Data Analysis Practice Questions — GRE General Test — Page 2

Question 21Question →

An environmental monitoring station recorded the daily nitrogen dioxide ( $\text{NO}_2$ ) concentrations, in parts per billion (ppb), for 6 consecutive days: $19, 28, 30, 35, 42,$ and $42$ . On the 7th day, a measurement of $x$ ppb was recorded, where $x > 45$ . If the arithmetic mean of the 7 daily concentrations is equal to their median, what is the value of $x$ ?

Show answer & explanation

Answer: 49

Answer

The value of

x

is

49

.

Sorting the first 6 recorded concentrations gives

19, 28, 30, 35, 42, 42

. Because

x > 45

,

x

is strictly greater than all existing values, so the full set in ascending order is

19, 28, 30, 35, 42, 42, x

. The median of a 7-element set is the 4th value, which is

35

. The sum of the 7 concentrations is

19 + 28 + 30 + 35 + 42 + 42 + x = 196 + x

, making the arithmetic mean

\frac{196 + x}{7}

. Setting the mean equal to the median gives

\frac{196 + x}{7} = 35 \implies 196 + x = 245 \implies x = 49

.

Step-by-Step Solution

1

Order the first 6 data points in ascending order

The sorted list is

19, 28, 30, 35, 42, 42

.

Establishing ordered positions is necessary to determine the median.

2

Determine the median of the 7-element dataset

Since

x > 45

, the complete ordered dataset is

19, 28, 30, 35, 42, 42, x

, making the 4th element,

35

, the median.

The median of an odd number of sorted values is the middle element.

3

Express the arithmetic mean in terms of

x

and set it equal to the median

\frac{196 + x}{7} = 35

The problem states that the arithmetic mean equals the median.

4

Solve the linear equation for

x

196 + x = 245 \implies x = 49

Multiplying both sides by 7 and subtracting 196 isolates

x

.

Key Concept

Measures of Central Tendency (Mean and Median)

Question 22Question →

A specialty tea shop sells five distinct sizes of loose-leaf tea canisters. The table below shows the weight per canister and the number of canisters sold during a promotional week:

Canister Size	Weight per Canister (ounces)	Number of Canisters Sold
Sampler	5	16
Small	12	18
Medium	15	$n$
Large	20	6
Executive	25	4

If the mean weight of all canisters sold during the week is $13$ ounces, what is the median weight, in ounces, of the canisters sold?

12

ounces

13

ounces

15

ounces

15.4

ounces

16

ounces

Show answer & explanation

Answer: $15$ ounces

Answer

15 ounces

The correct answer is 15 ounces. Calculating the total weight as a function of

n

gives

516 + 15n

, and the total number of canisters as

44 + n

. Setting the mean

\frac{516 + 15n}{44 + n} = 13

yields

n = 28

. The total number of canisters sold is

44 + 28 = 72

. For 72 items sorted in ascending order, the median is the average of the 36th and 37th values. Cumulative frequencies show that the 1st through 16th items are 5 ounces, the 17th through 34th items are 12 ounces, and the 35th through 62nd items are 15 ounces. Thus, both the 36th and 37th items weigh 15 ounces, making the median 15 ounces.

Step-by-Step Solution

1

Set up an equation for the total number of canisters and total weight.

Total canisters =

16 + 18 + n + 6 + 4 = 44 + n

. Total weight =

5(16) + 12(18) + 15(n) + 20(6) + 25(4) = 516 + 15n

ounces.

The mean of a weighted dataset is the total sum of all values divided by the total number of items.

2

Solve for the missing frequency

n

using the given mean of 13 ounces.

\frac{516 + 15n}{44 + n} = 13 \implies 516 + 15n = 13(44 + n) \implies 516 + 15n = 572 + 13n \implies 2n = 56 \implies n = 28

.

Equating the weighted average expression to 13 allows us to determine the exact number of Medium canisters sold.

3

Determine the total number of items and locate the position of the median.

Total canisters =

44 + 28 = 72

. The median is the average of the

36^{\text{th}}

and

37^{\text{th}}

values when ordered.

For an even number of data points

N = 72

, the median is the arithmetic mean of the

(N/2)^{\text{th}}

and

(N/2 + 1)^{\text{th}}

items.

4

Compute the cumulative frequency distribution to find the values at the

36^{\text{th}}

and

37^{\text{th}}

positions.

5 oz: positions 1 to 16; 12 oz: positions 17 to 34; 15 oz: positions 35 to 62. Both

36^{\text{th}}

and

37^{\text{th}}

positions have a weight of 15 ounces, so the median is 15 ounces.

Cumulative counting reveals that items in positions 35 through 62 all correspond to 15-ounce canisters.

Key Concept

Weighted mean and finding the median of frequency distribution data

Question 23Question →

The distribution of scores on Exam X is normally distributed with a mean of $70$ and a standard deviation of $8$ . The distribution of scores on Exam Y is normally distributed with a mean of $150$ and a standard deviation of $20$ . Candidate Score $A$ is on Exam X and has a z-score of $+1.5$ . Candidate Score $B$ is on Exam Y and corresponds to the 84th percentile of Exam Y.

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

Select all that apply

Candidate Score

A

is equal to

82

.

The proportion of test-takers on Exam Y scoring higher than Candidate Score

B

is approximately

16\%

.

A score of

62

on Exam X corresponds to a higher percentile rank than Candidate Score

B

on Exam Y.

A score of

86

on Exam X corresponds to the 99.7th percentile of Exam X.

A score of

110

on Exam Y has a z-score of

-2.0

.

Show answer & explanation

Answer: Candidate Score $A$ is equal to $82$ .; The proportion of test-takers on Exam Y scoring higher than Candidate Score $B$ is approximately $16\%$ .; A score of $110$ on Exam Y has a z-score of $-2.0$ .

Answer

The statements asserting that Candidate Score

A

is equal to

82

, that approximately

16\%

of test-takers score higher than Candidate Score

B

, and that a score of

110

on Exam Y has a z-score of

-2.0

are all correct.

