Statistics and Data Analysis Practice Questions — GMAT — Page 4

Question 61Question →

A quality control inspector evaluates components from two separate production lines. Line 1 produces a batch of $12$ components containing $4$ defective and $8$ non-defective components. Line 2 produces a batch of $10$ components containing $2$ defective and $8$ non-defective components. The inspector randomly selects $2$ components from Line 1's batch without replacement, and then randomly selects $1$ component from Line 2's batch. What is the probability that exactly two of the three selected components are defective?

\frac{28}{165}

\frac{8}{45}

\frac{4}{33}

\frac{4}{55}

\frac{12}{77}

Show answer & explanation

Answer: $\frac{28}{165}$

Answer

The probability that exactly two of the three selected components are defective is

\frac{28}{165}

.

To find the probability of selecting exactly two defective components across both production lines, we sum the probabilities of two mutually exclusive events: (1) selecting two defective components from Line 1 (without replacement) and one non-defective component from Line 2, which has probability

\frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{12}{165}

; and (2) selecting one defective and one non-defective component from Line 1 and one defective component from Line 2, which has probability

2 \times \frac{4}{12} \times \frac{8}{11} \times \frac{2}{10} = \frac{16}{165}

. Adding these gives

\frac{12}{165} + \frac{16}{165} = \frac{28}{165}

.

Step-by-Step Solution

1

Identify the mutually exclusive cases that yield exactly two defective components.

Case 1: Draw 2 defective components from Line 1 AND 1 non-defective component from Line 2.
Case 2: Draw 1 defective and 1 non-defective component from Line 1 AND 1 defective component from Line 2.

The overall target outcome can be satisfied in two distinct ways depending on which line provides the defective components.

2

Calculate the probability for Case 1.

P(\text{Case 1}) = \left(\frac{4}{12} \times \frac{3}{11}\right) \times \frac{8}{10} = \frac{1}{11} \times \frac{4}{5} = \frac{4}{55} = \frac{12}{165}

.

Drawing without replacement from Line 1 gives a probability of

\frac{4}{12} \times \frac{3}{11} = \frac{1}{11}

. Multiplying by the probability of picking a non-defective component from Line 2 (

\frac{8}{10}

) yields

\frac{4}{55}

.

3

Calculate the probability for Case 2.

P(\text{Case 2}) = \left(2 \times \frac{4}{12} \times \frac{8}{11}\right) \times \frac{2}{10} = \frac{16}{33} \times \frac{1}{5} = \frac{16}{165}

.

The probability of getting 1 defective and 1 non-defective component from Line 1 in 2 draws without replacement is

2 \times \frac{4}{12} \times \frac{8}{11} = \frac{16}{33}

. Multiplying by the probability of picking a defective component from Line 2 (

\frac{2}{10}

) yields

\frac{16}{165}

.

4

Sum the probabilities of Case 1 and Case 2.

P(\text{Total}) = \frac{12}{165} + \frac{16}{165} = \frac{28}{165}

.

Since the two cases are mutually exclusive, their individual probabilities are added.

Key Concept

Probability of Independent and Dependent Events

Estimated Time:2m 0s

Question 62Question →

A panel discussion features 6 participants: 3 scientists, 2 economists, and 1 moderator. In how many different linear seating arrangements can these 6 participants be seated in a single row of 6 chairs if all 3 scientists must sit in adjacent seats?

24

36

120

144

720

Show answer & explanation

Answer: $144$

Answer

The total number of valid seating arrangements is 144.

To arrange participants under the condition that all 3 scientists sit together, treat the 3 scientists as a single block. This leaves 4 units to arrange: 1 block of scientists, 2 economists, and 1 moderator. These 4 units can be arranged in a row in

4! = 24

ways. Furthermore, within the scientist block, the 3 individual scientists can be arranged in

3! = 6

distinct orders. By the fundamental counting principle, multiplying the arrangements of the main units by the internal arrangements of the block gives

24 \times 6 = 144

total valid seating arrangements.

Step-by-Step Solution

1

Group the restricted items into a single block.

Treat the 3 scientists as 1 single block unit.

Because the 3 scientists must sit together in adjacent seats, they move as a single block along with the other participants.

2

Calculate the number of arrangements for the distinct units.

There are 4 units to arrange (1 scientist block + 2 economists + 1 moderator), giving

4! = 24

arrangements.

