Data Analysis Practice Questions — GRE General Test — Page 8

Question 141Question →

A cybersecurity system generates 5-character identification codes consisting of 3 distinct letters followed by 2 distinct digits. The letters must be selected from the set $\{A, B, C, D, E, F\}$ and the digits from the set $\{1, 2, 3, 4\}$ . If the first character of the code must be a vowel ( $A$ or $E$ ), how many such distinct 5-character identification codes can be formed?

120

240

480

720

1,440

Show answer & explanation

Answer: 480

Answer

480 distinct 5-character identification codes can be formed.

To construct the 5-character code, break the process into sequential choices for each position: Position 1 must be a vowel (

A

or

E

), giving 2 options. Position 2 can be any of the remaining 5 distinct letters. Position 3 can be any of the remaining 4 distinct letters. Position 4 (the first digit) can be any of the 4 available digits. Position 5 (the second digit) can be any of the remaining 3 distinct digits. Multiplying these independent choices together yields

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Step-by-Step Solution

1

Calculate the number of ways to choose the first letter (must be a vowel)

There are 2 choices (

A

or

E

).

The problem restricts the first character of the code to a vowel.

2

Calculate the number of choices for the second and third letters

5 choices for the second letter and 4 choices for the third letter.

The letters must be distinct, leaving 5 remaining letters from the original pool of 6 for the second position, and 4 for the third position.

3

Calculate the number of choices for the two distinct digits

4 choices for the first digit and 3 choices for the second digit.

There are 4 digits available in

\{1, 2, 3, 4\}

and they must be distinct.

4

Apply the Fundamental Counting Principle to find total codes

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Multiply the number of independent choices for each sequential position.

Key Concept

Fundamental Counting Principle with Permutations and Restrictions

Question 142Question →

A container holds $n$ spheres, exactly 5 of which are blue and the remaining $n - 5$ are green. Two spheres are drawn at random from the container one after another without replacement. If the probability that at least one of the selected spheres is green is $\frac{14}{15}$ , what is the total number of spheres $n$ in the container?

15

20

25

30

75

Show answer & explanation

Answer: 25

Answer

The total number of spheres

n

in the container is 25.

The probability of at least one green sphere is complementary to drawing zero green spheres (meaning both spheres drawn are blue). Subtracting

\frac{14}{15}

from 1 yields

P(\text{both blue}) = \frac{1}{15}

. Since the selection is without replacement,

P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}

. Equating this to

\frac{1}{15}

gives

n(n-1) = 300

. Solving the quadratic equation

n^2 - n - 300 = 0

gives

n = 25

, which correctly represents the total number of spheres.

Step-by-Step Solution

1

Use the complement rule to determine the probability that both selected spheres are blue.

P(\text{both blue}) = 1 - P(\text{at least one green}) = 1 - \frac{14}{15} = \frac{1}{15}

.

The event that at least one sphere is green is the complement of the event that both drawn spheres are blue.

2

Set up the joint probability equation for drawing two blue spheres sequentially without replacement.

P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)}

.

There are 5 blue spheres initially out of

n

. After drawing one blue sphere, 4 blue spheres remain out of

n - 1

total spheres.

3

Equate the expressions and solve for

n

.

\begin{aligned} \frac{20}{n(n - 1)} &= \frac{1}{15} \\ n(n - 1) &= 300 \\ n^2 - n - 300 &= 0 \\ (n - 25)(n + 12) &= 0 \end{aligned}

Cross-multiplying gives a quadratic equation in terms of

n

.

4

Select the valid positive integer solution for

n

.

n = 25

(since

n > 0

).

The total number of spheres must be a positive integer.

Key Concept

Probability of Complementary Events and Dependent Sequential Events

Estimated Time:2m 0s

Question 143Question →

A dataset $S$ consists of 7 distinct positive integers. The arithmetic mean of the numbers in $S$ is 20, and the median is 18. If the largest integer in $S$ is 35, what is the maximum possible value for the second-largest integer in $S$ ?

34

33

35

62

19

Show answer & explanation

Answer: 34

Answer

34

To find the maximum possible value of the second-largest integer, we arrange the 7 distinct positive integers in ascending order:

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

. The median

x_4 = 18

and the largest element

x_7 = 35

. The total sum of all 7 elements is

7 \times 20 = 140

. Since

x_6

must be strictly less than

x_7 = 35

, the maximum potential integer value for

x_6

is 34. If

x_6 = 34

, the remaining four elements (

x_1, x_2, x_3, x_5

) must sum to

87 - 34 = 53

. Choosing

x_5 = 19

(the smallest integer greater than 18) leaves a sum of 34 for

x_1 + x_2 + x_3

, which can be satisfied by distinct positive integers such as 1, 16, and 17. Thus, 34 is achievable.

Step-by-Step Solution

1

Calculate the total sum of the 7 integers in the dataset.

Total sum =

7 \times 20 = 140

.

The sum of a dataset equals the number of elements multiplied by its arithmetic mean.

2

Identify the positions of known elements when the dataset is ordered in ascending order

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

.

Median

x_4 = 18

and largest element

x_7 = 35

.

For a 7-element dataset, the median is the 4th element.

3

Calculate the combined sum of the remaining five unknown elements.

x_1 + x_2 + x_3 + x_5 + x_6 = 140 - (18 + 35) = 87

.

Subtracting the median and the largest element from the total sum gives the sum of the remaining five elements.

4

Determine the theoretical upper bound for the second-largest integer

x_6

.

