Statistics and Probability Practice Questions — JAMB UTME — Page 4

Question 61Question →

The table below shows the cumulative frequency distribution of the masses (in grams) of $100$ cocoa beans sampled from an agricultural yield:

Mass Class Interval (g)	Cumulative Frequency
$10 - 19$	$10$
$20 - 29$	$30$
$30 - 39$	$65$
$40 - 49$	$90$
$50 - 59$	$100$

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile ( $Q_3$ ) mass of the cocoa beans in grams.

Show answer & explanation

Answer: 43.5

Answer

The 75th percentile (

Q_3

) mass of the cocoa beans is

43.5\text{ g}

.

To find the 75th percentile (

Q_3

) from cumulative frequency data, calculate the rank

\frac{75}{100} \times 100 = 75

. This falls into the

40 - 49

class interval (boundaries

39.5 - 49.5

). Applying

Q_3 = L + \left(\frac{75 - F}{f}\right)c

yields

39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}

.

Step-by-Step Solution

1

Calculate the percentile position rank

Rank position is

75

The 75th percentile corresponds to

75\%

of the total sample size

N = 100

, giving

\frac{75}{100} \times 100 = 75

.

2

Locate the 75th percentile class interval and its boundaries

Class interval is

40 - 49

, with lower boundary

L = 39.5

and upper boundary

49.5

Cumulative frequency before

40 - 49

is

65

, and up to

40 - 49

is

90

. Since

65 < 75 \leq 90

, the 75th item falls in this interval.

3

Identify class parameters for interpolation

L = 39.5

,

F = 65

,

f = 25

,

c = 10

Lower boundary

L = 39.5

, previous cumulative frequency

F = 65

, class frequency

f = 90 - 65 = 25

, class width

c = 49.5 - 39.5 = 10

.

4

Compute

Q_3

using the linear interpolation formula

Q_3 = 43.5\text{ g}

Q_3 = 39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 39.5 + 4 = 43.5

.

Key Concept

Linear Interpolation of Percentiles from Cumulative Frequency Data

Question 62Question →

An environmental research station recorded the daily particulate matter concentration (in $\mu\text{g/m}^3$ ) near an urban center over a period of $50$ days. The observations are summarized in the table below:

Particulate Matter ( $\mu\text{g/m}^3$ )	Number of Days ( $f$ )
$20 - 29$	$6$
$30 - 39$	$10$
$40 - 49$	$15$
$50 - 59$	$11$
$60 - 69$	$8$

Find the estimated mean particulate matter concentration, in $\mu\text{g/m}^3$ , for the 50-day period.

Show answer & explanation

Answer: 45.5

Answer

The estimated mean particulate matter concentration is

45.5\text{ }\mu\text{g/m}^3

.

To calculate the mean of grouped data, determine the midpoint (

x

) of each class interval:

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

. Multiply each midpoint by its frequency (

f

) to get the products

147

,

345

,

667.5

,

599.5

, and

516

. The sum of these products (

\sum fx

) is

2275

. Dividing

\sum fx

by the total frequency (

\sum f = 50

) yields the estimated mean of

45.5\text{ }\mu\text{g/m}^3

.

Step-by-Step Solution

1

Calculate the class midpoints (

x

) for each class interval

Midpoints are

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

.

Grouped data uses the midpoint of each interval to represent all values falling within that interval.

2

Compute the product of frequency and midpoint (

fx

) for each class

6 \times 24.5 = 147

,

10 \times 34.5 = 345

,

15 \times 44.5 = 667.5

,

11 \times 54.5 = 599.5

,

8 \times 64.5 = 516

.

This determines the estimated sum of data values within each group.

3

Sum all products

\sum fx

and total frequency

\sum f

\sum fx = 147 + 345 + 667.5 + 599.5 + 516 = 2275

and

\sum f = 50

.

These totals are required for the mean formula.

4

Calculate the mean concentration using

\bar{x} = \frac{\sum fx}{\sum f}

\bar{x} = \frac{2275}{50} = 45.5

.

Dividing the total estimated sum by the total number of days gives the estimated mean.

Key Concept

Calculation of Mean for Grouped Frequency Data

Question 63Question →

The grouped frequency table below shows the distribution of marks obtained by candidates in a Mathematics examination:

Class Interval	Frequency
10 – 19	15
20 – 29	25
30 – 39	$k$
40 – 49	20
50 �� 59	10

When this data is represented on a pie chart, the sector corresponding to the score range 30 – 39 has a central angle of $108^\circ$ . Based on a cumulative frequency curve (ogive) constructed for this distribution, what is the score corresponding to the 75th percentile ( $Q_3$ ) of the candidates?