The Fundamental Counting Principle states that

n

distinct items can be linearly arranged in

n!

ways.

3

Calculate the internal arrangements within the block.

The 3 scientists can be arranged among themselves within their block in

3! = 6

ways.

The order of the individual scientists inside the block matters.

4

Multiply the unit arrangements by the internal arrangements.

Total arrangements =

4! \times 3! = 24 \times 6 = 144

.

By the multiplication principle, the total ways to complete both independent arrangement tasks is the product of their individual possibilities.

Key Concept

Permutations with Adjacency Restrictions (Block Method)

Estimated Time:1m 30s

Question 63Question →

A container holds $6$ red tokens and $n$ blue tokens, where $n$ is a positive integer. Two tokens are drawn at random from the container one after another without replacement. If the probability that the first token drawn is red and the second token drawn is blue is equal to $\frac{9}{35}$ , what is the value of $n$ ?

Show answer & explanation

Answer: 9

Answer

The value of

n

is

9

.

By applying the multiplication rule for dependent events without replacement, the probability is

\frac{6}{6+n} \times \frac{n}{5+n} = \frac{9}{35}

. Simplifying and solving the resulting quadratic equation

3n^2 - 37n + 90 = 0

yields

n = 9

as the only integer solution.

Step-by-Step Solution

1

Formulate the joint probability for dependent events.

The probability of drawing a red token first then a blue token second is

\frac{6}{6+n} \times \frac{n}{5+n}

.

Because sampling is done without replacement, the total number of remaining tokens decreases from

6+n

to

5+n

for the second draw.

2

Set up the algebraic equation with the given probability.

\frac{6n}{(6+n)(5+n)} = \frac{9}{35}

Equate the theoretical probability expression to the given value

\frac{9}{35}

.

3

Simplify and expand the equation.

Divide both sides of the numerator by

3

: \frac{2n}{n^2 + 11n + 30} = \frac{3}{35}. Cross-multiplying yields

70n = 3n^2 + 33n + 90

.

Simplifying ratios makes polynomial expansion easier.

4

Solve the quadratic equation for

n

.

Rearrange to

3n^2 - 37n + 90 = 0

, which factors into

(3n - 10)(n - 9) = 0

.

Find the roots of the quadratic equation.

5

Select the valid integer solution.

Since

n

must be an integer count of tokens,

n = 9

(rejecting

n = \frac{10}{3}

).

The number of tokens

n

must be a whole number.

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

Question 64Question →

A researcher recorded six test measurements arranged in ascending order: $1, 2, 8, x, y, 15$ . If the unique mode of the data set is $8$ and the arithmetic mean of the measurements is equal to their median, what is the value of $y$ ?

Show answer & explanation

Answer: 14

Answer

14

Because the data set is arranged in ascending order, the elements satisfy

1 \le 2 \le 8 \le x \le y \le 15

. For

8

to be the unique mode,

8

must appear more times than any other number, which forces

x = 8

. The median of a

6

-element set is the average of the

3^{\text{rd}}

and

4^{\text{th}}

elements, which are both

8

, making the median

8

. Setting the mean

\frac{1 + 2 + 8 + 8 + y + 15}{6} = \frac{34 + y}{6}

equal to

8

gives

34 + y = 48

, which yields

y = 14

. Since

8 \le 14 \le 15

, this maintains the ascending order and unique mode condition.

Step-by-Step Solution

1

Determine the value of x

x = 8

The set is given in ascending order (1 <= 2 <= 8 <= x <= y <= 15). For 8 to be the unique mode, 8 must appear at least twice, which requires x = 8.

2

Calculate the median of the data set

Median = 8

With 6 elements, the median is the average of the 3rd and 4th elements: (8 + 8) / 2 = 8.

3

Set up the equation for the arithmetic mean

(34 + y) / 6 = 8

The sum of the six numbers is 1 + 2 + 8 + 8 + y + 15 = 34 + y. Since the mean equals the median, the mean must equal 8.

4

Solve for y

y = 14

Multiplying both sides by 6 gives 34 + y = 48, so y = 48 - 34 = 14.