The correct statements correctly apply the z-score formula and standard normal curve benchmarks: Candidate Score

A

is

82

because

70 + 1.5(8) = 82

; the proportion of scores above the 84th percentile is

100\% - 84\% = 16\%

; and a score of

110

on Exam Y is

40

points below the mean of

150

, giving a z-score of

\frac{-40}{20} = -2.0

.

Step-by-Step Solution

1

Calculate Candidate Score

A

on Exam X using the z-score formula.

A = \mu + z \cdot \sigma = 70 + (1.5)(8) = 82

.

A z-score of

+1.5

means the score is

1.5

standard deviations above the mean.

2

Analyze Candidate Score

B

at the 84th percentile of Exam Y.

Percentile rank of 84th means

84\%

score below or equal to

B

, so

100\% - 84\% = 16\%

score higher than

B

. Under the standard normal curve (empirical rule), the 84th percentile corresponds to

z = +1.0

, which gives

B = 150 + 1.0(20) = 170

.

Percentile rank measures the proportion of values below a given point.

3

Evaluate the percentile rank of a score of

62

on Exam X.

z = \frac{62 - 70}{8} = -1.0

. A z-score of

-1.0

corresponds to the 16th percentile (

50\% - 34\% = 16\%

).

Since the 16th percentile is less than the 84th percentile (Candidate Score

B

), this statement is false.

4

Evaluate the percentile rank of a score of

86

on Exam X.

z = \frac{86 - 70}{8} = +2.0

. A z-score of

+2.0

corresponds to the 97.5th percentile (

50\% + 34\% + 13.5\% = 97.5\%

).

The 99.7th percentile corresponds to

z = +3.0

(score of

94

), making this statement false.

5

Calculate the z-score for a score of

110

on Exam Y.

z = \frac{110 - 150}{20} = -2.0

.

The raw score

110

is

40

units below the mean of

150

, which is exactly

2

standard deviations below the mean.

Key Concept

Normal Distribution, Z-Scores, and Percentile Benchmarks

Estimated Time:2m 0s

Question 24Question →

Two events $A$ and $B$ within a sample space have probabilities $P(A) = 0.60$ and $P(B) = 0.30$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If events

A

and

B

are independent, the probability that at least one of the two events occurs is

0.72

.

If events

A

and

B

are mutually exclusive, the joint probability

P(A \text{ and } B)

is

0

.

Events

A

and

B

can be both mutually exclusive and independent.

If events

A

and

B

are independent, the probability that both events occur is

0.90

.

If events

A

and

B

are mutually exclusive, the conditional probability

P(A \mid B)

is

0.30

.

Show answer & explanation

Answer: If events $A$ and $B$ are independent, the probability that at least one of the two events occurs is $0.72$ .; If events $A$ and $B$ are mutually exclusive, the joint probability $P(A \text{ and } B)$ is $0$ .

Answer

The correct statements are that if events are independent, the probability that at least one occurs is 0.72, and if events are mutually exclusive, their joint probability is 0.

The statement regarding independent events is correct because

P(A \cup B) = P(A) + P(B) - P(A)P(B) = 0.60 + 0.30 - 0.18 = 0.72

. The statement regarding mutually exclusive events is correct because by definition, mutually exclusive events cannot occur together, meaning

P(A \text{ and } B) = 0

.

Step-by-Step Solution

1

Analyze independence condition for

P(A \cup B)

P(A \cap B) = 0.60 \times 0.30 = 0.18

, so

P(A \cup B) = 0.60 + 0.30 - 0.18 = 0.72

.

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

2

Analyze mutual exclusivity definition

P(A \cap B) = 0

.

Mutually exclusive events cannot occur simultaneously.

3

Evaluate simultaneous independence and mutual exclusivity

Since

P(A)P(B) = 0.18 \neq 0

, the events cannot satisfy both conditions simultaneously.

Independence requires

P(A \cap B) = 0.18

, while mutual exclusivity requires

P(A \cap B) = 0

.

Key Concept

Probability rules for independent and mutually exclusive events

Question 25Question →

The frequency distribution table below shows the daily electricity consumption, in kilowatt-hours ( $\text{kWh}$ ), recorded for a sample of $40$ small retail stores over a monitoring period.

Daily Electricity Consumption ( $\text{kWh}$ )	Number of Stores
$10 \le e < 30$	$6$
$30 \le e < 50$	$14$
$50 \le e < 70$	$12$
$70 \le e < 90$	$8$

Based on the table, what is the estimated mean daily electricity consumption, in $\text{kWh}$ , per store, using the midpoint of each interval to represent all data within that interval?

Show answer & explanation

Answer: 51

Answer

51

To estimate the mean from grouped frequency data, each interval is represented by its midpoint. The midpoints for the four intervals are

20

,

40

,

60

, and

80

. Multiplying each midpoint by its frequency gives

20(6) = 120

,

40(14) = 560

,

60(12) = 720

, and

80(8) = 640

. The sum of these products is

120 + 560 + 720 + 640 = 2040

. Dividing this sum by the total frequency of

40

stores gives an estimated mean of

51\text{ kWh}

.

Step-by-Step Solution

1

Determine the class midpoint for each interval of electricity consumption.

Midpoints:

\frac{10+30}{2} = 20

,

\frac{30+50}{2} = 40

,

\frac{50+70}{2} = 60

, and

\frac{70+90}{2} = 80

.

When working with grouped frequency data, each interval is represented by its midpoint value to estimate total sums.

2

Multiply each midpoint by the number of stores in that interval to find the estimated sum of consumption per group.

20 \times 6 = 120

,

40 \times 14 = 560

,

60 \times 12 = 720

,

80 \times 8 = 640

.

Weighting each midpoint by its interval frequency yields the total estimated consumption across all stores in that bin.

3

Sum the total estimated consumption across all intervals.

120 + 560 + 720 + 640 = 2040 \text{ kWh}

.