42.0

42.5

41.5

44.5

Show answer & explanation

Answer: 42.0

Answer

42.0

The value 42.0 is obtained by first determining the missing frequency

k = 30

using the ratio

\frac{108^\circ}{360^\circ} = 0.3

, establishing

N = 100

. The 75th percentile position is at 75, which falls in the

40 – 49

class. Interpolating from the lower boundary of 39.5 yields

39.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 42.0

.

Step-by-Step Solution

1

Determine the unknown frequency

k

using the pie chart sector angle.

\frac{k}{\text{Total Frequency}} = \frac{108^\circ}{360^\circ} = 0.3

Total frequency

N = 15 + 25 + k + 20 + 10 = 70 + k

.

\frac{k}{70 + k} = 0.3 \implies k = 21 + 0.3k \implies 0.7k = 21 \implies k = 30

The sector angle in a pie chart is proportional to the category frequency relative to the total frequency across

360^\circ

.

2

Calculate cumulative frequencies and locate the 75th percentile position.

Total frequency

N = 100

.
Cumulative frequencies (

cf

):
-

10 – 19

(boundary

9.5 – 19.5

):

cf = 15

-

20 – 29

(boundary

19.5 – 29.5

):

cf = 40

-

30 – 39

(boundary

29.5 – 39.5

):

cf = 70

-

40 – 49

(boundary

39.5 – 49.5

):

cf = 90

-

50 – 59

(boundary

49.5 – 59.5

):

cf = 100

75th percentile position

= 0.75 \times 100 = 75\text{th candidate}

.

The 75th percentile corresponds to the value below which 75% of the total observations lie.

3

Apply linear interpolation on the percentile class interval

40 – 49

.

The 75th score lies in class interval

40 – 49

(boundaries

39.5 – 49.5

).
- Lower class boundary

L = 39.5

- Cumulative frequency prior to class

cf_b = 70

- Frequency of percentile class

f = 20

- Class width

c = 10

Q_3 = L + \left(\frac{0.75N - cf_b}{f}\right) \times c = 39.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 39.5 + 2.5 = 42.0

Linear interpolation along an ogive requires using exact class boundaries to compute specific percentile values.

Key Concept

Pie chart sector angles and linear interpolation on cumulative frequency distributions

Question 64Question →

The test scores of six students in a mathematics quiz are $12$ , $18$ , $25$ , $31$ , $47$ , and $50$ . What is the range of these scores?

38

50

30.5

28

Show answer & explanation

Answer: $38$

Answer

The range of the scores is

38

.

The range is a measure of dispersion defined as the difference between the maximum value (

50

) and the minimum value (

12

). Subtracting

12

from

50

gives

38

.

Step-by-Step Solution

1

Identify the maximum and minimum values in the dataset.

Maximum score =

50

, Minimum score =

12

.

Range measures the spread between the highest and lowest values in a distribution.

2

Calculate the range using the formula

\text{Range} = \text{Maximum} - \text{Minimum}

.

\text{Range} = 50 - 12 = 38

.

Subtracting the minimum value from the maximum value yields the measure of dispersion known as range.

Key Concept

Range of a Dataset

Question 65Question →

The mean score of $9$ students in a mathematics test was calculated as $12$ . It was later discovered that a score of $5$ was incorrectly recorded as $14$ , and an additional student's score of $31$ was omitted entirely. What is the correct mean score of all $10$ students?

13

13.9

14.4

14.8

Show answer & explanation

Answer: $13$

Answer

The correct mean score of all 10 students is

13

.

The original sum of

9

scores is

108

. Subtracting the incorrect value (

14

) and adding the true value (

5

) reduces the sum of the

9

scores to

99

. Adding the

10^{\text{th}}

score of

31

gives a total sum of

130

. Dividing

130

by

10

gives the correct mean of

13

.

Step-by-Step Solution

1

Calculate the initial total sum of the original 9 scores.

Initial sum

= 9 \times 12 = 108

.

Mean is defined as total sum divided by number of items, so total sum equals mean times count.

2

Adjust the total sum for the misread score.

Corrected sum of 9 scores

= 108 - 14 + 5 = 99

.

Subtract the incorrect value (

14

) and add the actual value (

5

).

3

Add the omitted 10th score to the sum.

New total sum

= 99 + 31 = 130

.

Including the omitted score increases the total score sum.

4

Divide the new total sum by the updated total number of students.

Correct mean

= \frac{130}{10} = 13

.

The total number of students increased from 9 to 10.

