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 representi

The table below displays the distribution of scores obtained by 50 candidates in a Mathema… — JAMB UTME

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.

Sector angle for the modal class in a pie chart representation $129.6^\circ$
Frequency density of the $30 - 39$ class interval in a histogram $1.8$
Lower class boundary of the class interval containing the median score $29.5$
Cumulative frequency corresponding to the upper class boundary of $29.5$ on an ogive $20$

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

8 + 12 = 20

Step-by-Step Solution

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.

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

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.

Identify the lower class boundary of the median class (

30 - 39

) and cumulative frequency at upper boundary

29.5

Lower boundary of

30 - 39

30 - 0.5 = 29.5

. Cumulative frequency up to

29.5

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.

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