Measures of Central Tendency Practice Questions — JAMB UTME

Question 1Question →

The daily agricultural output (in tonnes) of a cassava farm over five consecutive days was recorded as 12, 15, 18, 15, and 20. What is the mean daily output of the farm in tonnes?

Show answer & explanation

Answer: 16

Answer

The mean daily output of the farm is 16 tonnes.

The arithmetic mean is computed by adding all data values together (

12 + 15 + 18 + 15 + 20 = 80

) and dividing by the total number of data points (

5

), which gives

80 \div 5 = 16

tonnes.

Step-by-Step Solution

1

Calculate the total output by summing all daily figures.

12 + 15 + 18 + 15 + 20 = 80 tonnes

Finding the arithmetic mean requires calculating the aggregate total of all observations.

2

Divide the total output by the number of observations (days).

80 / 5 = 16 tonnes

The formula for the arithmetic mean of a sample is the sum of all values divided by the total number of values.

Key Concept

Arithmetic Mean of Ungrouped Data

Estimated Time:45s

Question 2Question →

The table below shows the distribution of weekly expenditures (in thousands of Naira, ₦'000) of a group of market traders:

Expenditure (₦'000)	Frequency ( $f$ )
10	5
20	8
30	$x$
40	4
50	3

If the mean weekly expenditure of the traders is ₦28,000 (represented as $28$ in the table units), find the value of $x$ .

Show answer & explanation

Answer: 20

Answer

The value of

x

is 20.

Using the arithmetic mean formula

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

, we set up the equation with the given mean of 28:

\frac{520 + 30x}{20 + x} = 28

. Cross-multiplying gives

520 + 30x = 560 + 28x

. Rearranging terms yields

2x = 40

, which gives

x = 20

.

Step-by-Step Solution

1

Sum all given frequencies including the unknown

x

to get the total frequency expression.

\sum f = 20 + x

The total number of observations is needed for the denominator of the arithmetic mean formula.

2

Multiply each expenditure value by its respective frequency and aggregate the terms.

\sum f x = 520 + 30x

The total weighted value of all expenditures is needed for the numerator of the mean formula.

3

Substitute the known mean value of 28 into the equation

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

and solve for

x

.

x = 20

Isolating

x

yields the exact missing frequency.

Key Concept

Finding a missing frequency from a frequency distribution given the arithmetic mean.

Question 3Question →

The table below shows the frequency distribution of prices (in ₦) for tubers of yam sold in an agricultural market:

Price (₦)	Frequency ( $f$ )
10 – 14	3
15 – 19	5
20 – 24	7
25 – 29	3
30 – 34	2

What is the mean price of a tuber of yam?

₦19

₦21

₦22

��23

Show answer & explanation

Answer: ₦21

Answer

The mean price of a tuber of yam is ₦21.

The mean price is calculated by multiplying each class interval midpoint by its frequency, adding these products together to obtain 420, and dividing by the total number of observations (20), giving ₦21.

Step-by-Step Solution

1

Determine the midpoint (

x

) for each price class interval.

Class 10–14:

x = 12

; Class 15–19:

x = 17

; Class 20–24:

x = 22

; Class 25–29:

x = 27

; Class 30–34:

x = 32

.

Midpoints represent the central value of grouped class intervals.

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to find

fx

, and sum all products (

\sum fx

).

\sum fx = (12 \times 3) + (17 \times 5) + (22 \times 7) + (27 \times 3) + (32 \times 2) = 36 + 85 + 154 + 81 + 64 = 420

.

Weighting midpoints by frequency gives the total aggregate value.

3

Sum the total frequencies (

\sum f

) and calculate the mean (

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

).

\sum f = 3 + 5 + 7 + 3 + 2 = 20

. Therefore,

\bar{x} = \frac{420}{20} = 21

.

The grouped mean is total aggregate value divided by total number of items.

Key Concept

Calculation of the Arithmetic Mean from Grouped Frequency Data

Question 4Question →

A sample survey of 20 small-scale enterprises in an industrial cluster recorded their daily profit (in thousands of Naira, $\text{₦'000}$ ) with the following frequency distribution:

Daily Profit ( $x$ in $\text{₦'000}$ )	Number of Enterprises ( $f$ )
10	3
15	5
20	7
25	3
30	2

What is the mean daily profit of these enterprises in thousands of Naira?