Key Concept

Correcting the mean of ungrouped data after data entry errors or additions

Estimated Time:1m 30s

Question 66Question →

The frequency distribution table below shows the daily rainfall (in mm) recorded across $80$ weather monitoring stations during a storm:

Daily Rainfall (mm)	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$14$
$30 - 39$	$26$
$40 - 49$	$20$
$50 - 59$	$12$

Using linear interpolation from the cumulative frequency distribution (ogive), what is the 75th percentile ( $P_{75}$ ) of the daily rainfall in mm?

Show answer & explanation

Answer: 45.5

Answer

The 75th percentile of the daily rainfall distribution is

45.5\text{ mm}

.

To find the 75th percentile (

P_{75}

), determine the

60^{\text{th}}

cumulative frequency position (

0.75 \times 80 = 60

). The value lies within the

40 - 49

class interval. Applying the lower class boundary

L = 39.5

, preceding cumulative frequency

cf_b = 48

, frequency

f = 20

, and class width

c = 10

, linear interpolation yields

P_{75} = 39.5 + \frac{60 - 48}{20} \times 10 = 45.5\text{ mm}

.

Step-by-Step Solution

1

Calculate cumulative frequencies across all class intervals.

Cumulative frequencies are

8

for

10-19

,

22

for

20-29

,

48

for

30-39

,

68

for

40-49

, and

80

for

50-59

. Total frequency

N = 80

.

Cumulative frequencies are necessary to locate percentile positions on an ogive.

2

Determine the position corresponding to the 75th percentile.

Position

= 0.75 \times 80 = 60^{\text{th}}

cumulative frequency item.

The 75th percentile represents

75\%

of the total sample size.

3

Identify the target class interval parameters containing the 60th observation.

The interval

40 - 49

contains cumulative frequencies from

49

to

68

. Parameters:

L = 39.5

,

c = 10

,

f = 20

,

cf_b = 48

.

Linear interpolation requires the exact boundaries and frequencies of the container class.

4

Apply the percentile interpolation formula

P_{75} = L + \left(\frac{60 - cf_b}{f}\right) \times c

.

P_{75} = 39.5 + \left(\frac{60 - 48}{20}\right) \times 10 = 39.5 + 6 = 45.5\text{ mm}

.

Calculates the exact rainfall value corresponding to the 75th percentile.

Key Concept

Linear interpolation for percentiles using cumulative frequency distribution (ogive)

Estimated Time:1m 30s

Question 67Question →

A library recorded the number of books borrowed over five consecutive days as $2$ , $4$ , $5$ , $6$ , and $8$ . What is the standard deviation of the number of books borrowed?

2

4

5

6

Show answer & explanation

Answer: $2$

Answer

The standard deviation of the number of books borrowed is

2

.

The mean of the five data points is

5

. The sum of the squared deviations from the mean is

9 + 1 + 0 + 1 + 9 = 20

. Dividing by

5

gives a variance of

4

. Taking the square root of

4

yields the standard deviation of

2

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the given dataset.

\bar{x} = \frac{2 + 4 + 5 + 6 + 8}{5} = \frac{25}{5} = 5

The mean is required to determine the deviations of each data point.

2

Find the deviations from the mean

(x - \bar{x})

and square each deviation

(x - \bar{x})^2

.

(2-5)^2 = 9, \quad (4-5)^2 = 1, \quad (5-5)^2 = 0, \quad (6-5)^2 = 1, \quad (8-5)^2 = 9

Squaring ensures all deviation values are positive before summation.

3

Calculate the variance (mean of squared deviations).

\text{Variance } (\sigma^2) = \frac{9 + 1 + 0 + 1 + 9}{5} = \frac{20}{5} = 4

Variance measures the average squared spread around the mean.

4

Take the square root of the variance to find the standard deviation.

\text{Standard Deviation } (\sigma) = \sqrt{4} = 2

Standard deviation expresses dispersion in the original unit of measurement.

Key Concept

Standard Deviation of Ungrouped Data

Question 68Question →

The table below shows the distribution of masses (in kg) of $50$ bags of cement inspected at a building construction site:

Mass (kg)	Frequency ( $f$ )
$40 - 44$	$5$
$45 - 49$	$12$
$50 - 54$	$18$
$55 - 59$	$15$

Calculate the mean mass of the bags of cement in kg.

Show answer & explanation

Answer: 51.3

Answer

The mean mass of the cement bags is 51.3 kg.

The mean mass of grouped data is computed using the formula

\bar{x} = \frac{\sum fx}{\sum f}

, where

x

represents the class midpoints and

f

represents the frequency of each class. The class midpoints are

42

,

47

,

52

, and

57

. Multiplying these midpoints by their respective frequencies yields

210

,

564

,

936

, and

855

. Summing these values gives

\sum fx = 2565

. Dividing by the total frequency

\sum f = 50

produces a mean mass of

\frac{2565}{50} = 51.3\text{ kg}

.

