SPM Mathematics Trial 2021 (Selangor), Paper 2 (Question 13)

Question 13:
Data in Table 1 shows the height, in cm, of a group of 10 volleyball players.

142	168	162	154	172
162	157	158	169	181

Table 1

(a) Determine 
(i) the first quartile
(ii) median
(iii) the third quartile
(iv) the interquartile range
[4 marks]


(b) Table 2.1 shows the number of words typed by a group of participants in a Shorthand and Secretarial Workshop within the duration of 5 minutes.

Number of words	Frequency
121 – 125	6
126 – 130	9
131 – 135	13
136 – 140	18
141 – 145	14
146 – 150	10

Table 2.1

Complete Table 2.2 in the answer space 13(b).
Calculate standard deviation of the data.
[6 marks]

Answer:
(b)

f	Midpoint	fx	x2	fx2
6	123
9	128
13	133
18	138
14	143
10	148
Σ f = 70		Σ fx =	Σ x2 =	Σ fx2 =

Solution:

142, 154, 157 (Q₁), 158, 162, | 162, 168, 169 (Q₃), 172, 183

(a)(i) the first quartile, Q₁ = 157

(a)(ii) $\text{median}=\frac{162+162}{2}=162$

(a)(iii) the third quartile, Q₃ = 169

(a)(iv) the interquartile range = Q₃ – Q₁ = 169 – 157 = 12


(b)

f	Midpoint	fx	x2	fx2
6	123	738	15 129	90 774
9	128	1 152	16 384	147 456
13	133	1 729	17 689	229 957
18	138	2 484	19 044	342 792
14	143	2 002	20 449	286 286
10	148	1 480	21 904	219 040
Σ f = 70		Σ fx = 9 585	Σ x2 = 110 599	Σ fx2 = 1 316 305

$\begin{array}{l}\text{Mean, }\overline{x}=\frac{\sum fx}{\sum f}\\ =\frac{9585}{70}\\ =136.9286\\ \\ \text{Standard deviation, }\sigma \\ =\sqrt{\frac{\sum f{x}^2}{\sum f}-{\left(\overline{x}\right)}^2}\\ =\sqrt{\frac{1316305}{70}-{\left(136.9286\right)}^2}\\ =\sqrt{54.9156}\\ =7.41\end{array}$