Show answer & explanation

Answer: 19

Answer

The mean daily profit of the enterprises is 19 thousand Naira.

The arithmetic mean for a frequency distribution is calculated using

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

. Multiplying each daily profit by its frequency yields a total sum of

380

. Dividing by the total frequency of

20

enterprises gives a mean daily profit of

19

thousand Naira.

Step-by-Step Solution

1

Multiply each profit level by its corresponding frequency to get the total profit contribution per group

f \cdot x

values are 30, 75, 140, 75, and 60

Each profit value must be weighted by how many enterprises earned that amount

2

Sum all weighted profit values to find total combined profit

\sum fx = 380

The sum of

f \cdot x

gives the grand total daily profit for all surveyed enterprises

3

Sum all frequencies to obtain total count of enterprises

\sum f = 20

The mean requires dividing total profit by total sample size

4

Divide total weighted profit by total number of enterprises

\bar{x} = \frac{380}{20} = 19

Formula for discrete grouped mean is

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

Key Concept

Calculation of Mean from a Discrete Frequency Distribution

Question 5Question →

The weekly milk yields (in liters) of six dairy cows on a commercial farm were recorded as follows: $8$ , $10$ , $14$ , $16$ , $22$ , and $26$ . What is the median weekly milk yield for the cows?

14 liters

15 liters

16 liters

17 liters

Show answer & explanation

Answer: 15 liters

Answer

15 liters

For an even number of observations (

n = 6

), the median is defined as the mean of the two middle values after arranging the dataset in order. The middle values in

8, 10, 14, 16, 22, 26

are

14

and

16

. Computing their average yields

\frac{14 + 16}{2} = 15

liters.

Step-by-Step Solution

1

Arrange the data in ascending order and count the number of observations

Ordered dataset:

8, 10, 14, 16, 22, 26

. Total observations (

n

) =

6

(an even number).

The median requires ordered data. For an even number of observations, the median is the average of the two central terms.

2

Identify the two central observations

The 3rd term is

14

and the 4th term is

16

.

The central positions correspond to

\frac{n}{2} = 3\text{rd}

and

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

values.

3

Calculate the arithmetic average of the two central terms

\text{Median} = \frac{14 + 16}{2} = 15

liters.

Taking the midpoint of the two central values gives the exact median of the dataset.

Key Concept

Calculation of median for an even number of ungrouped observations

Estimated Time:45s

Question 6Question →

The monthly revenue collection (in millions of Naira, $\text{₦'million}$ ) from six regional revenue offices of a state government was recorded as $14$ , $18$ , $12$ , $x$ , $22$ , and $16$ . If the arithmetic mean of the revenue collected across all six offices is $\text{₦}17\text{ million}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 20

Answer

The value of

x

is

20\text{ million Naira}

(or simply

20

).

The arithmetic mean is defined as

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

. For

n = 6

offices with a mean of

17\text{ million}

, the total sum must be

6 \times 17 = 102\text{ million}

. Summing the known values gives

14 + 18 + 12 + 22 + 16 = 82

. Subtracting

82

from

102

yields

x = 20\text{ million Naira}

.

Step-by-Step Solution

1

Calculate the sum of all observations in terms of

x

Sum

= 14 + 18 + 12 + x + 22 + 16 = 82 + x

The mean formula requires the total sum of all values divided by the number of observations.

2

Set up the mean equation using

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

\frac{82 + x}{6} = 17

$

The question specifies that the arithmetic mean across the 6 regional offices is 17.

3

Solve for the unknown value

x

82 + x = 102 \implies x = 20$

Subtracting the sum of the five known values (82) from the total required sum (102) yields the missing observation.

Key Concept

Calculation of a Missing Value from the Arithmetic Mean

Question 7Question →

The table below shows the distribution of monthly fuel consumption (in liters) for a sample of $50$ commercial transport vehicles operated by a logistics firm in Lagos:

Fuel Consumption (Liters)	Frequency ( $f$ )
10 – 19	8
20 – 29	13
30 – 39	16
40 – 49	9
50 – 59	4

Calculate the median fuel consumption (in liters) for this sample of vehicles.