Step-by-Step Solution

1

Determine the midpoint (x) of each class interval

Midpoints are 42, 47, 52, and 57

For grouped data, the class midpoint represents the average value of all observations falling within that class interval.

2

Calculate the product of each midpoint and its corresponding frequency (fx)

Products are 210, 564, 936, and 855

Multiplying the midpoint by frequency gives the total estimated mass contributed by that class interval.

3

Sum all frequencies and all fx products

Total frequency sum = 50, Total product sum = 2565

These totals are required to calculate the weighted average across all intervals.

4

Divide the sum of fx by the total frequency

Mean = 51.3 kg

Applying the formula for grouped mean: Mean = (sum of fx) / (sum of f).

Key Concept

Grouped Data Mean Calculation

Question 69Question →

The table below shows the frequency distribution of marks obtained by $25$ students in a mathematics quiz:

Score ( $x$ )	1	3	5	7	9	11
Frequency ( $f$ )	2	$p$	6	$q$	3	2

If the mean score of the distribution is $5.48$ , what is the median score?

5

6

3

7

Show answer & explanation

Answer: 5

Answer

The median score is 5.

By using the total student count of 25 and the mean formula, we obtain the simultaneous equations

p + q = 12

and

3p + 7q = 56

, which yield

p = 7

and

q = 5

. Computing cumulative frequencies shows that items 1 to 2 have score 1, items 3 to 9 have score 3, and items 10 to 15 have score 5. The 13th item lies in this third group, so the median score is 5.

Step-by-Step Solution

1

Set up an equation for total frequency

2 + p + 6 + q + 3 + 2 = 25 \implies p + q + 13 = 25 \implies p + q = 12

The sum of all frequencies equals the total number of students (

25

).

2

Set up an equation for the mean score

\sum fx = 1(2) + 3(p) + 5(6) + 7(q) + 9(3) + 11(2) = 3p + 7q + 81.
\text{Mean} = \frac{3p + 7q + 81}{25} = 5.48 \implies 3p + 7q + 81 = 137 \implies 3p + 7q = 56.

The mean of an ungrouped frequency distribution is calculated using \bar{x} = \frac{\sum fx}{N}.

3

Solve the system of linear equations for

p

and

q

Substitute

p = 12 - q

into

3p + 7q = 56

:
3(12 - q) + 7q = 56 \implies 36 + 4q = 56 \implies 4q = 20 \implies q = 5.
Then

p = 12 - 5 = 7

.

Finding the missing frequencies is necessary to determine cumulative frequencies.

4

Determine the position and value of the median score

Position of median = \frac{N + 1}{2} = \frac{25 + 1}{2} = 13\text{th position}.
Cumulative frequencies:
- Score 1: 2
- Score 3: 2 + 7 = 9
- Score 5: 9 + 6 = 15
Since the 13th value lies in the cumulative frequency interval up to 15, the median score is 5.

The median of

N=25

items is the score corresponding to the

\frac{N+1}{2}

th item when ordered.

Key Concept

Measures of Central Tendency for Ungrouped Data

Question 70Question →

The table below shows the distribution of heights (in cm) of $20$ potted plants recorded during a biology experiment:

Height (cm)	Frequency ( $f$ )
10 – 14	4
15 – 19	6
20 – 24	8
25 – 29	2

What is the mean height of the potted plants?

17\text{ cm}

19\text{ cm}

19.5\text{ cm}

21\text{ cm}

Show answer & explanation

Answer: $19\text{ cm}$

Answer

The mean height of the potted plants is

19\text{ cm}

.

The mean height is calculated by multiplying each class interval's midpoint by its frequency, summing these products (

380

), and dividing by the total number of plants (

20

), giving

19\text{ cm}

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval.

Class midpoints are

12

,

17

,

22

, and

27

.

The midpoint represents the estimated mean value of data items in a grouped class interval.

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to find

f \cdot x

.

4 \times 12 = 48

,

6 \times 17 = 102

,

8 \times 22 = 176

, and

2 \times 27 = 54

.

This determines the total estimated sum of values for each class.

3

Sum all

f \cdot x

values and calculate the total frequency

\sum f

.

\sum fx = 48 + 102 + 176 + 54 = 380

and

\sum f = 4 + 6 + 8 + 2 = 20$.

These totals are required for the grouped mean formula.

4

Compute the mean using the formula

\bar{x} = \frac{\sum fx}{\sum f}

.

\bar{x} = \frac{380}{20} = 19\text{ cm}$.

Dividing total sum of values by total frequency yields the mean.