Key Concept

Using statistical properties (mean, median, mode) of an ordered set to determine unknown values

Estimated Time:1m 15s

Question 65Question →

A dataset $X$ consists of five positive integers with a mean of 12, a median of 12, a unique mode of 12, and a range of 8. If $\sigma^2$ represents the variance (the square of the standard deviation) of dataset $X$ , what is the maximum possible value of $\sigma^2$ ?

6.4

7.2

7.6

8.0

8.4

Show answer & explanation

Answer: $7.6$

Answer

The maximum possible value of the variance

\sigma^2

is

7.6

.

To maximize variance

\sigma^2 = \frac{1}{5}\sum (x_i - 12)^2

, data points must be placed as far from the mean of 12 as possible. For five ordered positive integers

a \le b \le c \le d \le e

, the mean of 12 gives

a+b+c+d+e=60

, the median of 12 gives

c=12

, and the range of 8 gives

e=a+8

. Substituting yields

2a+b+d=40

. Since

b \le 12

and

d \le a+8

,

b+d \le a+20

, leading to

3a \ge 20

, so

a \ge 7

. Testing

a=7

gives

e=15

and

b+d=26

. To keep 12 as the unique mode,

b=12

and

d=14

, forming dataset

\{7, 12, 12, 14, 15\}

. The squared deviations from 12 are

(-5)^2 + 0^2 + 0^2 + 2^2 + 3^2 = 25 + 0 + 0 + 4 + 9 = 38

. Dividing by 5 yields the maximum variance of

7.6

.

Step-by-Step Solution

1

Express the five integers in ascending order and set up equations from given statistics.

Let the integers be

a \le b \le c \le d \le e

. Mean = 12 implies

a + b + c + d + e = 60

. Median = 12 implies

c = 12

. Range = 8 implies

e = a + 8

.

Translating statistical measures into algebraic constraints provides the boundary conditions for the dataset.

2

Substitute known values into the sum equation and determine constraints on

a

.

a + b + 12 + d + (a + 8) = 60 \implies 2a + b + d = 40

. Since

b \le 12

and

d \le e = a + 8

, we have

b + d \le a + 20

. Thus

2a + (a + 20) \ge 40 \implies 3a \ge 20 \implies a \ge 7

.

Determining the minimum value of

a

is critical because smaller values of

a

allow larger deviations from the mean of 12.

3

Evaluate valid dataset configurations for

a = 7

that satisfy the unique mode condition.

If

a = 7

, then

e = 15

, and

b + d = 26

. With

b \le 12

and

d \le 15

, and requiring 12 to be the unique mode, the only valid integer set is

\{7, 12, 12, 14, 15\}

.

The set must maintain 12 as the unique mode while maximizing the dispersion of values.

4

Calculate the variance

\sigma^2

for the set

\{7, 12, 12, 14, 15\}

.

Deviations from 12 are

-5, 0, 0, 2, 3

. Squared deviations are

25, 0, 0, 4, 9

. Sum of squared deviations = 38. Variance

\sigma^2 = \frac{38}{5} = 7.6

.

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

Key Concept

Properties of Variance and Standard Deviation under Range and Central Tendency Constraints

Question 66Question →

A committee of $10$ members consists of $5$ senior directors and $5$ junior associates. If $3$ committee members are selected at random one after another without replacement, what is the probability that exactly $2$ of the selected members are senior directors?

\frac{5}{36}

\frac{3}{8}

\frac{5}{12}

\frac{25}{144}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{5}{12}$

Answer

The probability that exactly

2

of the selected members are senior directors is

\frac{5}{12}

.

The correct answer is found by taking into account that the selections are dependent events without replacement. The probability of selecting senior directors on the first two draws and a junior associate on the third is

\frac{5}{10} \times \frac{4}{9} \times \frac{5}{8} = \frac{5}{36}

. Because there are

3

possible mutually exclusive orders to select

2

senior directors and

1

junior associate, multiplying

\frac{5}{36}

by

3

gives the final probability of

\frac{5}{12}

.

Step-by-Step Solution

1

Calculate the probability of drawing Senior, Senior, Junior in that specific order without replacement.

P(\text{Senior}_1) \times P(\text{Senior}_2 \mid \text{Senior}_1) \times P(\text{Junior}_3 \mid \text{Senior}_1 \cap \text{Senior}_2) = \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8} = \frac{100}{720} = \frac{5}{36}

Because draws are made without replacement, the sample space and favorable outcomes decrease with each draw.

