3

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

13

. With

k = 4

, the total number of items is

N = 23 + 4 = 27

. For an odd dataset of size

27

, the median is the

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

th item. Since the first

13

items are less than

3

, the

14

th item is

3

, making

3

the median. Any value of

k \le 3

results in a median less than

3

.

Step-by-Step Solution

1

Calculate the cumulative frequency of values below 3 days.

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

5 + 8 = 13

.

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

2

Express the total number of requests

N

in terms of

k

.

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

.

The total number of requests determines whether

N

is odd or even and where the median position lies.

3

Determine the condition for 3 days to be the median.

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

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

4

Test minimum integer values for

k

.

If

k = 3

,

N = 26

. The median is the average of the

13

th value (

2

) and the

14

th value (

3

), which is

2.5

. If

k = 4

,

N = 27

. The median is the

14

th value, which is

3

.

Testing

k = 4

yields

N = 27

where the

14

th value is

3

, satisfying the median requirement.

Key Concept

Median of a Frequency Distribution Table

Estimated Time:1m 30s

Question 8Question →

A survey of $120$ university students evaluated course enrollment in Computer Science and Mathematics. Of the students surveyed, $70$ are enrolled in Computer Science, $65$ are enrolled in Mathematics, and $25$ are enrolled in neither course. How many students are enrolled in both Computer Science and Mathematics?

15

25

40

50

95

Show answer & explanation

Answer: 40

Answer

40 students are enrolled in both Computer Science and Mathematics.

The correct answer is 40. From the total population of 120 students, 25 take neither subject, meaning

120 - 25 = 95

students take at least one subject. According to the Principle of Inclusion-Exclusion,

|C \cup M| = |C| + |M| - |C \cap M|

. Substituting the known values yields

95 = 70 + 65 - |C \cap M|

, which simplifies to

95 = 135 - |C \cap M|

, giving

|C \cap M| = 40

.

Step-by-Step Solution

1

Calculate the total number of students enrolled in at least one of the two courses.

Total in at least one course =

120 - 25 = 95

.

Subtracting the students enrolled in neither course from the total surveyed gives the union of the two sets,

|C \cup M|

.

2

Apply the Principle of Inclusion-Exclusion for two sets.

|C \cup M| = |C| + |M| - |C \cap M|

, so

95 = 70 + 65 - |C \cap M|

.

Adding

|C|

and

|M|

double-counts the students taking both courses, so subtracting the union yields the intersection.

3

Solve for the intersection

|C \cap M|

.

|C \cap M| = 135 - 95 = 40

.

Direct arithmetic evaluation gives the required count of students enrolled in both subjects.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:1m 0s

Question 9Question →

In a department of $90$ employees, $55$ speak French, $45$ speak Spanish, and $10$ speak neither French nor Spanish. How many employees speak both French and Spanish?

Show answer & explanation

Answer: 20

Answer

20

Subtracting the 10 employees who speak neither language from the total of 90 leaves 80 employees who speak at least one language. By the principle of inclusion-exclusion, Total(At least one) = French + Spanish - Both. Substituting the known values gives 80 = 55 + 45 - Both, which simplifies to 80 = 100 - Both, so Both = 20.

Step-by-Step Solution

1

Find the number of employees who speak at least one of the languages

80 employees

Subtract the 10 employees who speak neither language from the total department size of 90.

2

Sum the total counts for each language group

100

Add the number of French speakers (55) to Spanish speakers (45).

3

Calculate the overlap (intersection) using inclusion-exclusion

20 employees

Subtract the count of employees speaking at least one language (80) from the sum of the individual language groups (100).

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Estimated Time:45s

Question 10Question →

Dataset $D$ consists of $40$ distinct real numbers arranged in strictly increasing order: $x_1 < x_2 < \dots < x_{40}$ . Let $R_D$ , $Q_D$ , and $\sigma_D$ denote the range, interquartile range, and standard deviation of Dataset $D$ , respectively.