Key Concept

Grouped Mean Calculation using Class Midpoints

Estimated Time:1m 0s

Question 71Question →

Four different Mathematics textbooks and three different Physics textbooks are to be arranged in a line on a shelf. In how many distinct ways can the books be arranged if all four Mathematics textbooks must be kept together?

144

576

840

5,040

Show answer & explanation

Answer: 576

Answer

576 distinct ways

Treating the four Mathematics textbooks as a single unit gives 4 items to arrange on the shelf (the Mathematics unit and the three individual Physics textbooks). These 4 items can be arranged in

4! = 24

ways. Furthermore, the four distinct Mathematics textbooks within the unit can be arranged among themselves in

4! = 24

ways. By the multiplication principle of counting, the total number of distinct arrangements is

24 \times 24 = 576

.

Step-by-Step Solution

1

Group the Mathematics textbooks into a single block

1 Mathematics block + 3 individual Physics textbooks = 4 items to arrange.

Because all four Mathematics textbooks must remain together, they act as a single composite unit.

2

Calculate the arrangements of the 4 main items

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

There are 4 distinct items (the Mathematics block and 3 separate Physics books) to place in linear order.

3

Calculate internal permutations of the Mathematics textbooks inside their block

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

The 4 Mathematics textbooks can be arranged in different orders among themselves.

4

Apply the fundamental counting principle

Total arrangements =

24 \times 24 = 576

.

Multiply the number of block arrangements by the internal arrangements of the Mathematics textbooks.

Key Concept

Permutations with Restricted Grouping (Block Method)

Question 72Question →

The set of four numbers $\{x, 6, 8, 10\}$ has a variance of $5$ . Given that $x < 6$ , find the value of $x$ .

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

By expressing the mean as

\bar{x} = \frac{x+24}{4}

and setting the sum of squared deviations divided by 4 equal to 5, we arrive at the quadratic equation

x^2 - 16x + 48 = 0

. Solving this equation gives two possible values,

x = 4

and

x = 12

. Applying the restriction

x < 6

uniquely identifies

x = 4

as the correct solution.

Step-by-Step Solution

1

Calculate the mean of the dataset in terms of

x

\bar{x} = \frac{x + 24}{4}

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

2

Write the variance equation using the formula

\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}

\frac{(x - \bar{x})^2 + (6 - \bar{x})^2 + (8 - \bar{x})^2 + (10 - \bar{x})^2}{4} = 5

The given variance is 5 and the number of observations

N = 4

.

3

Substitute

\bar{x} = \frac{x + 24}{4}

and expand the squared deviations

\frac{3}{16}x^2 - 3x + 14 = 5 \implies x^2 - 16x + 48 = 0

Expanding the squared terms and multiplying through by

\frac{16}{3}

yields a standard quadratic equation.

4

Solve the quadratic equation

x^2 - 16x + 48 = 0

for

x

(x - 4)(x - 12) = 0 \implies x = 4 \text{ or } x = 12

Factoring the quadratic equation gives two possible roots.

5

Apply the given condition

x < 6

x = 4

Since

x < 6

, we reject

x = 12

and select

x = 4

.

Key Concept

Variance of Ungrouped Data and Algebraic Problem Solving

Question 73Question →

The frequency distribution table below shows the mass, in grams, of $50$ industrial steel bearings measured during a precision manufacturing audit:

Mass (g)	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$12$
$30 - 39$	$x$
$40 - 49$	$y$
$50 - 59$	$8$

If the mean mass of the bearings is $34.7\text{ g}$ , what is the value of the missing frequency $x$ ?

Show answer & explanation

Answer: 18

Answer

18

The value of

x

is 18 because setting up the total frequency sum gives

x + y = 25

, and using class midpoints to compute the mean yields

1915 - 10x = 1735

, which solves to

x = 18

.

Step-by-Step Solution

1

Express the relationship between the missing frequencies using total frequency.

x + y = 25

or

y = 25 - x

The total number of industrial steel bearings is 50, so

5 + 12 + x + y + 8 = 50

.

2

Determine the midpoint (

m

) of each class interval.

Midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5 respectively.

The class midpoint is calculated as

\frac{\text{lower limit} + \text{upper limit}}{2}

.

3

Formulate the equation for the sum of products of frequencies and midpoints.

\sum fm = 802.5 + 34.5x + 44.5y

Multiply each class midpoint by its corresponding frequency and sum the results.

4

Substitute

y = 25 - x

and solve for

x

using the mean formula.

x = 18

Setting

\frac{1915 - 10x}{50} = 34.7

yields

1915 - 10x = 1735

, which gives

10x = 180

and thus

x = 18

.