Since

x_6 < x_7 = 35

and all integers are distinct,

x_6 \le 34

.

The second-largest integer must be strictly less than the largest integer.

5

Verify if

x_6 = 34

can produce a valid dataset of distinct positive integers.

If

x_6 = 34

, then

x_1 + x_2 + x_3 + x_5 = 87 - 34 = 53

. Setting

x_5 = 19

,

x_3 = 17

,

x_2 = 16

, and

x_1 = 1

gives

1 + 16 + 17 + 19 = 53

, forming the valid set

\{1, 16, 17, 18, 19, 34, 35\}

.

Since a valid set of distinct positive integers exists satisfying all constraints, 34 is the maximum possible value.

Key Concept

Measures of Central Tendency and Data Constraints

Question 144Question →

An automated risk-management system uses three independent algorithms—Algorithm X, Algorithm Y, and Algorithm Z—to detect fraudulent transactions. The probability that Algorithm X detects a given fraudulent transaction is $\frac{3}{5}$ , the probability that Algorithm Y detects it is $\frac{2}{3}$ , and the probability that Algorithm Z detects it is $\frac{3}{4}$ . If a fraudulent transaction occurs, what is the probability that it will be detected by at least two of these three algorithms?

Show answer & explanation

Answer: 0.75

Answer

The probability that the transaction is detected by at least two of the three algorithms is 0.75 (or 3/4).

Because the algorithms operate independently, the event 'at least two algorithms detect the transaction' consists of four mutually exclusive outcomes: exactly X and Y detect (probability 6/60 = 0.10), exactly X and Z detect (probability 9/60 = 0.15), exactly Y and Z detect (probability 12/60 = 0.20), and all three detect (probability 18/60 = 0.30). Summing these four probabilities gives 0.10 + 0.15 + 0.20 + 0.30 = 0.75.

Step-by-Step Solution

1

Determine the complementary probabilities of non-detection for each algorithm.

P(X does not detect) = 2/5, P(Y does not detect) = 1/3, and P(Z does not detect) = 1/4.

The probability of an event's complement is 1 minus the probability of the event.

2

Calculate the probability for each scenario where exactly two algorithms detect the transaction.

P(X and Y only) = 6/60, P(X and Z only) = 9/60, P(Y and Z only) = 12/60.

Since the algorithms operate independently, joint probabilities are calculated by multiplying individual probabilities.

3

Calculate the probability that all three algorithms detect the transaction.

P(X and Y and Z) = 18/60.

Multiplying the individual detection probabilities of all three independent algorithms.

4

Sum the probabilities of all qualifying mutually exclusive outcomes.

(6/60) + (9/60) + (12/60) + (18/60) = 45/60 = 0.75.

The events representing different combinations of detections are mutually exclusive, so their probabilities add directly.

Key Concept

Independent Events and Addition Rule for Mutually Exclusive Outcomes

Question 145Question →

A university department tracked the number of research articles published by its 8 faculty members over a five-year period. The numbers of publications for 7 of the faculty members were $4, 7, 9, 12, 15, 18,$ and $23$ . If the arithmetic mean of the number of publications for all 8 faculty members is equal to $1.25$ times their median, what is the number of publications for the 8th faculty member?

Show answer & explanation

Answer: 47

Answer

47

The sum of the 7 known values is

88

, making the total sum

88 + x

and the arithmetic mean

\frac{88 + x}{8}

. Assuming

x \ge 15

, the 4th and 5th numbers when sorted are

12

and

15

, yielding a median of

13.5

. Setting the mean equal to

1.25 \times 13.5 = 16.875

gives

\frac{88 + x}{8} = 16.875

, which simplifies to

88 + x = 135

and yields

x = 47

.

Step-by-Step Solution

1

Calculate the sum of the 7 known data points and express the mean in terms of the unknown 8th value

x

.

Sum of 7 known values =

4 + 7 + 9 + 12 + 15 + 18 + 23 = 88

. Total mean =

\frac{88 + x}{8}

.

The arithmetic mean of

n

values is the sum of all values divided by

n

.

2

Analyze the position of

x

in sorted order to determine the median.

Assuming

x \ge 15

, the 4th and 5th values in ascending order are

12

and

15

, giving a median of

\frac{12 + 15}{2} = 13.5

.

For an even number of data points (

8

), the median is the average of the 4th and 5th terms in sorted order.

3

Formulate and solve the equation linking the mean and median.

\frac{88 + x}{8} = 1.25 \times 13.5 = 16.875 \implies 88 + x = 135 \implies x = 47

.

The problem specifies that the mean is equal to

1.25

times the median.

Key Concept

Calculating mean and median of a dataset containing an unknown value.

Estimated Time:1m 30s

Question 146Question →

A committee of $3$ members is to be selected at random without replacement from a group of $5$ data scientists and $5$ software engineers. What is the conditional probability that at least two data scientists are selected in total, given that the first person selected is a data scientist?

\frac{5}{18}

\frac{1}{2}

\frac{5}{9}

\frac{13}{18}

\frac{3}{4}

Show answer & explanation

Answer: $\frac{13}{18}$

Answer

The conditional probability is

\frac{13}{18}

.

The conditional probability is calculated by adjusting the sample space after the first draw. With 1 data scientist already selected, there are 4 data scientists and 5 software engineers left (9 total). Selecting at least 1 more data scientist in the next 2 draws has a probability complementary to selecting 0 more data scientists. The probability of selecting 0 additional data scientists is

\frac{\binom{5}{2}}{\binom{9}{2}} = \frac{10}{36} = \frac{5}{18}

. Therefore, the required conditional probability is

1 - \frac{5}{18} = \frac{13}{18}

.