2

Determine the number of distinct orderings of

2

Senior directors and

1

Junior associate.

\binom{3}{2} = 3

orderings: (Senior, Senior, Junior), (Senior, Junior, Senior), and (Junior, Senior, Senior).

The question asks for exactly

2

senior directors regardless of the order in which they are selected.

3

Multiply the single-sequence probability by the total number of valid orderings.

3 \times \frac{5}{36} = \frac{15}{36} = \frac{5}{12}

Since each distinct ordering is mutually exclusive and has the same probability, the total probability is the sum across all

3

orderings.

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

Question 67Question →

A data set consists of $7$ positive integers. The smallest integer in the set is $4$ , the range of the set is $19$ , the median is $12$ , and the set has a unique mode of $8$ . What is the maximum possible arithmetic mean of the $7$ integers?

13

13.5

14

14.5

15

Show answer & explanation

Answer: 14

Answer

The maximum possible arithmetic mean of the 7 integers is 14.

The correct answer of 14 is obtained by setting up the 7 ordered integers as

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

. We know

x_1 = 4

, median

x_4 = 12

, and

x_7 = 4 + 19 = 23

. Since 8 is the unique mode and

8 < 12

, 8 must fill

x_2

and

x_3

, giving it a frequency of 2. For 8 to remain the unique mode, no other element can appear more than once. Thus,

x_5, x_6, x_7

must all be distinct. To maximize the mean, we maximize

x_5

and

x_6

under

12 < x_5 < x_6 < 23

, yielding

x_5 = 21

and

x_6 = 22

. The maximum sum is

4 + 8 + 8 + 12 + 21 + 22 + 23 = 98

, giving a mean of

98 / 7 = 14

.

Step-by-Step Solution

1

Determine the fixed values in the ordered 7-element data set.

Let the 7 integers in non-decreasing order be

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

. Given

x_1 = 4

, median

x_4 = 12

, and range

x_7 - x_1 = 19

, we find

x_7 = 4 + 19 = 23

.

The median of a 7-element set is the 4th element, and range is maximum minus minimum.

2

Determine the frequency and values of the mode.

Since the unique mode is

8

and

8 < 12

, the value

8

must occupy positions

x_2

and

x_3

, so

x_2 = 8

and

x_3 = 8

. Thus,

8

appears exactly twice.

Position 1 is 4 and position 4 is 12, leaving only positions 2 and 3 for the mode value of 8.

3

Apply the unique mode constraint to maximize remaining terms.

Because

8

appears twice and is the unique mode, no other number can appear 2 or more times. Hence, all other elements must be distinct. To maximize the sum, we choose the largest distinct integers for

x_5

and

x_6

such that

12 < x_5 < x_6 < 23

, giving

x_5 = 21

and

x_6 = 22

.

Allowing any other value to repeat would create a second mode or a new unique mode, violating the problem conditions.

4

Calculate the maximum sum and arithmetic mean.

The maximal set is

\{4, 8, 8, 12, 21, 22, 23\}

. The sum is

4 + 8 + 8 + 12 + 21 + 22 + 23 = 98

. The maximum mean is

98 / 7 = 14

.

Dividing the maximum possible sum by the total number of elements yields the maximum mean.

Key Concept

Mean, Median, and Mode Constraints

Question 68Question →

A bakery display case contains $8$ fruit tarts and $12$ chocolate croissants. If a customer randomly selects $2$ pastry items one after another without replacement, what is the probability that the customer chooses exactly $1$ fruit tart and $1$ chocolate croissant?

\frac{24}{95}

\frac{12}{25}

\frac{48}{95}

\frac{47}{95}

\frac{6}{25}

Show answer & explanation

Answer: $\frac{48}{95}$

Answer

The probability that the customer chooses exactly

1

fruit tart and

1

chocolate croissant is

\frac{48}{95}

.

Selecting one fruit tart and one chocolate croissant can occur in two mutually exclusive orderings: (Fruit Tart, Chocolate Croissant) or (Chocolate Croissant, Fruit Tart). Since the selection is without replacement, the total number of remaining items drops to

19

for the second draw. The probability of the first order is

\frac{8}{20} \times \frac{12}{19} = \frac{96}{380}

, and the probability of the second order is

\frac{12}{20} \times \frac{8}{19} = \frac{96}{380}

. Summing these gives

\frac{192}{380}

, which simplifies to

\frac{48}{95}

.