Key Concept

Measures of Central Tendency for Grouped Data - Mean with Unknown Frequencies

Estimated Time:3m 0s

Question 74Question →

A set of $7$ numbers: $8, 12, 14, 15, 17, 18, 21$ has a mean of $15$ . When two additional numbers, $x$ and $y$ (where $x < y$ ), are included in the dataset, the mean of all $9$ numbers becomes $17$ . Given that the mode of the $9$ numbers is $18$ , what is the median of the combined set of $9$ numbers?

15

17

18

24

Show answer & explanation

Answer: $17$

Answer

The median of the combined set of

9

numbers is

17

.

The sum of the original 7 numbers is 105. For 9 numbers with a mean of 17, the sum is 153, meaning the two new numbers add up to 48. Since 18 is the mode, it must appear more than once, so one of the new numbers is 18 and the other is 30. Arranging the 9 numbers in order yields 8, 12, 14, 15, 17, 18, 18, 21, 30. The middle (5th) term is 17.

Step-by-Step Solution

1

Calculate the sum of the original 7 numbers and the total sum required for 9 numbers

Sum of 7 numbers =

8 + 12 + 14 + 15 + 17 + 18 + 21 = 105

. Sum of 9 numbers =

9 \times 17 = 153

.

The mean formula

\text{Mean} = \frac{\sum x}{n}

gives total sum =

\text{Mean} \times n

.

2

Determine the values of the two added numbers

x

and

y

x + y = 153 - 105 = 48

. Since

18

is the mode,

x = 18

and

y = 30

.

The original set has all distinct numbers. For

18

to be the mode,

18

must repeat, so one of the added values must be

18

.

3

Order all 9 numbers and find the median

Ordered set:

8, 12, 14, 15, 17, 18, 18, 21, 30

. Median (

5^{\text{th}}

term) =

17

.

For an odd number of items

n = 9

, the median is the

\frac{n+1}{2} = 5^{\text{th}}

term in ascending order.

Key Concept

Combining mean, mode, and median properties for ungrouped data

Estimated Time:1m 30s

Question 75Question →

Five packages delivered by a courier service have masses of $5\text{ kg}$ , $8\text{ kg}$ , $11\text{ kg}$ , $12\text{ kg}$ , and $14\text{ kg}$ . What is the mean deviation of the masses of these packages?

0.0\text{ kg}

2.8\text{ kg}

10.0\text{ kg}

14.0\text{ kg}

Show answer & explanation

Answer: $2.8\text{ kg}$

Answer

2.8\text{ kg}

The mean of the data set is

10\text{ kg}

. The distances of each data value from the mean are

5

,

2

,

1

,

2

, and

4

. The sum of these distances is

14

, and dividing by

5

yields a mean deviation of

2.8\text{ kg}

.

Step-by-Step Solution

1

Calculate the arithmetic mean

(\bar{x})

of the dataset

\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10\text{ kg}

Mean deviation requires the central mean value as a reference point for all deviations.

2

Find the absolute deviation

|x - \bar{x}|

for each value in the dataset

|5 - 10| = 5, |8 - 10| = 2, |11 - 10| = 1, |12 - 10| = 2, |14 - 10| = 4

Mean deviation measures the average distance of values from the mean regardless of sign.

3

Sum the absolute deviations and divide by the sample size

(n = 5)

\text{Mean Deviation} = \frac{5 + 2 + 1 + 2 + 4}{5} = \frac{14}{5} = 2.8\text{ kg}

Dividing total absolute deviation by the total count yields the mean deviation.

Key Concept

Mean Deviation

Question 76Question →

The cumulative frequency distribution table below shows the completion times (in minutes) for $120$ software engineers during a technical skill test:

Time Interval (min)	Frequency ( $f$ )	Cumulative Frequency ( $cf$ )
$10 - 19$	$12$	$12$
$20 - 29$	$28$	$40$
$30 - 39$	$40$	$80$
$40 - 49$	$24$	$104$
$50 - 59$	$16$	$120$

Using linear interpolation from the cumulative frequency data, what is the $65^{\text{th}}$ percentile completion time?

39.0\text{ minutes}

29.0\text{ minutes}

39.5\text{ minutes}

34.5\text{ minutes}

Show answer & explanation

Answer: $39.0\text{ minutes}$

Answer

39.0\text{ minutes}

To find the

65^{\text{th}}

percentile (

P_{65}

), we calculate

\frac{65}{100} \times 120 = 78

. The cumulative frequency table shows that rank

78

falls within the class interval

30 - 39

. The true lower boundary for this class is

L = 29.5

, the cumulative frequency of the preceding class is

F = 40

, the frequency of the percentile class is

f = 40

, and the class width is

c = 10

. Substituting into

P_{65} = L + \left(\frac{78 - F}{f}\right)c

yields

29.5 + \left(\frac{38}{40}\right) \times 10 = 39.0\text{ minutes}

.