A new dataset, Dataset $E$ , is created by replacing each of the $10$ largest numbers ( $x_{31}, x_{32}, \dots, x_{40}$ ) with the value of $x_{30}$ , which represents the $75\text{th}$ percentile of Dataset $D$ . Let $R_E$ , $Q_E$ , and $\sigma_E$ denote the range, interquartile range, and standard deviation of Dataset $E$ , respectively.

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

Select all that apply

The range of Dataset

E

is strictly less than the range of Dataset

D

(

R_E < R_D

).

The interquartile range of Dataset

E

is equal to the interquartile range of Dataset

D

(

Q_E = Q_D

).

The standard deviation of Dataset

E

is strictly less than the standard deviation of Dataset

D

(

\sigma_E < \sigma_D

).

The median of Dataset

E

is strictly less than the median of Dataset

D

.

The interquartile range of Dataset

E

is strictly less than the interquartile range of Dataset

D

(

Q_E < Q_D

).

Show answer & explanation

Answer: The range of Dataset $E$ is strictly less than the range of Dataset $D$ ( $R_E < R_D$ ).; The interquartile range of Dataset $E$ is equal to the interquartile range of Dataset $D$ ( $Q_E = Q_D$ ).; The standard deviation of Dataset $E$ is strictly less than the standard deviation of Dataset $D$ ( $\sigma_E < \sigma_D$ ).

Answer

The statements asserting that the range of Dataset E is strictly less than the range of Dataset D (

R_E < R_D

), the interquartile range of Dataset E is equal to the interquartile range of Dataset D (

Q_E = Q_D

), and the standard deviation of Dataset E is strictly less than the standard deviation of Dataset D (

\sigma_E < \sigma_D

) are all correct.

The range of Dataset E decreases because its maximum value is reduced from

x_{40}

to

x_{30}

while its minimum value

x_1

remains the same. The interquartile range remains unchanged because the boundary values defining the

25\text{th}

percentile (

x_{10}

) and

75\text{th}

percentile (

x_{30}

) are not altered. The standard deviation strictly decreases because reducing the

10

largest values brings them closer to the rest of the distribution, decreasing total variation.

Step-by-Step Solution

1

Analyze the change in the maximum and minimum values to evaluate the Range.

In Dataset

D

,

\text{Range } R_D = x_{40} - x_1

. In Dataset

E

, the maximum element becomes

x_{30}

while the minimum remains

x_1

. Since all numbers are distinct and ordered,

x_{30} < x_{40}

, so

R_E = x_{30} - x_1 < R_D

. Thus,

R_E < R_D

is true.

Range is defined as

\text{Maximum} - \text{Minimum}

.

2

Analyze the

25\text{th}

percentile (

Q_1

) and

75\text{th}

percentile (

Q_3

) to evaluate the Interquartile Range.

For

N=40

,

Q_1 = x_{10}

and

Q_3 = x_{30}

. Dataset

E

modifies only the values strictly above

x_{30}

(

x_{31}

through

x_{40}

). Thus

x_{10}

and

x_{30}

retain their exact values in Dataset

E

. Therefore,

Q_E = x_{30} - x_{10} = Q_D

.

Interquartile range depends solely on

Q_3 - Q_1

, which relies on position cutoffs that were unaffected.

3

Analyze the effect of pulling extreme upper values inward on the Standard Deviation.

Standard deviation measures overall spread around the mean and can be calculated via pairwise differences

\text{Var}(X) = \frac{1}{2N^2} \sum_{i,j} (x_i - x_j)^2

. Replacing

x_{31}, \dots, x_{40}

with

x_{30}

strictly reduces the distances between these upper points and all other points, ensuring

\sigma_E < \sigma_D

.

Reducing the distance of tail observations from the rest of the dataset strictly decreases overall variance and standard deviation.

Key Concept

Impact of Data Transformations on Measures of Dispersion and Position

Question 11Question →

The frequency distribution table below summarizes the processing times, in minutes, for a sample of 100 health insurance claims processed by an administrative team.