Step-by-Step Solution

1

Determine the remaining pool of candidates after the first selection.

Since 1 data scientist is selected first, 4 data scientists and 5 software engineers remain (total of 9 candidates). Two more candidates must be selected.

The conditional statement fixes the first selection, changing the sample space for the remaining 2 selections.

2

Identify the condition required for the target event to occur.

Since 1 data scientist is already selected, having at least two data scientists in total means selecting at least 1 additional data scientist in the remaining 2 draws.

Total data scientists = 1 (first draw) + (number of data scientists in next 2 draws).

3

Calculate the complementary probability (selecting 0 additional data scientists).

The number of ways to pick 2 software engineers from 5 is

\binom{5}{2} = 10

. The total ways to pick 2 people from 9 is

\binom{9}{2} = 36

. The probability of 0 additional data scientists is

\frac{10}{36} = \frac{5}{18}

.

Selecting 0 additional data scientists is equivalent to selecting 2 software engineers from the remaining group.

4

Subtract the complementary probability from 1.

1 - \frac{5}{18} = \frac{13}{18}

.

The sum of the probability of an event and its complement equals 1.

Key Concept

Conditional Probability without Replacement

Question 147Question →

A museum display curator is arranging $6$ distinct historical coins— $4$ silver coins and $2$ gold coins—in a single row inside a display case. If the $2$ gold coins cannot be placed adjacent to each other, how many different linear arrangements of the $6$ coins are possible?

240

360

480

576

720

Show answer & explanation

Answer: $480$

Answer

The total number of valid linear arrangements is

480

.

Arranging the

4

distinct silver coins yields

4! = 24

orderings. These

4

coins create

5

available gaps (including both ends). To guarantee the

2

distinct gold coins are not adjacent, each gold coin must occupy a separate gap. The number of ways to assign

2

distinct gold coins to

5

gaps is

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

. By the Fundamental Counting Principle, the total number of valid linear arrangements is

24 \times 20 = 480

.

Step-by-Step Solution

1

Calculate the total number of ways to arrange the unrestricted silver coins.

The

4

distinct silver coins can be arranged in

4! = 24

ways.

The relative positions of the silver coins matter because each coin is distinct.

2

Determine the number of available gap positions created by the silver coins to separate the gold coins.

Arranging

4

silver coins creates

5

possible gaps (one before the first coin, three between adjacent silver coins, and one after the last coin: \_ S \_ S \_ S \_ S \_).

Placing at most one gold coin per gap guarantees that the two gold coins will not be adjacent.

3

Calculate the number of ways to place the

2

distinct gold coins into the

5

available gaps.

The number of ways to place the

2

distinct gold coins into

5

distinct slots is

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

.

Order matters because the gold coins are distinct objects.

4

Apply the Fundamental Counting Principle to find the total arrangements.

Total arrangements =

24 \times 20 = 480

.

The arrangement of the silver coins and the placement of the gold coins are independent sequential choices.

Key Concept

Slot method for non-adjacent arrangements using permutations and the Fundamental Counting Principle

Question 148Question →

A healthcare clinic logged the waiting times, in minutes, for 7 patients on Monday: $12, 41, 24, 19, 33, 28,$ and $49$ . On Tuesday, 3 additional patients were logged. The arithmetic mean waiting time for all 10 patients combined was $30$ minutes, and the median waiting time of the 3 patients logged on Tuesday was $34$ minutes. If one of the patients logged on Tuesday had a waiting time of $22$ minutes and another had a waiting time of $34$ minutes, what was the waiting time, in minutes, of the third patient logged on Tuesday?

28 minutes

32 minutes

34 minutes

38 minutes

48 minutes

Show answer & explanation

Answer: 38 minutes

Answer

38 minutes

The option stating 38 minutes is correct because the total sum for all 10 patients combined is

10 \times 30 = 300

minutes. The sum of Monday's 7 patient waiting times is

12 + 41 + 24 + 19 + 33 + 28 + 49 = 206

minutes. Therefore, Tuesday's 3 patients must sum to

300 - 206 = 94

minutes. Since two of Tuesday's patients waited 22 minutes and 34 minutes, the third patient's waiting time is

94 - (22 + 34) = 38

minutes. Ordering the three Tuesday times

(22, 34, 38)

confirms the median is indeed 34 minutes.

Step-by-Step Solution

1

Calculate the total sum of waiting times for all 10 patients.

Total sum =

10 \times 30 = 300

minutes.

The arithmetic mean of

n

values is

\text{Sum} / n

, so

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

.

2

Calculate the sum of waiting times for the 7 patients logged on Monday.

Monday sum =

12 + 41 + 24 + 19 + 33 + 28 + 49 = 206

minutes.

Adding the individual values gives the total Monday waiting time.

3

Find the total sum of waiting times for the 3 patients logged on Tuesday.

Tuesday sum =

300 - 206 = 94

minutes.

Subtracting Monday's total from the combined total yields Tuesday's total.

4

Determine the third patient's waiting time on Tuesday using the given values.

Third patient's waiting time =

94 - (22 + 34) = 94 - 56 = 38

minutes.

The sum of the 3 Tuesday patients is 94 minutes, and two known patients waited 22 and 34 minutes. Note that the sorted set

(22, 34, 38)

has a median of 34 minutes, consistent with the problem statement.