This calculation aggregate gives the total estimated electricity consumed by all 40 stores combined.

4

Divide the total estimated electricity consumption by the total number of stores.

\frac{2040}{40} = 51 \text{ kWh}

.

Dividing the aggregate total by the sample size gives the estimated arithmetic mean per store.

Key Concept

Calculating the estimated mean of grouped data using class midpoints and weighted frequencies.

Estimated Time:1m 30s

Question 26Question →

A survey of $200$ financial executives analyzed their personal investment portfolios across three asset classes: Equities ( $E$ ), Real Estate ( $R$ ), and Commodities ( $C$ ). Exactly $20$ executives hold no investments in any of these three asset classes. Among the remaining executives, $115$ hold Equities, $95$ hold Real Estate, and $80$ hold Commodities. Furthermore, $45$ executives hold both Equities and Real Estate, $35$ hold both Real Estate and Commodities, and $40$ hold both Equities and Commodities. How many of the surveyed executives hold investments in exactly two of these three asset classes?

80

90

100

110

120

Show answer & explanation

Answer: 90

Answer

90

To find the number of executives holding investments in exactly two asset classes, we first determine the total number of executives invested in at least one asset class:

200 - 20 = 180

. Applying the three-set inclusion-exclusion formula,

180 = 115 + 95 + 80 - (45 + 35 + 40) + |E \cap R \cap C|

, which simplifies to

180 = 170 + |E \cap R \cap C|

, giving

|E \cap R \cap C| = 10

. The number of executives holding exactly two asset classes is calculated by subtracting the triple intersection (10) from each pairwise overlap:

(45 - 10) + (35 - 10) + (40 - 10) = 35 + 25 + 30 = 90

.

Step-by-Step Solution

1

Calculate the total number of executives holding at least one asset class.

The total number in the union

|E \cup R \cup C| = 200 - 20 = 180

.

Subtracting executives holding no investments gives the cardinality of the union.

2

Use the Principle of Inclusion-Exclusion for three sets to find the triple intersection

|E \cap R \cap C|

.

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

, so

180 = 115 + 95 + 80 - (45 + 35 + 40) + |E \cap R \cap C|

, yielding

180 = 290 - 120 + |E \cap R \cap C| = 170 + |E \cap R \cap C|

, hence

|E \cap R \cap C| = 10

.

The inclusion-exclusion formula relates the union, individual set sizes, pairwise overlaps, and the triple overlap.

3

Determine the number of executives in each pairwise intersection that belong ONLY to those two sets.

Equities and Real Estate only:

45 - 10 = 35

; Real Estate and Commodities only:

35 - 10 = 25

; Equities and Commodities only:

40 - 10 = 30

.

Subtracting the triple intersection count from each pairwise intersection isolates those holding exactly two asset classes.

4

Sum the counts for executives holding exactly two asset classes.

35 + 25 + 30 = 90

.

Adding these disjoint sets yields the total number of executives in exactly two categories.

Key Concept

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

Estimated Time:2m 0s

Question 27Question →

An agricultural research station evaluated $300$ soil plots for the presence of three specific mineral deficiencies: Nitrogen ( $N$ ), Phosphorus ( $P$ ), and Potassium ( $K$ ). The survey revealed the following data:
- Exactly $60$ plots exhibited none of the three deficiencies.
- $140$ plots exhibited Nitrogen deficiency ( $N$ ).
- $130$ plots exhibited Phosphorus deficiency ( $P$ ).
- $120$ plots exhibited Potassium deficiency ( $K$ ).
- Exactly $70$ plots exhibited exactly two of the three deficiencies.

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

Select all that apply

Exactly

40

plots exhibited all three mineral deficiencies.

Exactly

130

plots exhibited exactly one mineral deficiency.

The number of plots exhibiting at least two mineral deficiencies is

110

.

Exactly

170

plots exhibited at most two mineral deficiencies.

The number of plots with Nitrogen deficiency only is

70

.

Show answer & explanation

Answer: Exactly $40$ plots exhibited all three mineral deficiencies.; Exactly $130$ plots exhibited exactly one mineral deficiency.; The number of plots exhibiting at least two mineral deficiencies is $110$ .

Answer

The statements confirming that exactly 40 plots exhibited all three deficiencies, exactly 130 plots exhibited exactly one deficiency, and 110 plots exhibited at least two deficiencies are all correct.

The system of set equations shows that

n_3 = 40

plots have all three deficiencies,

n_1 = 130

plots have exactly one deficiency, and

n_2 + n_3 = 70 + 40 = 110

plots have at least two deficiencies. Therefore, the statements asserting

40

plots for all three deficiencies,

130

plots for exactly one deficiency, and

110

plots for at least two deficiencies are all guaranteed to be true.

Step-by-Step Solution

1

Determine the total number of plots exhibiting at least one deficiency.

The total number of plots with at least one deficiency is

300 - 60 = 240

.

Plots with no deficiencies are excluded from the set union

|N \cup P \cup K|

.

2

Set up the inclusion-exclusion equations for set membership.

Let

n_1

be the number of plots with exactly one deficiency,

n_2 = 70

be the number of plots with exactly two deficiencies, and

n_3

be the number of plots with all three deficiencies.
Equation 1 (Total elements in union):

n_1 + n_2 + n_3 = 240 \implies n_1 + 70 + n_3 = 240 \implies n_1 + n_3 = 170

.
Equation 2 (Sum of individual set cardinalities):

|N| + |P| + |K| = n_1 + 2n_2 + 3n_3 \implies 140 + 130 + 120 = n_1 + 2(70) + 3n_3 \implies 390 = n_1 + 140 + 3n_3 \implies n_1 + 3n_3 = 250

.

Each element in an individual set sum is counted once for single-set membership, twice for double-set membership, and three times for triple-set membership.

3

Solve the system of linear equations for

n_1

and

n_3

.