Claim Processing Time (minutes)	Frequency
$10 \le t < 20$	15
$20 \le t < 30$	25
$30 \le t < 40$	35
$40 \le t < 50$	20
$50 \le t < 60$	5

Which of the following statements regarding this frequency distribution must be true? Select all that apply.

Select all that apply

The interval

30 \le t < 40

contains the median processing time for the dataset.

Exactly 60 percent of the claims required 30 minutes or more to process.

The lower bound estimate for the overall mean processing time is 31.0 minutes.

The number of claims in the interval

20 \le t < 30

represents 25 percent of all claims.

The ratio of the number of claims completed in under 30 minutes to the number of claims taking 40 minutes or more is 2 to 1.

Show answer & explanation

Answer: The interval $30 \le t < 40$ contains the median processing time for the dataset.; Exactly 60 percent of the claims required 30 minutes or more to process.; The number of claims in the interval $20 \le t < 30$ represents 25 percent of all claims.

Answer

The true statements are: the median processing time lies in the interval

30 \le t < 40

; exactly 60 percent of the claims required 30 minutes or more to process; and the number of claims in the interval

20 \le t < 30

represents 25 percent of all claims.

The statement regarding the median is correct because the 50th and 51st values lie within the cumulative frequency span of 41 to 75, corresponding to

30 \le t < 40

. The statement regarding claims requiring 30 minutes or more is correct because

35 + 20 + 5 = 60

out of 100 claims is 60 percent. The statement regarding the

20 \le t < 30

interval is correct because 25 out of 100 claims is exactly 25 percent.

Step-by-Step Solution

1

Calculate cumulative frequencies to locate the median class interval.

Cumulative frequencies are 15 (

10 \le t < 20

), 40 (

20 \le t < 30

), 75 (

30 \le t < 40

), 95 (

40 \le t < 50

), and 100 (

50 \le t < 60

). For 100 observations, the 50th and 51st values fall in the

30 \le t < 40

interval.

The median of an even number

N=100

of sorted values is the average of the 50th and 51st values.

2

Calculate the percentage of claims taking 30 minutes or more.

Sum the frequencies for intervals

\ge 30

:

35 + 20 + 5 = 60

. As a percentage of 100 total claims, this is

\frac{60}{100} \times 100\% = 60\%

.

To find the proportion meeting or exceeding a threshold, sum all class frequencies with lower limits at or above 30 minutes.

3

Evaluate the percentage representation of the

20 \le t < 30

interval.

Frequency is 25 out of 100 total claims, which equals

\frac{25}{100} = 25\%

.

The relative frequency of a grouped class is its frequency divided by the total frequency.

4

Determine the lower bound for the estimated mean processing time to evaluate incorrect claims.

Minimum sum =

15(10) + 25(20) + 35(30) + 20(40) + 5(50) = 2750

. Lower bound mean =

2750 / 100 = 27.5

minutes.

The absolute minimum mean occurs when every observation equals the lower boundary of its respective interval.

5

Calculate the ratio of claims under 30 minutes to claims 40 minutes or more.

Claims

< 30

min =

15 + 25 = 40

. Claims

\ge 40

min =

20 + 5 = 25

. Ratio =

40 : 25 = 8 : 5

.

Must include all relevant intervals (including

50 \le t < 60

) when computing the denominator for

\ge 40

minutes.

Key Concept

Grouped Frequency Distributions, Median Class Identification, and Relative Frequencies

Question 12Question →

A university research department needs to form a 4-person delegation consisting of exactly 2 senior researchers and 2 junior researchers. The pool of available candidates contains 5 senior researchers and 6 junior researchers. However, 2 specific senior researchers refuse to serve on the delegation together. How many different 4-person delegations can be formed under these conditions?

120

135

150

270

300

Show answer & explanation

Answer: 135

Answer

135 different 4-person delegations can be formed.

To calculate the number of valid delegations, first find the unrestricted number of delegations by multiplying the combinations of senior researchers

\binom{5}{2} = 10

and junior researchers

\binom{6}{2} = 15

, which equals 150. Next, calculate the number of invalid delegations where both restricted senior researchers are chosen together, which is

\binom{2}{2} \times \binom{6}{2} = 15

. Subtracting the invalid delegations from the total gives

150 - 15 = 135

.