Key Concept

Combined Mean and Median Properties

Question 149Question →

A technology committee has a pool of $8$ available guest speakers consisting of $5$ computer scientists and $3$ data privacy experts. A $4$ -person panel is to be formed from this pool. Which of the following statements regarding the possible panel selections must be true? Select all such statements.

Select all that apply

The total number of different

4

-person panels that can be formed without any restriction on specialty is equal to

70

.

The number of different

4

-person panels that consist of exactly

2

computer scientists and

2

data privacy experts is equal to

30

.

The number of different

4

-person panels containing at least

1

data privacy expert is equal to

65

.

The number of ways to select and arrange

4

computer scientists in a specific speaking order on stage is equal to

20

.

The number of different

4

-person panels containing no computer scientists is equal to

1

.

Show answer & explanation

Answer: The total number of different $4$ -person panels that can be formed without any restriction on specialty is equal to $70$ .; The number of different $4$ -person panels that consist of exactly $2$ computer scientists and $2$ data privacy experts is equal to $30$ .; The number of different $4$ -person panels containing at least $1$ data privacy expert is equal to $65$ .

Answer

The correct statements are those asserting that total unrestricted panels equal 70, panels with exactly 2 computer scientists and 2 privacy experts equal 30, and panels with at least 1 privacy expert equal 65.

The total unrestricted 4-person panels from 8 speakers is 8C4 = 70. Selecting 2 computer scientists (5C2 = 10) and 2 privacy experts (3C2 = 3) yields 10 × 3 = 30 panels. Using complementary counting, panels with at least 1 privacy expert equal total panels (70) minus panels composed entirely of computer scientists (5C4 = 5), giving 70 - 5 = 65.

Step-by-Step Solution

1

Calculate unrestricted combinations of 4 speakers out of 8

\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70

Selection order does not matter for committee membership, so we use combinations.

2

Calculate combinations with restricted counts (2 CS and 2 PE)

\binom{5}{2} \times \binom{3}{2} = 10 \times 3 = 30

By the Fundamental Counting Principle, independent choices for each subgroup are multiplied.

3

Calculate combinations with 'at least 1' condition using the complementary counting method

Total panels minus panels with 0 PE:

70 - \binom{5}{4} = 70 - 5 = 65

Subtracting outcomes that violate the constraint from total possible outcomes simplifies 'at least 1' calculations.

4

Evaluate the permutation and impossible constraint statements

Arranging 4 CS out of 5 gives

P(5,4) = 120 \neq 20

. A 4-person panel with 0 CS requires 4 PE out of 3, which gives

0 \neq 1

.

Arrangement requires permutations, and panel requirements exceeding available pool members yield zero valid groups.

Key Concept

Combinations, Permutations, and Complementary Counting Principle

Question 150Question →

A quality control analyst evaluates a manufacturing process in which two specific types of flaws, Flaw $X$ and Flaw $Y$ , can occur on produced glass panels. The probability that a randomly selected panel has Flaw $X$ is $P(X) = 0.20$ , and the probability that it has Flaw $Y$ is $P(Y) = 0.30$ . The analyst confirms that Flaw $X$ and Flaw $Y$ are mutually exclusive events.

Which of the following statements MUST be true regarding these two flaw types? Select all such statements.

Select all that apply

The probability that a randomly selected panel has at least one of the two flaws is

0.50

.

The conditional probability of a panel having Flaw

X

given that it has Flaw

Y

,

P(X \mid Y)

, is equal to

0

.

The probability that a randomly selected panel has both Flaw

X

and Flaw

Y

is

0.06

.

Flaw

X

and Flaw

Y

are independent events.

The probability that a randomly selected panel has neither Flaw

X

nor Flaw

Y

is

0.94

.

Show answer & explanation

Answer: The probability that a randomly selected panel has at least one of the two flaws is $0.50$ .; The conditional probability of a panel having Flaw $X$ given that it has Flaw $Y$ , $P(X \mid Y)$ , is equal to $0$ .

Answer

The statement that the probability of having at least one flaw is 0.50 and the statement that the conditional probability P(X | Y) is 0 are both true.

Because Flaw

X

and Flaw

Y

are mutually exclusive, their intersection

P(X \cap Y) = 0

. By the addition rule, the probability of at least one flaw is

P(X \cup Y) = P(X) + P(Y) = 0.20 + 0.30 = 0.50

. Furthermore, the conditional probability

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

. Thus, both the statement claiming the union probability is

0.50

and the statement claiming the conditional probability is

0

are correct.

Step-by-Step Solution

1

Analyze the definition of mutually exclusive events

Since Flaw

X

and Flaw

Y

are mutually exclusive, they cannot occur simultaneously on the same panel. Therefore,

P(X \cap Y) = 0

.

By definition, mutually exclusive events have an intersection probability of zero.

2

Calculate the union probability P(X or Y)

Using the addition rule for mutually exclusive events:

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.20 + 0.30 - 0 = 0.50

.

The probability of at least one event occurring is the sum of their individual probabilities when the intersection is zero.

3

Calculate the conditional probability P(X | Y)

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

.

If Flaw

Y

is known to occur, Flaw

X

cannot occur due to mutual exclusivity.

4

Evaluate independence between the two events

For independence,

P(X \cap Y)

must equal

P(X) \times P(Y) = 0.20 \times 0.30 = 0.06

. Since

0 \neq 0.06

, the events are dependent.

Two events with non-zero probabilities that are mutually exclusive are always dependent because the occurrence of one guarantees the non-occurrence of the other.