Subtract Equation 1 from Equation 2:

(n_1 + 3n_3) - (n_1 + n_3) = 250 - 170 \implies 2n_3 = 80 \implies n_3 = 40

.
Substitute

n_3 = 40

into Equation 1:

n_1 + 40 = 170 \implies n_1 = 130

.

This yields

40

plots with all three deficiencies and

130

plots with exactly one deficiency.

4

Evaluate each option statement against the computed set cardinalities.

1) All three deficiencies (

n_3

) =

40

(True).
2) Exactly one deficiency (

n_1

) =

130

(True).
3) At least two deficiencies (

n_2 + n_3

) =

70 + 40 = 110

(True).
4) At most two deficiencies = Total -

n_3 = 300 - 40 = 260 \neq 170

(False).
5) Nitrogen deficiency only depends on how the

70

dual-deficiency plots are split between

N \cap P

,

N \cap K

, and

P \cap K

, which is not uniquely determined by the given data (False).

Only statements meeting the exact numerical bounds and logical guarantees must be true.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Partitioning into Disjoint Regions

Question 28Question →

An agricultural research station evaluated a sample of $120$ organic farm plots for contamination by three types of crop fungi: *Fusarium* ( $F$ ), *Rhizoctonia* ( $R$ ), and *Sclerotinia* ( $S$ ). The audit revealed the following data:

- $55$ plots were contaminated with *Fusarium*.
- $50$ plots were contaminated with *Rhizoctonia*.
- $45$ plots were contaminated with *Sclerotinia*.
- $20$ plots were contaminated with both *Fusarium* and *Rhizoctonia*.
- $15$ plots were contaminated with both *Rhizoctonia* and *Sclerotinia*.
- $18$ plots were contaminated with both *Fusarium* and *Sclerotinia*.
- $8$ plots were contaminated with all three fungi.

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

Select all that apply

Exactly 15 farm plots were free from all three types of crop fungi.

Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Exactly 37 farm plots were contaminated by *Fusarium* only.

Exactly 53 farm plots were contaminated by at least two of the three crop fungi.

Exactly 50 percent of all plots contaminated by *Rhizoctonia* were also contaminated by *Fusarium*.

Show answer & explanation

Answer: Exactly 15 farm plots were free from all three types of crop fungi.; Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Answer

The correct statements are that exactly 15 farm plots were free from all three types of crop fungi, and exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

The statement regarding 15 farm plots being free from all three fungi is correct because the total number of plots contaminated by at least one fungus is calculated using inclusion-exclusion: 55 + 50 + 45 - 20 - 15 - 18 + 8 = 105, leaving 120 - 105 = 15 plots clean. The statement regarding 29 farm plots contaminated by exactly two fungi is correct because summing the mutually exclusive regions containing exactly two fungi yields (20 - 8) + (15 - 8) + (18 - 8) = 12 + 7 + 10 = 29.

Step-by-Step Solution

1

Calculate the non-overlapping regions for each intersection.

Determining exact two-set regions requires removing elements that belong to all three sets.

2

Calculate single-fungus-only regions.

|F only| = 55 - (12 + 10 + 8) = 25.
|R only| = 50 - (12 + 7 + 8) = 23.
|S only| = 45 - (10 + 7 + 8) = 20.

Subtract all double and triple overlaps from total individual set counts.

3

Determine union of all three sets and the neither region.

Total contaminated = 25 + 23 + 20 + 12 + 7 + 10 + 8 = 105.
Plots free from all fungi = 120 - 105 = 15.

Subtracting the total contaminated plots from the overall sample size gives the plots outside all sets.

4

Verify statement validity.

Plots free of all fungi = 15 (Valid).
Plots with exactly two fungi = 12 + 7 + 10 = 29 (Valid).

Match calculated regional counts against the presented options.

Key Concept

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

Question 29Question →

An urban planning department surveyed a total of $360$ commercial buildings regarding three sustainability certifications: LEED ( $L$ ), BREEAM ( $B$ ), and WELL ( $W$ ). Exactly $50$ of the surveyed buildings held none of the three certifications. The survey revealed the following data:

• $180$ buildings held the LEED certification.
• $140$ buildings held the BREEAM certification.
• $130$ buildings held the WELL certification.
• $45$ buildings held both the LEED and BREEAM certifications.
• $40$ buildings held both the BREEAM and WELL certifications.
• $25$ buildings held all three certifications.

How many of the surveyed buildings held EXACTLY TWO of the three certifications?

Show answer & explanation

Answer: 90

Answer

90

To find the number of buildings with exactly two certifications, we first determine the total number of buildings holding at least one certification:

360 - 50 = 310

. Applying the three-set inclusion-exclusion formula,

310 = 180 + 140 + 130 - (45 + 40 + |L \cap W|) + 25

, which simplifies to

310 = 390 - |L \cap W|

, giving

|L \cap W| = 80

. Next, we isolate the regions holding exactly two certifications by subtracting the 25 triple-certified buildings from each pairwise intersection: LEED & BREEAM only =

45 - 25 = 20

, BREEAM & WELL only =

40 - 25 = 15

, and LEED & WELL only =

80 - 25 = 55

. Summing these three exclusive regions gives

20 + 15 + 55 = 90

.

Step-by-Step Solution

1

Determine the total size of the union of all three sets

|L \cup B \cup W| = 360 - 50 = 310

Buildings holding at least one certification represent the entire surveyed population minus those holding no certifications.

2

Solve for the unknown overlap of LEED and WELL certifications using the inclusion-exclusion formula

|L \cap W| = 80

Using

|L \cup B \cup W| = |L| + |B| + |W| - |L \cap B| - |B \cap W| - |L \cap W| + |L \cap B \cap W|

, we have

310 = 180 + 140 + 130 - 45 - 40 - |L \cap W| + 25 = 390 - |L \cap W|

.

3

Calculate the number of buildings in each region corresponding to exactly two certifications

LEED and BREEAM only:

45 - 25 = 20

; BREEAM and WELL only:

40 - 25 = 15

; LEED and WELL only:

80 - 25 = 55

Each total pairwise intersection includes the 25 buildings that hold all three certifications, so subtracting 25 isolates those holding exclusively two certifications.