Step-by-Step Solution

1

Determine the rank of the

65^{\text{th}}

percentile (

P_{65}

).

Rank

= \frac{65}{100} \times 120 = 78

.

The percentile rank identifies the position of the data point within the total frequency

N = 120

.

2

Identify the percentile class and its boundaries.

Percentile class is

30 - 39

, with lower boundary

L = 29.5

, upper boundary

= 39.5

, and class width

c = 10

.

The cumulative frequency increases from

40

to

80

across this interval, containing rank

78

.

3

Apply the linear interpolation formula for percentiles on grouped data.

P_{65} = L + \left(\frac{\frac{65N}{100} - F}{f}\right) \times c = 29.5 + \left(\frac{78 - 40}{40}\right) \times 10 = 29.5 + 9.5 = 39.0\text{ minutes}

.

Where

F = 40

is the cumulative frequency prior to the class and

f = 40

is the frequency of the class.

Key Concept

Linear Interpolation of Percentiles from Cumulative Frequency Distributions

Question 77Question →

The table below presents the distribution of weekly expenditure (in thousands of Naira) of $20$ small-scale farmers in a rural community:

Weekly Expenditure (\text{N}'000)	Frequency ( $f$ )
10 – 19	3
20 – 29	7
30 – 39	6
40 – 49	4

What is the mean weekly expenditure of the farmers?

\text{N}25.5\text{ thousand}

\text{N}29.5\text{ thousand}

\text{N}30.0\text{ thousand}

\text{N}34.5\text{ thousand}

Show answer & explanation

Answer: $\text{N}30.0\text{ thousand}$

Answer

The mean weekly expenditure is

\text{N}30.0\text{ thousand}

(or

\text{N}30,000

).

The mean of grouped data is computed using the formula

\bar{x} = \frac{\sum fx}{\sum f}

. Finding the midpoints (

14.5, 24.5, 34.5, 44.5

), multiplying by their respective frequencies (

3, 7, 6, 4

), and dividing the sum (

600

) by the total frequency (

20

) yields

\text{N}30.0\text{ thousand}

.

Step-by-Step Solution

1

Calculate the midpoint (

x

) for each class interval.

Midpoints are:

14.5

for 10–19;

24.5

for 20–29;

34.5

for 30–39; and

44.5

for 40–49.

Grouped data mean calculations require representing each interval by its class mark (midpoint).

2

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to get

f \cdot x

, and sum these values.

\sum (f \cdot x) = (3 \times 14.5) + (7 \times 24.5) + (6 \times 34.5) + (4 \times 44.5) = 43.5 + 171.5 + 207.0 + 178.0 = 600

.

This yields the total estimated sum of all expenditures across all observations.

3

Divide the total sum

\sum (f \cdot x)

by the total frequency

\sum f

.

Mean

\bar{x} = \frac{600}{20} = 30.0

.

The mean formula for grouped data is

\bar{x} = \frac{\sum fx}{\sum f}

.

Key Concept

Calculation of Mean for Grouped Frequency Data

Estimated Time:1m 0s

Question 78Question →

The frequency distribution table below shows the number of books checked out daily at a public library over a period of $40$ days:

Number of Books	Number of Days ( $f$ )
$1 - 5$	$4$
$6 - 10$	$10$
$11 - 15$	$16$
$16 - 20$	$10$

What is the mean number of books checked out per day?

10.0

10.5

12.0

14.0

Show answer & explanation

Answer: $12.0$

Answer

The mean number of books checked out per day is

12.0

.

The correct answer of

12.0

is found by calculating the midpoint (

x

) for each class interval (

3, 8, 13, 18

), multiplying each midpoint by its respective frequency to find

f \cdot x

, summing these products to obtain

\sum fx = 480

, and dividing by the total frequency

\sum f = 40

.

Step-by-Step Solution

1

Determine the midpoint (

x

) for each class interval.

Midpoints:

x_1 = \frac{1+5}{2} = 3

,

x_2 = \frac{6+10}{2} = 8

,

x_3 = \frac{11+15}{2} = 13

,

x_4 = \frac{16+20}{2} = 18

.

Grouped data calculations require a single representative value (the midpoint) for each class interval.

2

Calculate the product of frequency and midpoint (

f \cdot x

) for each class interval.

f_1 x_1 = 4 \times 3 = 12

,

f_2 x_2 = 10 \times 8 = 80

,

f_3 x_3 = 16 \times 13 = 208

,

f_4 x_4 = 10 \times 18 = 180

.