Step-by-Step Solution

1

Determine the total number of items and individual counts.

Total items =

8 + 12 = 20

. Fruit tarts =

8

, Chocolate croissants =

12

.

Establishing the initial sample space is necessary for calculating single-event probabilities.

2

Calculate the probability for each distinct order of selection without replacement.

Probability of (Fruit Tart then Chocolate Croissant) =

\frac{8}{20} \times \frac{12}{19} = \frac{96}{380}

. Probability of (Chocolate Croissant then Fruit Tart) =

\frac{12}{20} \times \frac{8}{19} = \frac{96}{380}

.

Because sampling is done without replacement, the total pool decreases from

20

to

19

after the first selection.

3

Sum the probabilities of the mutually exclusive outcomes and simplify.

\frac{96}{380} + \frac{96}{380} = \frac{192}{380} = \frac{48}{95}

.

The desired event occurs if either sequence happens, so the probabilities of the two sequences are added together.

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

Estimated Time:1m 30s

Question 69Question →

A dataset consists of 7 integers listed in ascending order: $3, 4, 4, x, 9, y, 15$ . If the unique mode of the dataset is $4$ and the arithmetic mean of the dataset is equal to its median, what is the value of $y$ ?

Show answer & explanation

Answer: 13

Answer

The value of

y

is 13.

Because the dataset is in ascending order, the median of the 7 terms is the 4th term,

x

. Setting the mean

\frac{35 + x + y}{7}

equal to

x

yields

y = 6x - 35

. Since

y

lies between

9

and

15

inclusive, solving

9 \le 6x - 35 \le 15

gives

x = 8

. Substituting

x = 8

into

y = 6x - 35

gives

y = 13

.

Step-by-Step Solution

1

Determine the median of the 7-element ordered set.

The median is the 4th term,

x

.

For an odd number of ordered terms (

n = 7

), the median is the

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

term.

2

Set up the equation equating the arithmetic mean to the median.

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

, which simplifies to

y = 6x - 35

.

The mean of a dataset is the sum of all elements divided by the total number of elements.

3

Apply the ascending order constraint to find the value of

x

.

Since

9 \le y \le 15

, we have

9 \le 6x - 35 \le 15

, giving

44 \le 6x \le 50

. The only integer value for

x

in this range is

8

.

The dataset is listed in ascending order, so the element

y

after

9

and before or equal to

15

must satisfy

9 \le y \le 15

.

4

Calculate the value of

y

.

y = 6(8) - 35 = 13

.

Substitute

x = 8

back into the simplified linear equation relating

x

and

y

.

Key Concept

Relating arithmetic mean and median in an ordered set using inequalities derived from ascending order constraints.

Estimated Time:1m 40s

Question 70Question →

A commercial bakery produces two batches of bread dough. Batch X consists of $30$ kilograms of dough with a sugar concentration of $12\%$ by weight. Batch Y consists of $45$ kilograms of dough with a sugar concentration of $20\%$ by weight. If Batch X and Batch Y are mixed together completely, what is the sugar concentration, as a percentage by weight, of the combined mixture?

Show answer & explanation

Answer: 16.8

Answer

The sugar concentration of the combined mixture is 16.8 percent by weight.

The overall concentration is calculated by taking the weighted average of the two concentrations. Since Batch X and Batch Y have weights in the ratio of 2 to 3, the overall percentage is [2(12%) + 3(20%)] / (2 + 3) = 84 / 5 = 16.8%. Alternatively, the total amount of sugar is 3.6 kg + 9.0 kg = 12.6 kg out of a total mass of 75 kg, yielding 12.6 / 75 = 16.8%.

Step-by-Step Solution

1

Calculate the total mass of sugar contributed by Batch X

3.6 kg of sugar (

30 \times 0.12 = 3.6

)

To find the total sugar in the combined mixture, compute the sugar content from each component batch.

2

Calculate the total mass of sugar contributed by Batch Y

9.0 kg of sugar (

45 \times 0.20 = 9.0

)

Compute the sugar mass for the second batch.

3

Sum the total sugar mass and total batch mass

Total sugar =

12.6

kg (

3.6 + 9.0

); Total mass =

75

kg (

30 + 45

)

The combined concentration equals total sugar divided by total mixture mass.