4

Sum the three isolated regions

20 + 15 + 55 = 90

The total number of buildings holding exactly two certifications is the sum of the three non-overlapping regions representing two certifications.

Key Concept

Three-Set Inclusion-Exclusion Principle and Partitioning Venn Diagrams

Estimated Time:2m 0s

Question 30Question →

Events $E$ and $F$ are two events in a sample space such that $P(E) = 0.30$ and $P(F) = 0.40$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

.

If

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

If

E

and

F

are independent, then

P(E \text{ or } F) = 0.70

.

If

E

and

F

are mutually exclusive, then

E

and

F

are independent.

If

E

and

F

are independent, then

P(E \text{ given } F) = 0.12

.

Show answer & explanation

Answer: If $E$ and $F$ are mutually exclusive, then $P(E \text{ and } F) = 0$ .; If $E$ and $F$ are independent, then $P(E \text{ and } F) = 0.12$ .

Answer

The correct statements are that if

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

, and if

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

The statement asserting that

P(E \text{ and } F) = 0

for mutually exclusive events is correct because mutually exclusive events by definition cannot occur together. The statement asserting that

P(E \text{ and } F) = 0.12

for independent events is correct because independent events satisfy the multiplication rule

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

Step-by-Step Solution

1

Evaluate the statement regarding mutually exclusive events.

P(E \text{ and } F) = 0

.

By definition, mutually exclusive events cannot both happen at the same time, so

P(E \text{ and } F) = 0

is true.

2

Evaluate the joint probability for independent events.

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

The multiplication rule for independent events states that

P(E \text{ and } F) = P(E) \cdot P(F)

.

3

Evaluate the union probability

P(E \text{ or } F)

for independent events.

P(E \text{ or } F) = P(E) + P(F) - P(E \text{ and } F) = 0.30 + 0.40 - 0.12 = 0.58

.

Simply adding

P(E) + P(F) = 0.70

fails to subtract the intersection that is counted twice.

4

Evaluate conditional probability for independent events.

P(E \text{ given } F) = P(E) = 0.30

.

Independence implies that knowing event

F

occurred does not change the probability of event

E

.

Key Concept

Probability rules for mutually exclusive and independent events

Question 31Question →

The frequency distribution table below summarizes the scores of 20 students on a statistics test.

Score	Number of Students
5	3
6	5
7	$x$
8	$y$
9	4

If the arithmetic mean score of the 20 students is $7.15$ , what is the median score of the 20 students?

6.5

7.0

7.15

7.5

8.0

Show answer & explanation

Answer: 7.5

Answer

The median score of the 20 students is 7.5.

To find the median score of 20 students, we first determine the missing student counts

x

and

y

. From the total count of 20 students,

x + y = 20 - (3 + 5 + 4) = 8

. Using the given mean of 7.15, the sum of all scores is

20 \times 7.15 = 143

. The sum expressed in terms of

x

and

y

is

3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Setting

81 + 7x + 8y = 143

gives

7x + 8y = 62

. Solving

x + y = 8

and

7x + 8y = 62

yields

x = 2

and

y = 6

. Listing scores in non-decreasing order, the 10th student scored 7 and the 11th student scored 8. Because the total number of students is even, the median is the average of these two middle values:

(7 + 8) / 2 = 7.5

.

Step-by-Step Solution

1

Set up an equation for the total number of students to find the sum of

x

and

y

.

3 + 5 + x + y + 4 = 20 \implies x + y + 12 = 20 \implies x + y = 8

.

The total frequency of all score categories must equal the total number of students (20).

2

Set up an equation for the sum of all scores using the given arithmetic mean.

Total sum

= 20 \times 7.15 = 143

. Also, total sum

= 3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Therefore,

81 + 7x + 8y = 143 \implies 7x + 8y = 62

.

The sum of all data values equals the mean multiplied by the total number of observations.

3

Solve the system of linear equations for

x

and

y

.

Multiply

x + y = 8

by 7 to get

7x + 7y = 56

. Subtract this from

7x + 8y = 62

to get

y = 6

. Then

x = 8 - 6 = 2

.

Determining the exact frequencies

x=2

and

y=6

is required to locate the middle values.

4

Determine the cumulative frequencies and find the 10th and 11th values.

Score 5: positions 1 to 3 (3 students); Score 6: positions 4 to 8 (5 students); Score 7: positions 9 to 10 (2 students); Score 8: positions 11 to 16 (6 students); Score 9: positions 17 to 20 (4 students). The 10th value is 7 and the 11th value is 8.

For an even number of data points (

N=20

), the median is the arithmetic mean of the

(N/2)

-th and

(N/2 + 1)

-th values.

5

Calculate the median score.

\text{Median} = \frac{7 + 8}{2} = 7.5

.

Average the two middle values.

Key Concept

Calculating the median from a frequency distribution table with missing counts by combining system of linear equations with cumulative frequency analysis.

Question 32Question →

A dataset consists of 15 numerical values with a mean of 50 and a standard deviation of 8. If a constant value of 5 is added to every number in the dataset, what is the standard deviation of the resulting dataset?

8

13

40

55

64

Show answer & explanation

Answer: 8

Answer

The standard deviation of the resulting dataset is 8.

Adding a constant to every data point in a set shifts the central location (the mean and median) by that constant, but leaves the distances between points unchanged. Because standard deviation depends solely on the distances between each point and the mean, the standard deviation remains 8.

Step-by-Step Solution

1

Recall the effect of adding a constant to data points on measures of dispersion.

Measures of dispersion (range, IQR, standard deviation) describe the spread or scatter of data relative to its center.

When a constant

k

is added to every value

x_i

, the new mean becomes

\text{mean} + k

.

2

Evaluate the distance of each transformed data point from the new mean.

(x_i + k) - (\text{mean} + k) = x_i - \text{mean}

.