Step-by-Step Solution

1

Calculate the total number of possible delegations without any restrictions.

\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150

Choosing 2 senior researchers out of 5 requires combinations,

\binom{5}{2} = 10

. Choosing 2 junior researchers out of 6 requires combinations,

\binom{6}{2} = 15

. By the Fundamental Counting Principle, the total number of unrestricted delegations is

10 \times 15 = 150

.

2

Calculate the number of invalid delegations that include both restricted senior researchers.

\binom{2}{2} \times \binom{6}{2} = 1 \times 15 = 15

If both conflicting senior researchers are placed on the delegation, there is only

\binom{2}{2} = 1

way to select the senior members. The 2 junior researchers can still be selected in

\binom{6}{2} = 15

ways, giving 15 invalid delegations.

3

Subtract the number of invalid delegations from the total possible delegations.

150 - 15 = 135

Subtracting the 15 restricted delegations from the 150 total delegations leaves 135 valid delegations.

Key Concept

Combinations with Exclusion Restrictions and Fundamental Counting Principle

Question 13Question →

In a reliability study of a power station, Event $A$ is defined as an inverter system malfunction and Event $B$ is defined as a battery backup failure during peak hours. The probability that at least one of these two malfunctions occurs is $0.80$ , and the probability that an inverter system malfunction occurs without a battery backup failure is $0.35$ . If Event $A$ and Event $B$ are independent events, what is the probability of an inverter system malfunction, $P(A)$ ?

\frac{7}{20}

\frac{9}{20}

\frac{7}{11}

\frac{11}{20}

\frac{4}{5}

Show answer & explanation

Answer: $\frac{7}{11}$

Answer

The probability of an inverter system malfunction,

P(A)

, is

\frac{7}{11}

.

The probability of the union of two events can be decomposed as

P(A \cup B) = P(A \cap B^c) + P(B)

. Given

P(A \cup B) = 0.80

and

P(A \cap B^c) = 0.35

, solving yields

P(B) = 0.45

. Because Event

A

and Event

B

are independent,

P(A \cap B) = P(A)P(B) = 0.45 P(A)

. Since

P(A \cap B^c) = P(A) - P(A \cap B)

, we substitute to get

0.35 = P(A) - 0.45 P(A) = 0.55 P(A)

. Solving for

P(A)

gives

P(A) = \frac{0.35}{0.55} = \frac{7}{11}

.

Step-by-Step Solution

1

Relate the union probability to the difference event

A \cap B^c

and event

B

.

P(A \cup B) = P(A \cap B^c) + P(B)

.

The union of two events

A

and

B

can be partitioned into the region where only

A

occurs (

A \cap B^c

) and all outcomes in

B

.

2

Calculate

P(B)

using the given values.

P(B) = P(A \cup B) - P(A \cap B^c) = 0.80 - 0.35 = 0.45 = \frac{9}{20}

.

Subtracting

P(A \cap B^c)

from

P(A \cup B)

yields

P(B)

.

3

Apply the definition of independent events to express

P(A \cap B)

in terms of

P(A)

.

P(A \cap B) = P(A) \cdot P(B) = 0.45 P(A)

.

For independent events, the joint probability is the product of their individual probabilities.

4

Set up and solve the equation for

P(A)

.

P(A \cap B^c) = P(A) - P(A \cap B) \implies 0.35 = P(A) - 0.45 P(A) = 0.55 P(A) \implies P(A) = \frac{0.35}{0.55} = \frac{7}{11}

.

Dividing

0.35

by

0.55

yields the exact value of

P(A)

.

Key Concept

Probability rules for independent events and set operations on sample spaces

Estimated Time:2m 0s

Question 14Question →

A quality control inspector evaluates a batch of $16$ precision components. Exactly $10$ of the components meet all engineering specifications, while $6$ have minor surface defects. If the inspector randomly selects $2$ components from the batch one after another without replacement, what is the probability that both selected components meet all engineering specifications?