This yields the total value contributed by each class interval.

3

Calculate total frequency (

\sum f

) and total sum of products (

\sum fx

).

\sum f = 4 + 10 + 16 + 10 = 40

, and

\sum fx = 12 + 80 + 208 + 180 = 480

.

These sums are needed to compute the weighted mean.

4

Apply the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f}

.

\bar{x} = \frac{480}{40} = 12.0

.

Dividing the sum of all values by the total frequency yields the grouped mean.

Key Concept

Grouped Mean Calculation

Question 79Question →

The table below displays the distribution of scores obtained by $50$ candidates in a Mathematics assessment test:

Score Interval	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$12$
$30 - 39$	$18$
$40 - 49$	$7$
$50 - 59$	$5$

Match each statistical feature of the charts representing this data on the left with its corresponding numerical value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Sector angle for the modal class matches

129.6^\circ

; Frequency density of the

30 - 39

class interval matches

1.8

; Lower class boundary of the median class matches

29.5

; Cumulative frequency up to

29.5

matches

20

.

Each chart feature correctly aligns with its mathematically derived value: the modal class sector angle is

\frac{18}{50} \times 360^\circ = 129.6^\circ

, the frequency density is

\frac{18}{10} = 1.8

, the median class lower boundary is

29.5

, and the cumulative frequency up to

29.5

is

8 + 12 = 20

.

Step-by-Step Solution

1

Determine the total frequency and locate the modal and median classes.

Total frequency

N = 8 + 12 + 18 + 7 + 5 = 50

. The modal class is

30 - 39

(highest frequency =

18

). The median position is

\frac{50}{2} = 25^{\text{th}}

, which falls within the

30 - 39

class interval since cumulative frequency reaches

38

at the end of this class.

Identifying

N

, the modal class, and the median position is required for calculating chart parameters.

2

Calculate the pie chart sector angle for the modal class (

30 - 39

).

Sector angle

= \frac{\text{Frequency}}{N} \times 360^\circ = \frac{18}{50} \times 360^\circ = 129.6^\circ

.

Pie chart sectors represent relative frequencies scaled to

360^\circ

.

3

Compute the frequency density for the histogram bar of class

30 - 39

.

Class width

= 39.5 - 29.5 = 10

. Frequency density

= \frac{\text{Frequency}}{\text{Class width}} = \frac{18}{10} = 1.8

.

Histogram height represents frequency density, defined as frequency divided by class width.

4

Identify the lower class boundary of the median class (

30 - 39

) and cumulative frequency at upper boundary

29.5

.

Lower boundary of

30 - 39

is

30 - 0.5 = 29.5

. Cumulative frequency up to

29.5

is

8 + 12 = 20

.

Class boundaries eliminate gaps between intervals for continuous plots like ogives and histograms.

Key Concept

Calculating statistical chart parameters (pie chart sector angles, histogram frequency densities, class boundaries, and cumulative frequencies) from grouped frequency distributions.

Question 80Question →

The table below records the daily water consumption, in liters, of $40$ households in a residential community:

Daily Water Consumption (liters)	Frequency ( $f$ )
$10 - 19$	$6$
$20 - 29$	$10$
$30 - 39$	$14$
$40 - 49$	$7$
$50 - 59$	$3$

Calculate the mean daily water consumption for this community in liters.

Show answer & explanation

Answer: 32.25

Answer

The mean daily water consumption is

32.25

liters.

The mean of a grouped frequency distribution is computed using the formula

\bar{x} = \frac{\sum f x}{\sum f}

, where

x

represents the midpoint of each class interval and

f

is the class frequency. Calculating the midpoints yields

14.5, 24.5, 34.5, 44.5, 54.5

. Multiplying each midpoint by its frequency gives products of

87, 245, 483, 311.5, 163.5

, which sum to

1290

. Dividing

1290

by the total frequency of

40

yields

32.25

liters.

Step-by-Step Solution

1

Find the class midpoint (

x

) for each interval by taking the average of the upper and lower limits of each class.

Midpoints are

14.5

,

24.5

,

34.5

,

44.5

, and

54.5

.

For grouped data, the midpoint serves as the representative value for all data within that class interval.

2

Compute the product of frequency and midpoint (

f \cdot x

) for each class interval.

Products are

87

,

245

,

483

,

311.5

, and

163.5

.

This accounts for the total sum contributed by each group.

3

Sum all products

\sum f x

and divide by the total number of households

\sum f = 40

.

\text{Mean} = \frac{1290}{40} = 32.25.

The formula for the estimated mean of grouped data is

\bar{x} = \frac{\sum f x}{\sum f}

.

Key Concept

Grouped Mean Estimation using Class Midpoints

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