4

Compute the combined percentage concentration

16.8% (

\frac{12.6}{75} \times 100\% = 16.8\%

)

Divide total sugar mass by total mixture mass and multiply by 100.

Key Concept

Weighted Average of Combined Sets

Estimated Time:1m 30s

Question 71Question →

A software development company has two engineering divisions: Cloud Infrastructure and Mobile Applications. The $45$ engineers in the Cloud Infrastructure division completed code reviews with an average duration of $25$ minutes per review, while the engineers in the Mobile Applications division completed code reviews with an average duration of $40$ minutes per review. If the overall average code review duration for all engineers across both divisions combined was $31$ minutes, how many engineers are in the Mobile Applications division?

18

27

30

45

75

Show answer & explanation

Answer: $30$

Answer

30 engineers

The overall average of a combined set is weighted by the number of elements in each group. Using the distance-to-mean principle, the ratio of the number of engineers in Cloud Infrastructure (

N_c

) to Mobile Applications (

N_m

) is inversely proportional to their distances from the combined mean:

N_c : N_m = (40 - 31) : (31 - 25) = 9 : 6 = 3 : 2

. Given

N_c = 45

, setting up the proportion

\frac{45}{N_m} = \frac{3}{2}

yields

N_m = 30

.

Step-by-Step Solution

1

Set up the weighted average equation for the combined set.

\frac{45 \times 25 + N_m \times 40}{45 + N_m} = 31

The total time across both divisions divided by total engineers equals the combined average time.

2

Multiply both sides by

(45 + N_m)

and expand.

1125 + 40 N_m = 31(45) + 31 N_m = 1395 + 31 N_m

To clear the denominator and group like terms.

3

Solve for

N_m

.

40 N_m - 31 N_m = 1395 - 1125 \implies 9 N_m = 270 \implies N_m = 30

Isolating

N_m

gives the exact count of engineers in the Mobile Applications division.

Key Concept

Weighted Average and Combined Sets

Estimated Time:2m 0s

Question 72Question →

A university department offers a specialized statistics course to two groups of students: undergraduate students and graduate students. A total of $120$ undergraduate students took the final examination, scoring an average of $80$ points. The remaining students in the course were graduate students, who achieved an average score of $95$ points on the same examination. If the combined average score for all students in the course was $86$ points, how many graduate students were enrolled in the course?

72

80

90

120

180

Show answer & explanation

Answer: 80

Answer

80 graduate students were enrolled in the course.

The correct answer is obtained by applying the weighted average relationship:

N_{\text{undergrad}} \times (\text{Mean}_{\text{combined}} - \text{Mean}_{\text{undergrad}}) = N_{\text{grad}} \times (\text{Mean}_{\text{grad}} - \text{Mean}_{\text{combined}})

. Substituting the given values gives

120 \times (86 - 80) = N_{\text{grad}} \times (95 - 86)

, which simplifies to

720 = 9 N_{\text{grad}}

, yielding

N_{\text{grad}} = 80

.

Step-by-Step Solution

1

Determine the distances of each group's mean score from the combined mean score.

Undergraduate distance:

86 - 80 = 6

; Graduate distance:

95 - 86 = 9

.

In a weighted average of two groups, the total deviation below the combined mean must equal the total deviation above the combined mean.

2

Set up the balance equation using the group counts and their respective distances from the combined mean.

120 \times 6 = N \times 9

, where

N

represents the number of graduate students.

The weight of each group is inversely proportional to the distance of its mean from the overall combined mean.

3

Solve the linear equation for

N

.

720 = 9N \implies N = 80

.

Dividing the total undergraduate score deviation (

720

) by the graduate score deviation per student (

9

) gives the number of graduate students.

Key Concept

Weighted Average of Two Combined Sets

Question 73Question →

A regional healthcare provider operates two outpatient facilities, Facility X and Facility Y. Facility X manages $60$ patient appointments per day with an average consultation time of $15$ minutes per patient. Facility Y manages $40$ patient appointments per day with an average consultation time of $25$ minutes per patient. What is the overall average consultation time, in minutes per patient, for all $100$ appointments combined across both facilities?

Show answer & explanation

Answer: 19

Answer

The overall average consultation time across both facilities is

19

minutes per patient.