The deviation of each point from the mean remains identical to its original deviation.

3

Determine the standard deviation of the new dataset.

The standard deviation is unchanged and remains 8.

Because all individual deviations from the mean are preserved, the average distance from the mean does not change.

Key Concept

Invariance of standard deviation under addition of a constant

Question 33Question →

A dataset of $20$ numerical measurements has a range of $30$ and a standard deviation of $6.4$ . If every measurement in the dataset is multiplied by $-2$ and then increased by $5$ to construct a new dataset, what are the range and standard deviation of the new dataset, respectively?

Range is

65

and standard deviation is

17.8

Range is

-60

and standard deviation is

-12.8

Range is

60

and standard deviation is

12.8

Range is

60

and standard deviation is

25.6

Range is

30

and standard deviation is

6.4

Show answer & explanation

Answer: Range is $60$ and standard deviation is $12.8$

Answer

The range of the new dataset is

60

and the standard deviation is

12.8

.

Under a linear transformation

Y = aX + b

, all measures of dispersion (range, interquartile range, standard deviation) are multiplied by

|a|

and are unaffected by the additive constant

b

. Here,

a = -2

and

b = 5

. The absolute multiplier is

|-2| = 2

. Therefore, the new range is

2 \times 30 = 60

and the new standard deviation is

2 \times 6.4 = 12.8

.

Step-by-Step Solution

1

Analyze the general transformation formula for measures of dispersion under linear transformations of the form

Y = aX + b

.

For any constant multiplier

a

and constant shift

b

, the new range is

|a| \times \text{Range}_X

and the new standard deviation is

|a| \times \sigma_X

.

Measures of dispersion reflect spread; multiplying each data value by

a

scales the spread by

|a|

, while adding a constant shift

b

translates all values equally without expanding or contracting their relative distance.

2

Calculate the new range using

a = -2

and

b = 5

.

\text{New Range} = |-2| \times 30 = 2 \times 30 = 60.

The range scales by the absolute value of the multiplier

|-2| = 2

, while the additive constant

5

has no impact on the spread.

3

Calculate the new standard deviation using

a = -2

and

b = 5

.

\text{New Standard Deviation} = |-2| \times 6.4 = 2 \times 6.4 = 12.8.

The standard deviation scales by

|-2| = 2

, and the shift of

+5

does not change the dispersion around the mean.

Key Concept

Effect of Linear Transformations on Measures of Dispersion

Estimated Time:1m 30s

Question 34Question →

An original dataset $X$ consists of $20$ distinct positive numerical values with standard deviation $\sigma$ , range $R$ , and interquartile range $IQR$ . A new dataset $Y$ is created by transforming each data value $x_i$ in $X$ according to the rule $y_i = -3x_i + 5$ . Which of the following statements about the dispersion metrics of dataset $Y$ compared to dataset $X$ must be true? Select all such statements.

Select all that apply

The standard deviation of dataset

Y

is

3\sigma

.

The interquartile range of dataset

Y

is

3 \times IQR

.

The standard deviation of dataset

Y

is

3\sigma + 5

.

The range of dataset

Y

is

-3R + 5

.

The interquartile range of dataset

Y

is equal to

-3 \times IQR

.

Show answer & explanation

Answer: The standard deviation of dataset $Y$ is $3\sigma$ .; The interquartile range of dataset $Y$ is $3 \times IQR$ .

Answer

The statements asserting that the standard deviation of dataset Y is 3 times the standard deviation of dataset X, and that the interquartile range of dataset Y is 3 times the interquartile range of dataset X, are both correct.

For any data set undergoing a linear transformation

y = ax + b

, all measures of dispersion (including standard deviation, range, and interquartile range) are scaled by the absolute value of the multiplier

|a|

and remain completely unaffected by the additive constant

b

. Since

a = -3

,

|a| = |-3| = 3

. Therefore, both the standard deviation and the interquartile range scale by a factor of 3.

Step-by-Step Solution

1

Recall the general rule for linear transformation of measures of dispersion.

For any linear transformation

y = ax + b

, measures of dispersion (Range, IQR, Standard Deviation) scale by

|a|

and are unaffected by

b

.

Measures of dispersion evaluate the spread/distance between points, which shifts unchanged when a constant is added, and scales non-negatively when multiplied by a constant.

2

Identify the values of the multiplicative constant

a

and additive constant

b

.

a = -3

and

b = 5

, giving

|a| = |-3| = 3

.

The transformation equation is

y_i = -3x_i + 5

.

3

Apply the scaling factor

|a| = 3

to each measure of dispersion.

Standard deviation of

Y

is

3\sigma

, IQR of

Y

is

3 \times IQR

, and Range of

Y

is

3R

.

All measures of dispersion scale by a factor of 3 regardless of the negative sign of the multiplier or the addition of 5.

Key Concept

Linear Transformations on Measures of Dispersion

Question 35Question →

A store manager recorded the number of orders processed online each day during a one-week period: $14$ , $7$ , $25$ , $11$ , $18$ , $2$ , and $16$ . What is the median number of daily online orders processed during this week?

11

14

16

18

93

Show answer & explanation

Answer: $14$

Answer

The median number of daily online orders is

14

.

To find the median, the seven numbers must first be placed in ascending order:

2, 7, 11, 14, 16, 18, 25

. Since there is an odd number of data points (

7

), the median is the single middle value located at the 4th position, which is

14

.

Step-by-Step Solution

1

Arrange the given dataset of seven numbers in ascending order.

The sorted dataset is

2, 7, 11, 14, 16, 18, 25

.

To determine the median of any dataset, the numerical values must first be ordered from smallest to largest.

2

Find the position of the middle element for a dataset with

n = 7

values.

The median position is

\frac{7 + 1}{2} = 4

.

When a dataset contains an odd number of elements

n

, the median is located at the

\frac{n+1}{2}

-th position.

3

Identify the 4th element in the sorted list.

The 4th element is

14

.