Show answer & explanation

Answer: 32

Answer

The median fuel consumption for the sample of vehicles is

32

liters.

The median of a grouped frequency distribution is computed using continuous class boundaries. With

N=50

, the median position is

25

. The

30–39

class has lower boundary

29.5

, frequency

16

, preceding cumulative frequency

21

, and class interval

10

. Substituting these values gives

29.5 + \frac{25 - 21}{16} \times 10 = 32

liters.

Step-by-Step Solution

1

Calculate the total frequency (

N

) and construct cumulative frequencies (

cf

).

Total frequency

N = 8 + 13 + 16 + 9 + 4 = 50

. Cumulative frequencies are: 10–19 (

cf = 8

), 20–29 (

cf = 21

), 30–39 (

cf = 37

), 40–49 (

cf = 46

), 50–59 (

cf = 50

).

Cumulative frequencies are required to identify the class containing the median value.

2

Locate the median position and determine the median class parameters.

Position

= \frac{N}{2} = \frac{50}{2} = 25

. The

25^{\text{th}}

item falls in the

30 - 39

class. Lower boundary (

L

)

= 29.5

, preceding cumulative frequency (

cf_p

)

= 21

, median class frequency (

f_m

)

= 16

, class width (

c

)

= 10

.

The median class is the first class whose cumulative frequency meets or exceeds

N/2

.

3

Substitute the parameters into the grouped median formula.

\text{Median} = L + \left(\frac{\frac{N}{2} - cf_p}{f_m}\right) \times c = 29.5 + \left(\frac{25 - 21}{16}\right) \times 10 = 29.5 + \left(\frac{4}{16}\right) \times 10 = 29.5 + 2.5 = 32

.

Linear interpolation within the median class yields the precise median measurement.

Key Concept

Median of Grouped Data using Class Boundaries and Linear Interpolation

Estimated Time:2m 0s

Question 8Question →

The reported mean monthly wage of $50$ employees in a manufacturing firm was recorded as $42$ (in thousands of Naira, $\text{₦'000}$ ). During an internal audit, two transcription errors were discovered: a wage of $\text{₦}64,000$ was incorrectly recorded as $\text{₦}46,000$ , and a wage of $\text{₦}28,000$ was incorrectly recorded as $\text{₦}82,000$ . Additionally, $10$ new workers were recruited at an average monthly wage of $\text{₦}57,000$ (recorded as $57$ in $\text{₦'000}$ ). What is the corrected mean monthly wage (in $\text{₦'000}$ ) for the entire workforce of $60$ employees?

Show answer & explanation

Answer: 43.9

Answer

The corrected mean monthly wage for the entire workforce of 60 employees is 43.9 in thousands of Naira (₦'000), which represents ₦43,900.

The initial reported wage sum of ₦2,100,000 (

2100

in

\text{₦'000}

) is adjusted by adding

\text{₦}18,000

for the under-recorded entry and subtracting

\text{₦}54,000

for the over-recorded entry, giving a corrected baseline sum of

\text{₦}2,064,000

(

2064

in

\text{₦'000}

). Adding the

\text{₦}570,000

(

570

in

\text{₦'000}

) earned by the

10

new recruits gives a total aggregate wage sum of

\text{₦}2,634,000

(

2634

in

\text{₦'000}

) across

60

total employees. Dividing

2634

by

60

gives an exact corrected mean of

43.9

in

\text{₦'000}

(or

\text{₦}43,900

).

Step-by-Step Solution

1

Find initial reported total wage expenditure

50 × 42 = 2100 (in ₦'000)

Total value equals sample size multiplied by reported arithmetic mean.

2

Calculate net error adjustment

(64 - 46) + (28 - 82) = +18 - 54 = -36 (in ₦'000)

Under-recorded item adds +18, while over-recorded item subtracts -54.

3

Adjust initial total wage expenditure

2100 - 36 = 2064 (in ₦'000)

Correcting errors adjusts the sum of the original 50 workers' wages.

4

Calculate wage expenditure of new workers

10 × 57 = 570 (in ₦'000)

Total earnings of additional workers equal number of recruits times their mean wage.

5

Compute total combined expenditure and workforce size

Total sum = 2064 + 570 = 2634 (in ₦'000); Total N = 50 + 10 = 60

Combine corrected original wage sum with new recruitment total.

