7.4.3 Measures of Dispersion for Grouped Data, SPM Paper (Long Questions)

Question 5:

Time (seconds)	Frequency
35 – 39	5
40 – 44	8
45 – 49	9
50 – 54	15
55 – 59	11
60 – 64	2

Solution:
(a) Modal class = time 50 – 54 seconds (highest frequency).

(b)

$\begin{array}{l}\text{Estimated mean}\\ \text{=}\frac{37\times 5+42\times 8+47\times 9+52\times 15+57\times 11+62\times 2}{50}\\ \text{=}\frac{2475}{50}=49.5\end{array}$

(c)
 

Upper Boundary	Cumulative Frequency
34.5	0
39.5	5
44.5	13
49.5	22
54.5	37
59.5	48
64.5	50

(d) 

Question 6:
The table below shows the frequency distribution of the age of 100 scouts in a camping spot.


(a) Based on table above, complete table in the answer space.

(b) Calculate the estimated age, in years, of a scout in the camping spot.

(c) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 10 scouts on the vertical axis, draw an ogive for the data.


(d) Any scout which is older than 18 years old is considered senior scout.
Based on the ogive drawn in part (c), find the percentage of senior scout in the camping spot.

Answer:




Solution:
(a)


(b)
$\begin{array}{l}\text{Estimated mean age}\\ \text{= }\frac{0\times 2+12\times 7+18\times 12+26\times 17+25\times 22+13\times 27+6\times 32}{100}\\ \text{= }\frac{1835}{100}\\ =18.35\text{ years old}\end{array}$

(c)



(d)
$\begin{array}{l}\text{Number of senior scouts}\\ =100-49\\ =51\\ \text{Percentage of senior scouts}\\ =\frac{51}{100}\times 100\%\\ =51\%\end{array}$