The total consultation time across both facilities is

60 \times 15 + 40 \times 25 = 900 + 1000 = 1900

minutes. Dividing this by the total number of appointments (

60 + 40 = 100

) gives a weighted average of

\frac{1900}{100} = 19

minutes.

Step-by-Step Solution

1

Calculate the total consultation minutes contributed by Facility X.

60 \times 15 = 900

minutes

Total time for a set is the number of items multiplied by the mean of that set.

2

Calculate the total consultation minutes contributed by Facility Y.

40 \times 25 = 1000

minutes

Total time for the second set is the number of items multiplied by its mean.

3

Find the grand total consultation time and total number of patients.

Total minutes =

900 + 1000 = 1900

; Total patients =

60 + 40 = 100

Combined sets require aggregating both total values and total counts.

4

Compute the combined weighted average.

\frac{1900}{100} = 19

minutes

The weighted average is total accumulated value divided by total count.

Key Concept

Weighted Average of Combined Sets

Estimated Time:1m 30s

Question 74Question →

A financial consultancy firm consists of two divisions: Risk Management and Wealth Advisory. The $40$ analysts in the Risk Management division earn an average annual bonus of $\$15,000$ , while the analysts in the Wealth Advisory division earn an average annual bonus of $\$25,000$ . If the average annual bonus across all analysts in both divisions combined is $\$21,000$ , how many analysts work in the Wealth Advisory division?

24

40

50

60

100

Show answer & explanation

Answer: 60

Answer

The Wealth Advisory division has 60 analysts.

The combined mean of two sets is closer to the mean of the larger set. Using the distance-ratio rule for weighted averages, the ratio of the number of Risk Management analysts to Wealth Advisory analysts is inversely proportional to their distances from the combined mean:

\text{Ratio} = \frac{25,000 - 21,000}{21,000 - 15,000} = \frac{4,000}{6,000} = \frac{2}{3}

. Given that there are 40 analysts in Risk Management (representing 2 parts), each part equals 20 analysts. Therefore, the 3 parts for Wealth Advisory equal

3 \times 20 = 60

analysts.

Step-by-Step Solution

1

Set up the weighted average equation with

N

representing the number of analysts in the Wealth Advisory division.

Total bonus for Risk Management =

40 \times 15,000 = 600,000

. Total bonus for Wealth Advisory =

N \times 25,000

. Total analysts =

40 + N

.

The combined mean is equal to the sum of all individual values divided by the total number of items.

2

Formulate the combined average equation and clear the fraction.

\frac{600,000 + 25,000N}{40 + N} = 21,000 \implies 600,000 + 25,000N = 21,000(40 + N)

Equate the weighted average expression to the given overall average of

\$21,000

.

3

Expand and solve the linear equation for

N

.

600,000 + 25,000N = 840,000 + 21,000N \implies 4,000N = 240,000 \implies N = 60

Isolate the variable

N

to find the exact number of analysts in Wealth Advisory.

Key Concept

Weighted Average and Combined Sets

Estimated Time:1m 30s

Question 75Question →

A citrus orchard harvested two varieties of oranges: Valencia and Navel. The harvest consisted of $800$ crates of Valencia oranges with an average weight of $45$ pounds per crate, and $1,200$ crates of Navel oranges with an average weight of $50$ pounds per crate. What is the average weight, in pounds per crate, for the combined harvest of all crates?

Show answer & explanation

Answer: 48

Answer

The average weight of the combined harvest is 48 pounds per crate.

The weighted average is found by calculating the total overall weight divided by the total number of crates. Since Valencia oranges account for 36,000 pounds (800 crates at 45 lbs/crate) and Navel oranges account for 60,000 pounds (1,200 crates at 50 lbs/crate), the total weight is 96,000 pounds across 2,000 crates. Dividing 96,000 by 2,000 gives exactly 48 pounds per crate.

Step-by-Step Solution

1

Calculate the total weight contributed by each variety of oranges.

Valencia weight =

800 \times 45 = 36,000

lbs; Navel weight =

1,200 \times 50 = 60,000

lbs.

Total weight equals the product of the number of units and the average weight per unit for each group.

2

Determine the total combined weight and the total number of crates.

Total weight =

36,000 + 60,000 = 96,000

lbs; Total crates =

800 + 1,200 = 2,000

crates.