In the ordered sequence

2, 7, 11, 14, 16, 18, 25

, the 4th entry is

14

.

Key Concept

The median is the middle value in a set of numerical data that has been arranged in ascending or descending order.

Question 36Question →

Dataset $A$ consists of $n$ numerical values with a standard deviation of $\sigma_A$ and an interquartile range of $\text{IQR}_A$ . Dataset $B$ is constructed by multiplying each value in Dataset $A$ by $4$ and then subtracting $10$ from each resulting value. Which of the following correctly expresses the standard deviation $\sigma_B$ and the interquartile range $\text{IQR}_B$ of Dataset $B$ in terms of $\sigma_A$ and $\text{IQR}_A$ ?

\sigma_B = 4\sigma_A - 10

and

\text{IQR}_B = 4\text{IQR}_A - 10

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

\sigma_B = 16\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

\sigma_B = 4\sigma_A

and

\text{IQR}_B = \text{IQR}_A

\sigma_B = 4\sigma_A - 10

and

\text{IQR}_B = 4\text{IQR}_A

Show answer & explanation

Answer: $\sigma_B = 4\sigma_A$ and $\text{IQR}_B = 4\text{IQR}_A$

Answer

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

When a dataset undergoes a linear transformation of the form

Y = cX + k

, any measure of dispersion

D

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

D_Y = |c| \cdot D_X

. The constant shift

k

shifts all points equally and does not affect spread. Therefore, multiplying by

4

multiplies both

\sigma_A

and

\text{IQR}_A

by

4

, and subtracting

10

has no effect on either measure.

Step-by-Step Solution

1

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

c = 4

.

Multiplying all data values by

c

scales all measures of spread (including standard deviation, range, and interquartile range) by

|c| = 4

. Thus, initial scaled values give

4\sigma_A

and

4\text{IQR}_A

.

Measures of dispersion reflect distances between data points, which scale linearly when all values are multiplied by a constant.

2

Analyze the effect of subtracting a constant

k = 10

from each data point.

Subtracting a constant shifts the entire distribution along the number line without altering the relative distances between any data points.

Because measures of dispersion measure relative spread rather than absolute location, adding or subtracting a constant has zero effect on standard deviation or interquartile range.

3

Combine the transformation effects to state

\sigma_B

and

\text{IQR}_B

.

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

.

The multiplicative factor scales both metrics by 4, and the additive shift of

-10

leaves both metrics unchanged.

Key Concept

Linear Transformations on Measures of Dispersion

Estimated Time:1m 15s

Question 37Question →

A research team measured the height, in centimeters, of 15 plants in a greenhouse. The dataset has a range of $24\text{ cm}$ , an interquartile range (IQR) of $10\text{ cm}$ , and a standard deviation of $5.2\text{ cm}$ . If the height of every plant is increased by exactly $8\text{ cm}$ , which of the following statements regarding the modified dataset must be true? Select all such statements.

Select all that apply

The range of the modified dataset is

32\text{ cm}

.

The range of the modified dataset is

24\text{ cm}

.

The interquartile range of the modified dataset is

18\text{ cm}

.

The standard deviation of the modified dataset is

5.2\text{ cm}

.

The standard deviation of the modified dataset is

13.2\text{ cm}

.

Show answer & explanation

Answer: The range of the modified dataset is $24\text{ cm}$ .; The standard deviation of the modified dataset is $5.2\text{ cm}$ .

Answer

The statement that the range of the modified dataset is

24\text{ cm}

and the statement that the standard deviation of the modified dataset is

5.2\text{ cm}

are both true.

When a constant is added to every value in a dataset, the entire distribution shifts by that amount. Consequently, the relative positions and distances between data points do not change, leaving measures of dispersion (range, IQR, standard deviation) unchanged. Therefore, the range remains

24\text{ cm}

and the standard deviation remains

5.2\text{ cm}

.

Step-by-Step Solution

1

Analyze the effect of adding a constant to a dataset on measures of dispersion.

Adding a constant

k

to every data value shifts the center (mean, median) by

k

, but all measures of spread (range, IQR, standard deviation) remain completely unchanged.

Measures of dispersion quantify the spread or distance between data values, which is invariant under a uniform translation.

2

Evaluate each statement against the unchanged measures of dispersion.

The range remains

24\text{ cm}

, the IQR remains

10\text{ cm}

, and the standard deviation remains

5.2\text{ cm}

.

Comparing the calculated original measures to the statements shows that the range of

24\text{ cm}

and standard deviation of

5.2\text{ cm}

are correct.

Key Concept

Effect of Linear Transformations (Addition of a Constant) on Measures of Dispersion

Question 38Question →

Dataset $S$ consists of 10 distinct numerical values. Dataset $T$ is created by replacing the maximum value of Dataset $S$ with a number that is strictly greater than that maximum value, while all other 9 values remain unchanged. Which of the following statements comparing Dataset $S$ and Dataset $T$ must be true? Select all that apply.

Select all that apply

The range of Dataset

T

is greater than the range of Dataset

S

.

The interquartile range (IQR) of Dataset

T

is equal to the interquartile range of Dataset

S

.

The standard deviation of Dataset

T

is equal to the standard deviation of Dataset

S

.

The median of Dataset

T

is strictly greater than the median of Dataset

S

.

The 75th percentile of Dataset

T

is strictly greater than the 75th percentile of Dataset

S

.

Show answer & explanation

Answer: The range of Dataset $T$ is greater than the range of Dataset $S$ .; The interquartile range (IQR) of Dataset $T$ is equal to the interquartile range of Dataset $S$ .

Answer

The statements asserting that the range of Dataset

T

is greater than the range of Dataset

S

and that the interquartile range (IQR) of Dataset

T

is equal to the interquartile range of Dataset

S

are correct.

The statement regarding the range is correct because increasing the maximum value increases the difference between the maximum and minimum values. The statement regarding the interquartile range is correct because the 25th percentile (

Q_1

) and 75th percentile (

Q_3

) depend only on the first 8 ordered elements of a 10-element dataset, so changing the 10th element leaves

Q_1

and

Q_3

identical.