Key Concept

Mutually Exclusive vs. Independent Events

Question 151Question →

A librarian is arranging $7$ distinct books on a display shelf: $4$ history books and $3$ science books. If all $3$ science books must be placed adjacent to one another, in how many different linear orders can the $7$ books be arranged?

720

144

5,040

120

35

Show answer & explanation

Answer: 720

Answer

720 linear arrangements

To arrange items with an adjacency restriction, bundle the restricted items (the 3 science books) into 1 block. This leaves 4 history books and 1 science block, totaling 5 units to arrange linearly in 5! = 120 ways. Within the science block, the 3 distinct books can be arranged in 3! = 6 ways. By the Fundamental Counting Principle, the total number of linear arrangements is 5! × 3! = 120 × 6 = 720.

Step-by-Step Solution

1

Group the constrained items into a single block.

Treat the

3

distinct science books as a single unit or block, denoted as

[S]

.

Since the science books must always stand next to each other, they move as a single combined item.

2

Count the total number of items to arrange linearly.

We have

4

individual history books plus

1

science block, giving

4 + 1 = 5

items.

The

5

units can be ordered among themselves in

5!

ways.

3

Calculate the outer arrangements of the 5 units.

5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

.

This accounts for all position permutations of the history books and the block.

4

Calculate internal permutations within the science block and multiply.

3! = 6

ways within the block. Total arrangements =

120 \times 6 = 720

.

By the Fundamental Counting Principle, total ways equals outer permutations multiplied by inner permutations.

Key Concept

Permutations with Adjacency Restrictions (Block Method)

Question 152Question →

A quality assurance auditor reviewed the processing times, in seconds, for a batch of 7 completed tasks: $22, 10, 24, 18, 30, 22,$ and $14$ . An 8th task with a processing time of $x$ seconds, where $x$ is a positive integer, is added to the batch. Which of the following statements must be true? Select all such statements.

Select all that apply

If

x = 22

, the median of the 8 processing times is equal to 22.

If the arithmetic mean of the 8 processing times is equal to 21, then

x = 28

.

If

x = 18

, the median of the 8 processing times is equal to 21.

If

x = 30

, the arithmetic mean of the dataset increases by 5.

If

x = 10

, the mode of the 8 processing times becomes 10.

Show answer & explanation

Answer: If $x = 22$ , the median of the 8 processing times is equal to 22.; If the arithmetic mean of the 8 processing times is equal to 21, then $x = 28$ .

Answer

The correct statements are those asserting that if

x=22

, the median of the 8 processing times is 22, and if the mean of the 8 processing times is 21, then

x=28

.

The statement regarding

x=22

is correct because inserting 22 places 22 at both the 4th and 5th positions of the 8-element ordered list, yielding a median of 22. The statement regarding a mean of 21 is correct because the required total sum for 8 items is

8 \times 21 = 168

, which requires

x = 168 - 140 = 28

.

Step-by-Step Solution

1

Order the original dataset and calculate initial metrics.

Sorted original list:

10, 14, 18, 22, 22, 24, 30

. Original sum = 140, original mean =

\frac{140}{7} = 20

, original median = 22 (4th term).

Establishing the initial baseline values is necessary to evaluate statements about changes in mean and median.

2

Evaluate the statement regarding median when

x=22

.

Inserting

x=22

yields sorted list:

10, 14, 18, 22, 22, 22, 24, 30

. The 4th and 5th items are both 22, giving median

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

.

Confirms the first statement is true.

3

Evaluate the statement regarding mean equal to 21.

Target sum for 8 items with mean 21 is

8 \times 21 = 168

. Setting

140 + x = 168

yields

x = 28

.

Confirms the second statement is true.

4

Evaluate remaining incorrect statements.

For

x=18

, ordered list is

10, 14, 18, 18, 22, 22, 24, 30

, giving median

\frac{18+22}{2} = 20 \neq 21

. For

x=30

, new mean is

21.25

, an increase of

1.25 \neq 5

. For

x=10

, 10 and 22 both appear twice, making it bimodal rather than having a single mode of 10.

Verifies that all other statements are false.

Key Concept

Measures of Central Tendency (Mean, Median, Mode) for modified datasets

Question 153Question →

A medical laboratory uses an automated analyzer to screen blood samples for two distinct markers, Marker A and Marker B. The probability that a randomly selected sample contains Marker A is $0.40$ , and the probability that it contains Marker B is $0.25$ . If the presence of Marker A and the presence of Marker B are independent events, what is the probability that a randomly selected sample contains at least one of these two markers?

0.10

0.45

0.55

0.65

0.70

Show answer & explanation

Answer: 0.55

Answer

The probability that a randomly selected sample contains at least one of the two markers is

0.55

.

To find the probability that a sample contains at least one marker, apply the general addition rule

P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B})

. Because the events are independent,

P(\text{A and B}) = P(\text{A}) \times P(\text{B}) = 0.40 \times 0.25 = 0.10

. Substituting the values gives

0.40 + 0.25 - 0.10 = 0.55

. Alternatively, using the complementary probability rule yields

1 - P(\text{neither}) = 1 - (1 - 0.40)(1 - 0.25) = 1 - (0.60 \times 0.75) = 1 - 0.45 = 0.55

.

Step-by-Step Solution

1

Identify the given probabilities and event relationship.

P(\text{A}) = 0.40

,

P(\text{B}) = 0.25

, and events A and B are independent.

Establishing the parameters is necessary to apply the appropriate probability formulas.

