SPM Mathematics Trial 2021 (Kelantan), Paper 2 (Question 11)

Question 11:
The mean for the set number k, k + 4, k + 8, 3k – 4, 2k + 3, k – 1 and 4k is 7
(a) Find the value of k [3 marks]
(b) Find the interquartile range [2 marks]
(c) Find the standard deviation [3 marks]


Solution:
(a)

$\begin{array}{l}\text{Mean, }\overline{x}=\frac{{\displaystyle \sum x}}{\text{N}}\\ \text{Given mean = 7}\\ \text{7}=\frac{k+k+4+k+8+3k-4+2k+3+k-1+4k}{\text{7}}\\ 49=13k+10\\ 13k=39\\ k=\frac{\text{39}}{\text{13}}\\ =3\end{array}$


(b)
Interquartile range = Third quartile (Q₃) – first quartile (Q₁)

Substitute k into each of the set number,
3, 3 + 4, 3 + 8, 3(3) – 4, 2(3) + 3, 3 – 1 and 4(3)
3, 7, 11, 5, 9, 2 and 12

Arrange the set number in ascending order,
2, 3 (Q₁), 5, 7 (Q₂), 9, 11 (Q₃), 12

Thus, the interquartile range = 11 – 3 = 8


(c)

$\begin{array}{}\begin{array}{l}\text{Standard deviation, }\sigma =\sqrt{\frac{{\displaystyle \sum x^2}}{\text{N}}-{\left(\overline{x}\right)}^2}\\ \\ =\sqrt{\frac{2^2+3^2+5^2+7^2+9^2+{11}^2+{12}^2}{\text{7}}-{\left(7\right)}^2}\\ \\ =\sqrt{\frac{433}{7}-49}\\ \\ =3.586\end{array}\end{array}$