Step-by-Step Solution

1

Analyze the impact on Range

\text{Range}(T) > \text{Range}(S)

Range is defined as

\text{Maximum} - \text{Minimum}

. Since the minimum is unchanged and the maximum increases, the range must increase.

2

Analyze the impact on Interquartile Range (IQR)

\text{IQR}(T) = \text{IQR}(S)

IQR is

Q_3 - Q_1

. For 10 ordered elements,

Q_1

and

Q_3

depend on the positions of the bottom 75% of the data. Modifying only the 10th (largest) element leaves

Q_1

and

Q_3

unchanged.

3

Analyze the impact on Median and Standard Deviation

Median is unchanged; Standard deviation increases.

The median depends on the 5th and 6th values, which are unaffected. Moving an extreme value further out increases distance from the mean, strictly increasing standard deviation.

Key Concept

Effect of outlier modification on measures of position (

Q_1, Q_3

, Median) versus measures of dispersion (Range, Standard Deviation, IQR).

Question 39Question →

The table below shows the frequency distribution of the weights, in grams, for a sample of 80 manufactured components.

Weight (grams)	Frequency
100–109	12
110–119	24
120–129	28
130–139	16

What percentage of the components in the sample have a weight of at least 120 grams?

Show answer & explanation

Answer: 55

Answer

55%

To calculate the percentage of components with a weight of at least 120 grams, sum the frequencies of all weight intervals corresponding to 120 grams or more:

28 + 16 = 44

. Next, divide this count by the total number of components (

80

) and multiply by

100

:

\frac{44}{80} \times 100\% = 55\%

.

Step-by-Step Solution

1

Identify the number of components with weight at least 120 grams

44 components

The weight intervals '120–129' and '130–139' contain 28 and 16 components respectively, giving

28 + 16 = 44

.

2

Calculate the percentage relative to the total sample size of 80 components

55%

Dividing the target count (44) by the total sample size (80) and multiplying by 100 yields

\frac{44}{80} \times 100 = 55\%

.

Key Concept

Calculating class percentages from grouped frequency tables

Question 40Question →

Dataset $S$ consists of $80$ distinct numerical scores arranged in ascending order, with an interquartile range denoted by $\text{IQR}_S$ . A new dataset, $T$ , is created by adding four scores to Dataset $S$ : two scores that are strictly smaller than the minimum value of Dataset $S$ , and two scores that are strictly greater than the maximum value of Dataset $S$ . If $\text{IQR}_T$ represents the interquartile range of Dataset $T$ , which of the following statements must be true?

\text{IQR}_T > \text{IQR}_S

\text{IQR}_T = \text{IQR}_S

\text{IQR}_T < \text{IQR}_S

\text{IQR}_T = \text{IQR}_S + 4

The relationship between

\text{IQR}_T

and

\text{IQR}_S

cannot be determined from the given information.

Show answer & explanation

Answer: $\text{IQR}_T > \text{IQR}_S$

Answer

The interquartile range of Dataset

T

is strictly greater than the interquartile range of Dataset

S

(

\text{IQR}_T > \text{IQR}_S

).

The statement that

\text{IQR}_T > \text{IQR}_S

is correct. Inserting two items below the minimum value shifts the rank position of the first quartile to lower-indexed elements of the original dataset, reducing

Q_1

. Inserting two items above the maximum value shifts the rank position of the third quartile to higher-indexed elements of the original dataset, increasing

Q_3

. Since the original numbers are strictly increasing,

Q_3

increases while

Q_1

decreases, resulting in a larger interquartile range for Dataset

T

.

Step-by-Step Solution

1

Determine the quartile positions for the original dataset

S

.

Dataset

S

has

n = 80

distinct elements

x_1 < x_2 < \dots < x_{80}

. Its first quartile

Q_{1,S}

is the average of

x_{20}

and

x_{21}

, and its third quartile

Q_{3,S}

is the average of

x_{60}

and

x_{61}

. Thus,

\text{IQR}_S = Q_{3,S} - Q_{1,S} = \frac{x_{60} + x_{61}}{2} - \frac{x_{20} + x_{21}}{2}

.

Quartiles partition sorted data into four equal parts based on element indices.

2

Determine the new element positions for dataset

T

.

Dataset

T

has

n = 84

elements ordered as

a, b, x_1, x_2, \dots, x_{80}, c, d

, where

a, b < x_1

and

c, d > x_{80}

. The elements of

S

are now shifted right by 2 index positions in

T

.

Inserting two values at the bottom of the ordered list increases the rank index of all original elements by 2.

3

Calculate the first and third quartiles for dataset

T

.

In Dataset

T

, the first quartile

Q_{1,T}

is the average of the 21st and 22nd elements of

T

, which correspond to

x_{19}

and

x_{20}

of

S

. The third quartile

Q_{3,T}

is the average of the 63rd and 64th elements of

T

, which correspond to

x_{61}

and

x_{62}

of

S

.

The 25th and 75th percentile positions in an 84-element dataset correspond to indices 21/22 and 63/64.

4

Compare

\text{IQR}_T

with

\text{IQR}_S

.

Since

S

consists of distinct values in strictly ascending order (

x_{19} < x_{20} < x_{21}

and

x_{60} < x_{61} < x_{62}

),

Q_{1,T} = \frac{x_{19} + x_{20}}{2} < \frac{x_{20} + x_{21}}{2} = Q_{1,S}

and

Q_{3,T} = \frac{x_{61} + x_{62}}{2} > \frac{x_{60} + x_{61}}{2} = Q_{3,S}

. Therefore,

\text{IQR}_T = Q_{3,T} - Q_{1,T} > Q_{3,S} - Q_{1,S} = \text{IQR}_S

.

Subtracting a smaller number from a larger number increases the overall difference.

Key Concept

Interquartile Range and Position Indices under Data Insertion

Data Analysis

Answer

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