2

Calculate the joint probability

P(\text{A and B})

.

P(\text{A and B}) = P(\text{A}) \times P(\text{B}) = 0.40 \times 0.25 = 0.10

.

For independent events, the probability of both events occurring simultaneously is the product of their individual probabilities.

3

Apply the general addition rule for probability to find

P(\text{A or B})

.

P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B}) = 0.40 + 0.25 - 0.10 = 0.55

.

The probability of at least one event occurring requires subtracting the overlapping joint probability to avoid double-counting.

Key Concept

Probability of Independent Events and the General Addition Rule

Question 154Question →

A regional library recorded the daily number of public computer reservations over a 7-day period: $14, 25, 18, 31, 25, 12,$ and $21$ . During the following week, 2 additional daily reservation values, $x$ and $y$ , were added to the dataset. If the median of the resulting 9-day dataset is equal to the mode of the original 7-day dataset, which of the following values could be the sum of $x$ and $y$ ?

36

42

46

48

54

Show answer & explanation

Answer: 54

Answer

54

The mode of the original 7 values is 25. When 2 values (

x

and

y

) are added to form a 9-value dataset, the median is the 5th value when ordered. In the original sorted dataset (

12, 14, 18, 21, 25, 25, 31

), there are only 3 values greater than or equal to 25. To make the 5th value of the combined 9-value list equal to 25, both added values must be at least 25 (

x \ge 25

and

y \ge 25

). Consequently, their sum

x + y

must be at least

50

. The only choice that satisfies

x + y \ge 50

is 54.

Step-by-Step Solution

1

Identify the mode of the original 7-day dataset.

The original values are

14, 25, 18, 31, 25, 12, 21

. The value

25

appears twice, while all other numbers appear once. Therefore, the mode is

25

.

The question states that the target median of the 9-day dataset must equal the mode of the original dataset.

2

Sort the original 7-day dataset in ascending order.

The sorted original dataset is

12, 14, 18, 21, 25, 25, 31

.

Determining position-based metrics like median requires ordering the data.

3

Determine the condition required for the median of the 9-day dataset to be 25.

In a sorted 9-element dataset, the median is the 5th element. For the 5th element to be

25

, there must be at least 5 elements in the set that are greater than or equal to

25

. The original set contains 3 elements

\ge 25

(

25, 25, 31

). Thus, both new values

x

and

y

must be

\ge 25

.

If only one of

x

or

y

were

\ge 25

, there would only be 4 elements

\ge 25

, making the 5th element (median) less than

25

.

4

Calculate the lower bound for the sum

x + y

and select the valid option.

Since

x \geq 25

and

y \geq 25

, the sum must satisfy

x + y \geq 25 + 25 = 50

. Among the given options, only

54

is greater than or equal to

50

.

Any sum less than

50

violates the condition that both

x

and

y

are at least

25

.

Key Concept

Determining positional metrics (median) after dataset expansion

Question 155Question →

A technology startup recorded the monthly software subscription costs, in dollars, for 6 department teams: $450, 120, 310, 120, 280,$ and $500$ . A seventh department team with a monthly subscription cost of $x$ dollars is added to the dataset, where $x$ is a positive integer. Which of the following statements regarding the measures of central tendency for the updated 7-team dataset must be true? Select all that apply.

Select all that apply

The median of the subscription costs for the 7 teams cannot be greater than 310.

If

x = 600

, the arithmetic mean of the subscription costs for the 7 teams is greater than the median.

If

x = 120

, the median of the subscription costs for the 7 teams is 120.

The arithmetic mean of the subscription costs for the 7 teams will equal the median if

x = 290

.

The mode of the subscription costs for the 7 teams is uniquely 120 for any positive integer value of

x

.

Show answer & explanation

Answer: The median of the subscription costs for the 7 teams cannot be greater than 310.; If $x = 600$ , the arithmetic mean of the subscription costs for the 7 teams is greater than the median.

Answer

The statements that must be true are that the median cannot exceed 310, and that if x = 600, the arithmetic mean is greater than the median.

Sorting the six known values gives 120, 120, 280, 310, 450, 500. In a dataset of 7 numbers, the median is the 4th value when sorted. If x is added, the 4th value will be 280 if x ≤ 280, x if 280 < x < 310, or 310 if x ≥ 310. Thus, the median is capped at 310. Furthermore, setting x = 600 yields a median of 310 and a mean of (1780 + 600) / 7 = 340, which is greater than the median.

Step-by-Step Solution

1

Sort the existing 6 dataset elements in ascending order and compute their sum.

Sorted list: 120, 120, 280, 310, 450, 500. Sum = 1780.

Establishing the baseline position of values helps determine how inserting x shifts the median, mean, and mode.

2

Analyze the position of the median for a 7-element dataset.

The median is the 4th element when sorted. If x ≤ 280, the 4th element is 280. If 280 < x < 310, the 4th element is x. If x ≥ 310, the 4th element is 310. Thus, 280 ≤ median ≤ 310.

This establishes upper and lower bounds for the median regardless of x.

3

Evaluate the statement regarding median upper bound (median ≤ 310).

Since the maximum possible 4th element in the sorted 7-element list is 310, the median can never be greater than 310.

Confirms the validity of the statement concerning the maximum median.

4

Evaluate the scenario where x = 600.

Sum = 1780 + 600 = 2380. Mean = 2380 / 7 = 340. Median = 310 (since 600 ≥ 310). 340 > 310.