Show answer & explanation

Answer: 0.375

Answer

The probability that both selected components meet all engineering specifications is

0.375

(or

\frac{3}{8}

).

Since the components are selected without replacement, the outcome of the second draw depends on the outcome of the first draw. The probability of selecting a qualifying component first is

\frac{10}{16}

. Given that a qualifying component was drawn first,

9

qualifying components remain out of

15

total components. The joint probability of both events occurring is

\frac{10}{16} \times \frac{9}{15} = \frac{90}{240} = \frac{3}{8} = 0.375

.

Step-by-Step Solution

1

Determine the probability of selecting a component meeting specifications on the first draw.

P(E_1) = \frac{10}{16} = \frac{5}{8}

There are

10

qualifying components out of

16

total components.

2

Determine the conditional probability of selecting a second component meeting specifications, given that the first component selected also met specifications.

P(E_2 \mid E_1) = \frac{9}{15} = \frac{3}{5}

Because sampling is done without replacement,

9

qualifying components remain out of

15

total remaining components.

3

Apply the multiplication rule for dependent events to calculate the probability of both events occurring.

P(E_1 \cap E_2) = P(E_1) \times P(E_2 \mid E_1) = \frac{5}{8} \times \frac{3}{5} = \frac{3}{8} = 0.375

For dependent events, the joint probability is the product of the first event's probability and the conditional probability of the second event.

Key Concept

Probability of Dependent Events without Replacement

Estimated Time:1m 15s

Question 15Question →

A constant value of $7$ is added to each number in a dataset to form a new dataset. Which of the following statistical measures of the new dataset are equal to the corresponding measures of the original dataset? Select all that apply.

Select all that apply

The range

The standard deviation

The interquartile range

The mean

The median

Show answer & explanation

Answer: The range; The standard deviation; The interquartile range

Answer

The range, the standard deviation, and the interquartile range of the new dataset are equal to those of the original dataset.

Adding a constant to every value in a dataset shifts the entire distribution uniformly without altering its overall shape or spread. Consequently, measures of dispersion—including the range, standard deviation, and interquartile range—remain completely unchanged.

Step-by-Step Solution

1

Evaluate the effect of adding a constant on measures of central tendency

If each data point

x_i

is replaced by

x_i + 7

, the mean

\mu

becomes

\mu + 7

and the median

M

becomes

M + 7

.

Measures of central tendency shift directly by the added constant, so they are not equal to the original measures.

2

Evaluate the effect of adding a constant on measures of dispersion

The new range is

(x_{\text{max}} + 7) - (x_{\text{min}} + 7) = x_{\text{max}} - x_{\text{min}}

. The new interquartile range is

(Q_3 + 7) - (Q_1 + 7) = Q_3 - Q_1

. Each deviation from the mean is

(x_i + 7) - (\mu + 7) = x_i - \mu

, so the standard deviation is unchanged.

Measures of dispersion reflect the relative distances between data points or distance from the mean, which are unaffected by shifting the whole distribution.

Key Concept

Effect of adding a constant on measures of central tendency versus measures of dispersion

Question 16Question →

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

18.5

19.0

19.5

20.0

20.5

Show answer & explanation

Answer: $19.5$

Answer

The maximum possible median of dataset

R

is

19.5

.

In the sorted set of

9

distinct integers, the median is the

5

th element, which is

20

. Removing the largest value leaves

8

integers, whose median is the average of the

4

th and

5

th elements. Because all integers are distinct, the

4

th element can be at most

19

. The maximum median is therefore

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

.

Step-by-Step Solution

1

Identify the position of the median in the original dataset

S

.

In a dataset of

9

sorted distinct integers

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

, the median is the

5

th element, so

x_5 = 20

.

For an odd number of ordered elements

n=9

, the median is at index

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

.

2

Determine the formula for the median of the modified dataset

R

.

Removing the largest element

x_9

leaves

8

ordered elements

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

. The median of

R

is

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

.