Combined weighted averages require summing total values across all subsets.

3

Divide total weight by total number of crates to find the overall weighted average.

Weighted average =

\frac{96,000}{2,000} = 48

lbs per crate.

Weighted mean formula:

\bar{x} = \frac{\sum (n_i \cdot x_i)}{\sum n_i}

.

Key Concept

Weighted Average of Combined Sets

Estimated Time:1m 30s

Question 76Question →

Data Set $P$ consists of five distinct integers with a range of $20$ and a standard deviation of $d_1$ , where $d_1 > 0$ . A new Data Set $Q$ is formed by multiplying each integer in Data Set $P$ by $-2$ and then adding $7$ to the product. What are the range and standard deviation of Data Set $Q$ in terms of $d_1$ ?

Range is

40

; standard deviation is

2d_1

Range is

40

; standard deviation is

-2d_1

Range is

-40

; standard deviation is

-2d_1

Range is

47

; standard deviation is

2d_1 + 7

Range is

20

; standard deviation is

2d_1

Show answer & explanation

Answer: Range is $40$ ; standard deviation is $2d_1$

Answer

Range is

40

; standard deviation is

2d_1

The correct answer accounts for both linear transformation rules for measures of dispersion: multiplying every value in a dataset by a scalar

c

multiplies both the range and standard deviation by

|c|

, and adding a constant

k

to every value does not alter either measure. Here, multiplying by

-2

scales the range from

20

to

40

and the standard deviation from

d_1

to

2d_1

. Adding

7

has no effect on either measure.

Step-by-Step Solution

1

Analyze the effect of multiplying each element by a constant

c = -2

.

The range scales by

|c| = |-2| = 2

, giving a new range of

20 \times 2 = 40

. The standard deviation scales by

|c| = |-2| = 2

, giving a new standard deviation of

2d_1

.

Measures of dispersion (range and standard deviation) are non-negative distance metrics, so they scale by the absolute value of any multiplicative constant.

2

Analyze the effect of adding a constant

k = 7

to each element.

The range remains

40

, and the standard deviation remains

2d_1

.

Adding a constant shifts every data point by the exact same distance, leaving all pairwise distances between data points unchanged.

Key Concept

Linear Transformations on Range and Standard Deviation

Question 77Question →

Data set $A$ consists of five consecutive even integers. What is the ratio of the standard deviation of data set $A$ to the range of data set $A$ ?

\frac{\sqrt{2}}{4}

-\frac{\sqrt{2}}{4}

\frac{\sqrt{10}}{8}

\frac{\sqrt{2}}{2}

1

Show answer & explanation

Answer: $\frac{\sqrt{2}}{4}$

Answer

\frac{\sqrt{2}}{4}

For any set of five consecutive even integers, the deviations from the mean are always

-4, -2, 0, 2, 4

, yielding a variance of 8 and a standard deviation of

2\sqrt{2}

. The range of any five consecutive even integers is always

8

. Taking the ratio of the standard deviation to the range gives

\frac{2\sqrt{2}}{8}

, which simplifies to

\frac{\sqrt{2}}{4}

.

Step-by-Step Solution

1

Represent the dataset algebraic terms

Let the five consecutive even integers be

x, x+2, x+4, x+6, x+8

.

Choosing symmetric terms simplifies calculating the mean and deviations.

2

Calculate the mean of the dataset

Mean

\mu = \frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} = x+4

.

The mean of an evenly spaced set of consecutive numbers is equal to the middle element.

3

Calculate the deviations from the mean and sum of squared deviations

Deviations:

-4, -2, 0, 2, 4

.
Squared deviations:

16, 4, 0, 4, 16

.
Sum of squared deviations

= 40

.

Standard deviation measures dispersion from the mean.

4

Calculate the variance and standard deviation

Variance

\sigma^2 = \frac{40}{5} = 8

.
Standard deviation

\sigma = \sqrt{8} = 2\sqrt{2}

.

Variance is the average of squared deviations, and standard deviation is its non-negative square root.

5

Calculate the range and the ratio of standard deviation to range

Range

= (x+8) - x = 8

.
Ratio

= \frac{\sigma}{\text{Range}} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}

.

Range is the difference between the maximum and minimum values.

Key Concept

Properties of Standard Deviation and Range for Consecutively Spaced Datasets

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