Confirms that when x = 600, mean (340) is strictly greater than median (310).

Key Concept

Measures of Central Tendency (Mean, Median, Mode)

Question 156Question →

A sports scientist records the recovery times, in minutes, for 5 subjects after an intense workout. Four of the times are $44, 16, 52,$ and $32$ minutes, and the fifth time is $x$ minutes. The 5 times are listed in no particular order. If $32 < x < 44$ and the arithmetic mean of all 5 recovery times is equal to the median of the 5 recovery times, what is the value of $x$ ?

26

29

34

36

48

Show answer & explanation

Answer: 36

Answer

The value of

x

is 36.

Because

32 < x < 44

, the 5 numbers listed in ascending order are

16, 32, x, 44, 52

. The median of these 5 numbers is the middle value,

x

. Setting the mean

\frac{16 + 32 + x + 44 + 52}{5} = \frac{144 + x}{5}

equal to the median

x

gives

\frac{144 + x}{5} = x

, which simplifies to

4x = 144

and yields

x = 36

.

Step-by-Step Solution

1

Order the dataset in ascending numerical order.

The five data values in sorted order are

16, 32, x, 44, 52

because it is given that

32 < x < 44

.

Finding the median of a dataset requires arranging all elements in ascending or descending order first.

2

Identify the median of the 5 values.

The median is the 3rd element in the sorted 5-element list, which is

x

.

For an odd number of items (

n = 5

), the median is the middle value at position

\frac{5+1}{2} = 3

.

3

Express the arithmetic mean in terms of

x

.

\text{Mean} = \frac{16 + 32 + x + 44 + 52}{5} = \frac{144 + x}{5}$.

The arithmetic mean is defined as the sum of all values divided by the total count of values (

n = 5

).

4

Set the mean equal to the median and solve for

x

.

\frac{144 + x}{5} = x \implies 144 + x = 5x \implies 4x = 144 \implies x = 36$.

The problem states that the arithmetic mean equals the median.

Key Concept

Measures of Central Tendency (Mean, Median, Mode)

Estimated Time:1m 45s

Question 157Question →

Two events $E$ and $F$ in a sample space have probabilities $P(E) = 0.40$ and $P(F) = 0.50$ . Which of the following statements must be true? Select all such statements.

Select all that apply

If events

E

and

F

are mutually exclusive, then

P(E \text{ or } F) = 0.90

.

If events

E

and

F

are independent, then

P(E \text{ and } F) = 0.20

.

Events

E

and

F

cannot be both mutually exclusive and independent.

If events

E

and

F

are independent, then

P(E \text{ or } F) = 0.90

.

If events

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0.90

.

Show answer & explanation

Answer: If events $E$ and $F$ are mutually exclusive, then $P(E \text{ or } F) = 0.90$ .; If events $E$ and $F$ are independent, then $P(E \text{ and } F) = 0.20$ .; Events $E$ and $F$ cannot be both mutually exclusive and independent.

Answer

The correct statements are: 'If events E and F are mutually exclusive, then P(E or F) = 0.90', 'If events E and F are independent, then P(E and F) = 0.20', and 'Events E and F cannot be both mutually exclusive and independent.'

The statement regarding mutually exclusive events correctly uses the addition rule P(E or F) = P(E) + P(F) = 0.90 because there is no overlap. The statement regarding independent events correctly applies the multiplication rule P(E and F) = P(E) * P(F) = 0.20. Finally, events with non-zero probabilities cannot be both mutually exclusive (requiring zero intersection) and independent (requiring positive intersection), making the impossibility statement correct.

Step-by-Step Solution

1

Evaluate the union probability for mutually exclusive events.

P(E or F) = P(E) + P(F) = 0.40 + 0.50 = 0.90.

By definition of mutually exclusive events, P(E and F) = 0.

2

Evaluate the joint probability for independent events.

P(E and F) = P(E) * P(F) = 0.40 * 0.50 = 0.20.

The multiplication rule applies directly to independent events.

3

Evaluate whether mutually exclusive events with non-zero probabilities can be independent.

They cannot be both mutually exclusive and independent.

Mutual exclusivity requires P(E and F) = 0, whereas independence requires P(E and F) = 0.20, which are contradictory.

4

Evaluate the union probability for independent events.

P(E or F) = 0.40 + 0.50 - 0.20 = 0.70.

The general addition rule P(E or F) = P(E) + P(F) - P(E and F) must account for the overlap.

Key Concept

Distinction between independent events and mutually exclusive events in probability

Question 158Question →

At an automated agricultural sorting facility, harvested apples are inspected for two independent quality conditions: surface bruising and internal core rot. The probability that a randomly chosen apple has surface bruising is $\frac{1}{6}$ , and the probability that it has internal core rot is $\frac{1}{5}$ . What is the probability that a randomly chosen apple has at least one of these two quality conditions?

\frac{1}{30}

\frac{2}{11}

\frac{1}{3}

\frac{11}{30}

\frac{2}{5}

Show answer & explanation

Answer: $\frac{1}{3}$

Answer

\frac{1}{3}

The probability of at least one of two independent events occurring is given by

P(B \cup R) = P(B) + P(R) - P(B \cap R)

. Since the events are independent,

P(B \cap R) = P(B) \times P(R) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}

. Substituting these values gives

\frac{1}{6} + \frac{1}{5} - \frac{1}{30} = \frac{5}{30} + \frac{6}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}

. Alternatively, using the complement rule,

P(\text{at least one}) = 1 - P(\text{neither}) = 1 - \left(1 - \frac{1}{6}\right)\left(1 - \frac{1}{5}\right) = 1 - \left(\frac{5}{6} \times \frac{4}{5}\right) = 1 - \frac{4}{6} = \frac{1}{3}

.