For an even number of ordered elements

n=8

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

4

and

5

.

3

Maximize the value of

x_4

.

Since all integers are distinct and

x_5 = 20

,

x_4

must be an integer strictly less than

20

. Thus, the maximum integer value

x_4

can take is

19

.

Maximizing

x_4

maximizes the average

\frac{x_4 + 20}{2}

.

4

Verify that

x_4 = 19

is achievable under the given mean constraint.

The sum of the smallest

4

integers is

4 \times 10 = 40

. With

x_4 = 19

, we require

x_1 + x_2 + x_3 = 21

. Choosing distinct integers such as

x_1 = 1

,

x_2 = 2

, and

x_3 = 18

satisfies

1 + 2 + 18 = 21

and

x_1 < x_2 < x_3 < x_4

.

Confirming feasibility ensures the upper bound

19

is valid.

5

Calculate the maximum median of dataset

R

.

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

.

Substitute the maximum value of

x_4 = 19

into the median expression.

Key Concept

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

Question 17Question →

In a group of 200 student test scores, a score of $S$ is located at the 75th percentile of the distribution. Which of the following statements must be true?

Score

S

is equal to 75.

Approximately 75 students scored at or below

S

.

Approximately 150 students scored at or below

S

.

Score

S

is 75 percent greater than the average score of the distribution.

The standard deviation of the test scores is equal to 75.

Show answer & explanation

Answer: Approximately 150 students scored at or below $S$ .

Answer

Approximately 150 students scored at or below

S

.

The 75th percentile of a distribution indicates that approximately

75\%

of the data points lie at or below that value. For a group of 200 students,

75\%

of 200 is calculated as

0.75 \times 200 = 150

. Therefore, approximately 150 students achieved a score at or below

S

.

Step-by-Step Solution

1

Understand the definition of a percentile rank

The 75th percentile means that approximately

75\%

of all observations in the dataset are less than or equal to that score.

By definition, the

p

-th percentile represents the value below which

p

percent of the observations fall.

2

Calculate 75% of the total number of students

0.75 \times 200 = 150

students.

The dataset contains 200 total student scores, so

75\%

of 200 yields the number of students at or below score

S

.

Key Concept

Percentile Position

Estimated Time:45s

Question 18Question →

Dataset $X$ consists of seven numerical values: $\{12, 16, 20, 24, 28, 32, 36\}$ . Dataset $Y$ is formed by replacing the value $36$ in Dataset $X$ with $56$ , while keeping all other six values the same. Which of the following statements correctly compares the interquartile range (IQR) and standard deviation of Dataset $Y$ to those of Dataset $X$ ?

The IQR remains unchanged, while the standard deviation increases.

Both the IQR and the standard deviation increase.

The IQR increases, while the standard deviation remains unchanged.

Both the IQR and the standard deviation remain unchanged.

The IQR decreases, while the standard deviation increases.

Show answer & explanation

Answer: The IQR remains unchanged, while the standard deviation increases.

Answer

The IQR remains unchanged, while the standard deviation increases.

The statement specifying that the IQR remains unchanged while the standard deviation increases is correct because the quartiles Q1 (16) and Q3 (32) are unaffected by changing the largest value from 36 to 56, maintaining an IQR of 16. In contrast, standard deviation measures spread relative to the mean, so pulling the maximum value further out increases overall deviation.

Step-by-Step Solution

1

Calculate the IQR of Dataset X

For Dataset X = {12, 16, 20, 24, 28, 32, 36}, the median is 24. The lower half is {12, 16, 20} with Q1 = 16. The upper half is {28, 32, 36} with Q3 = 32. Thus, IQR_X = 32 - 16 = 16.

IQR is calculated as the difference between the third quartile (Q3) and the first quartile (Q1).

2

Calculate the IQR of Dataset Y

For Dataset Y = {12, 16, 20, 24, 28, 32, 56}, the median is still 24. The lower half remains {12, 16, 20} (Q1 = 16). The upper half becomes {28, 32, 56}, so Q3 is still 32. Thus, IQR_Y = 32 - 16 = 16.