6

Calculate final corrected combined mean

2634 / 60 = 43.9 (in ₦'000)

Divide aggregate wage sum by aggregate total number of employees.

Key Concept

Corrected Mean and Weighted Combined Mean

Question 9Question →

The mean daily production cost of five small poultry farms in Ogun State is $\text{₦}45,000$ . If the daily production costs of four of the farms are $\text{₦}38,000$ , $\text{₦}42,000$ , $\text{₦}46,000$ , and $\text{₦}50,000$ , what is the daily production cost (in $\text{₦}$ ) of the fifth farm?

Show answer & explanation

Answer: 49000

Answer

The daily production cost of the fifth farm is

\text{₦}49,000

.

The arithmetic mean formula is

\bar{X} = \frac{\sum X}{N}

. For 5 farms with a mean of

\text{₦}45,000

, the total combined cost is

5 \times \text{₦}45,000 = \text{₦}225,000

. Summing the four given farm costs yields

\text{₦}38,000 + \text{₦}42,000 + \text{₦}46,000 + \text{₦}50,000 = \text{₦}176,000

. The cost for the fifth farm is

\text{₦}225,000 - \text{₦}176,000 = \text{₦}49,000

.

Step-by-Step Solution

1

Calculate total production cost of all 5 farms

Total cost =

\text{₦}225,000

Using the arithmetic mean formula

\bar{X} = \frac{\sum X}{N}

, the total sum

\sum X = N \times \bar{X} = 5 \times 45,000 = 225,000

.

2

Sum the daily production costs of the four known farms

Known total =

\text{₦}176,000

38,000 + 42,000 + 46,000 + 50,000 = 176,000

.

3

Determine the unknown fifth farm cost by subtraction

Fifth farm cost =

\text{₦}49,000

Subtracting the sum of the four known values from the overall total:

225,000 - 176,000 = 49,000

.

Key Concept

Calculating a missing value using the total sum property of the arithmetic mean

Estimated Time:1m 15s

Question 10Question →

The table below shows the frequency distribution of weekly cocoa bean output (in metric tonnes) produced by 40 agricultural cooperative societies in Southwestern Nigeria:

Output Range (Tonnes)	Number of Cooperatives ( $f$ )
$10 - 19$	$4$
$20 - 29$	$10$
$30 - 39$	$16$
$40 - 49$	$8$
$50 - 59$	$2$

What is the estimated median weekly cocoa output of the cooperatives?

33.25

tonnes

33.75

tonnes

32.88

tonnes

34.50

tonnes

Show answer & explanation

Answer: $33.25$ tonnes

Answer

The estimated median weekly cocoa output is

33.25

tonnes.

The option stating

33.25

tonnes is correct because the median class is

30 - 39

with lower boundary

L = 29.5

, cumulative frequency prior

F = 14

, class frequency

f_m = 16

, and class width

c = 10

. Substituting into

\text{Median} = L + \left(\frac{N/2 - F}{f_m}\right)c

yields

29.5 + \left(\frac{20 - 14}{16}\right) \times 10 = 33.25

.

Step-by-Step Solution

1

Determine the median position and identify the median class

Total frequency

N = 4 + 10 + 16 + 8 + 2 = 40

. Median position is

\frac{N}{2} = \frac{40}{2} = 20^{\text{th}}

item. Cumulative frequencies are:

10-19: 4

;

20-29: 14

;

30-39: 30

. The

20^{\text{th}}

item lies within the

30 - 39

interval.

The median class is the class interval containing the

\frac{N}{2}

position.

2

Identify the required statistical parameters for grouped median formula

Lower class boundary (

L

) =

29.5

; Cumulative frequency before median class (

F

) =

14

; Frequency of median class (

f_m

) =

16

; Class width (

c

) =

39.5 - 29.5 = 10

.

Grouped data estimation requires exact continuous boundaries rather than discrete limits.

3

Apply the grouped median interpolation formula

\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times c = 29.5 + \left(\frac{20 - 14}{16}\right) \times 10 = 29.5 + \left(\frac{6}{16}\right) \times 10 = 29.5 + 3.75 = 33.25

.

Interpolates the exact median position within the continuous interval of the median class.

Key Concept

Calculation of Median from Grouped Frequency Distribution

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