Step-by-Step Solution

1

Identify the given probabilities and event relationship.

Let

B

be the event that an apple has surface bruising, so

P(B) = \frac{1}{6}

. Let

R

be the event that an apple has internal core rot, so

P(R) = \frac{1}{5}

. The events are given as independent.

Establishing the mathematical definitions and given conditions is necessary before applying probability rules.

2

Calculate the joint probability of both events occurring,

P(B \cap R)

.

P(B \cap R) = P(B) \times P(R) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}

.

For independent events, the probability that both occur simultaneously is the product of their individual probabilities.

3

Apply the addition rule of probability (inclusion-exclusion principle) to find

P(B \cup R)

.

P(B \cup R) = P(B) + P(R) - P(B \cap R) = \frac{1}{6} + \frac{1}{5} - \frac{1}{30} = \frac{5}{30} + \frac{6}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}

.

The probability of at least one event occurring requires subtracting the intersection so that the overlapping outcome is not counted twice.

Key Concept

Probability of the Union of Independent Events

Estimated Time:1m 30s

Question 159Question →

A dataset consists of 7 integers: $5, 8, 12, 14, 16, 20,$ and $x$ . If the median of the dataset is strictly greater than the mean of the dataset, which of the following could be the value of $x$ ? Select all such values.

Select all that apply

4

10

13

18

25

Show answer & explanation

Answer: $4$ ; $13$ ; $18$

Answer

The values of

x

that make the median strictly greater than the mean are

4

,

13

, and

18

.

The median of a 7-element dataset is its 4th element when arranged in ascending order. Testing the values shows that for

4

, the median (

12

) exceeds the mean (

11.29

); for

13

, the median (

13

) exceeds the mean (

12.57

); and for

18

, the median (

14

) exceeds the mean (

13.29

). All three satisfy the condition.

Step-by-Step Solution

1

Express the mean of the dataset in terms of

x

.

The sum of the known 6 numbers is

5 + 8 + 12 + 14 + 16 + 20 = 75

. The total sum for 7 numbers is

75 + x

, making the mean

\text{Mean} = \frac{75 + x}{7}

.

The mean is calculated as the sum of all elements divided by the total count (7).

2

Analyze the median across different ranges of

x

.

For 7 numbers sorted in order, the median is the 4th number. If

x \le 12

, the sorted list starts with

x, 5, 8, 12

or similar, so the 4th element is

12

. If

12 < x < 14

, the 4th element is

x

. If

x \ge 14

, the 4th element is

14

.

The position of

x

relative to the known numbers determines which element falls into the middle (4th) spot.

3

Test the condition

\text{Median} > \text{Mean}

for each piecewise case.

Case 1 (

x \le 12

):

12 > \frac{75 + x}{7} \implies 84 > 75 + x \implies x < 9

. Thus,

x = 4

works, but

x = 10

does not.
Case 2 (

12 < x < 14

):

x > \frac{75 + x}{7} \implies 7x > 75 + x \implies 6x > 75 \implies x > 12.5

. Thus,

x = 13

works.
Case 3 (

x \ge 14

):

14 > \frac{75 + x}{7} \implies 98 > 75 + x \implies x < 23

. Thus,

x = 18

works, but

x = 25

does not.

Solving the inequality for each case yields all valid ranges for

x

:

x < 9

,

x = 13

, and

14 \le x < 23

.

Key Concept

Measures of Central Tendency (Mean vs. Median Analysis with Variables)

Question 160Question →

During a quality assurance test of a dual-sensor monitoring device, Sensor $A$ operates independently of Sensor $B$ . The probability that Sensor $A$ detects a target signal during a test trial is $0.80$ , and the probability that Sensor $B$ detects the target signal during the same trial is $0.75$ . What is the probability that exactly one of the two sensors detects the target signal during a test trial?

Show answer & explanation

Answer: 0.35

Answer

0.35

To find the probability that exactly one sensor detects the signal, we must evaluate two disjoint scenarios: Sensor

A

detects while Sensor

B

fails (

0.80 \times 0.25 = 0.20

), or Sensor

A

fails while Sensor

B

detects (

0.20 \times 0.75 = 0.15

). Summing these two probabilities gives

0.20 + 0.15 = 0.35

. Alternatively, subtracting the probability that both sensors detect (

0.80 \times 0.75 = 0.60

) from the probability that at least one detects (

1 - 0.20 \times 0.25 = 0.95

) yields

0.95 - 0.60 = 0.35

.

Step-by-Step Solution

1

Determine the probabilities of individual event non-occurrences

P(A') = 0.20

and

P(B') = 0.25

The probability of an event not occurring is

1

minus the probability that it occurs.

2

Calculate joint probability for each mutually exclusive event outcome

P(A \text{ and } B') = 0.80 \times 0.25 = 0.20

and

P(A' \text{ and } B) = 0.20 \times 0.75 = 0.15

Since the sensors operate independently,

P(X \text{ and } Y) = P(X) \times P(Y)

.

3

Combine the independent outcomes that satisfy the condition

0.20 + 0.15 = 0.35

The scenarios (only

A

detects, or only

B

detects) are mutually exclusive, so their probabilities are added.

Key Concept

Probability of Independent and Mutually Exclusive Events

Data Analysis

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