The third quartile is the middle number of the upper half, which remains 32 regardless of replacing 36 with 56.

3

Compare the standard deviation of Dataset X and Dataset Y

Replacing 36 with a significantly higher value 56 increases the distance of the maximum data point from the mean, increasing overall variance and thus increasing the standard deviation.

Standard deviation measures the average distance of data points from the mean and is highly sensitive to extreme values/outliers.

Key Concept

Sensitivity of Measures of Dispersion to Outliers

Question 19Question →

The high temperatures, in degrees Fahrenheit, recorded in a city over a 7-day period were 64, 58, 75, 67, 61, 83, and 72. What is the interquartile range of these temperatures, in degrees Fahrenheit?

Show answer & explanation

Answer: 14

Answer

The interquartile range of the recorded temperatures is 14.

To calculate the interquartile range, first arrange the dataset in ascending order: 58, 61, 64, 67, 72, 75, 83. The median of the 7 values is 67. The first quartile (

Q_1

) is the median of the lower three values (58, 61, 64), which is 61. The third quartile (

Q_3

) is the median of the upper three values (72, 75, 83), which is 75. Subtracting

Q_1

from

Q_3

yields an interquartile range of

75 - 61 = 14

.

Step-by-Step Solution

1

Order the dataset from least to greatest

58, 61, 64, 67, 72, 75, 83

Finding quartiles requires data to be arranged in ascending order.

2

Find the first quartile (

Q_1

) and third quartile (

Q_3

)

Q_1 = 61

and

Q_3 = 75

The median of the dataset (the 4th value) is 67. The lower half of the data consists of 58, 61, 64 (median 61), and the upper half consists of 72, 75, 83 (median 75).

3

Compute the difference between

Q_3

and

Q_1

75 - 61 = 14

The interquartile range is defined as

IQR = Q_3 - Q_1

.

Key Concept

Interquartile Range (IQR)

Question 20Question →

The table below shows the distribution of scores achieved by 20 students on a final exam:

Score	Frequency
60	3
70	5
80	6
90	4
100	2

What is the interquartile range (IQR) of the scores?

Show answer & explanation

Answer: 20

Answer

The interquartile range (IQR) of the scores is 20.

For a set of 20 ordered scores, the first quartile

Q_1

is the median of the first 10 scores (average of the 5th and 6th values), and the third quartile

Q_3

is the median of the last 10 scores (average of the 15th and 16th values). Using the cumulative frequencies, the 5th and 6th scores are both 70 (

Q_1 = 70

), and the 15th and 16th scores are both 90 (

Q_3 = 90

). Thus, the interquartile range is

IQR = Q_3 - Q_1 = 90 - 70 = 20

.

Step-by-Step Solution

1

Determine the cumulative frequency to locate quartile positions.

Score 60 occupies positions 1 to 3; Score 70 occupies positions 4 to 8; Score 80 occupies positions 9 to 14; Score 90 occupies positions 15 to 18; Score 100 occupies positions 19 to 20.

Tracking data positions in a frequency distribution allows efficient determination of medians and quartiles without expanding the raw list.

2

Calculate the first quartile (

Q_1

).

Q_1 = \frac{70 + 70}{2} = 70

.

With 20 total values, the lower half consists of the first 10 values (positions 1 through 10). The median of these 10 values is the average of the 5th and 6th values, which are both 70.

3

Calculate the third quartile (

Q_3

).

Q_3 = \frac{90 + 90}{2} = 90

.

The upper half consists of the last 10 values (positions 11 through 20). The median of these 10 values is the average of the 15th and 16th values, which are both 90.

4

Subtract

Q_1

from

Q_3

to find the interquartile range.

IQR = Q_3 - Q_1 = 90 - 70 = 20

.

The interquartile range represents the spread of the middle 50% of the dataset.

Key Concept

Interquartile Range (IQR) from a Frequency Distribution

Estimated Time:1m 30s

Data Analysis

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