Soalan Percubaan Matematik SPM 2021 (Kelantan), Kertas 2 – Soalan 11

Soalan 11:
Min bagi set nombor k, k + 4, k + 8, 3k – 4, 2k + 3, k – 1 dan 4k ialah 7
(a) Hitung nilai k [3 markah]
(b) Cari julat antara kuartil [2 markah]
(c) Cari sisihan piawai [3 markah]


Penyelesaian:
(a)

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

(b)
Julat antara kuartil = kuartil ketiga (Q₃) – kuartil pertama (Q₁)

Gantikan nilai k ke dalam setiap nombor,
3, 3 + 4, 3 + 8, 3(3) – 4, 2(3) + 3, 3 – 1 dan 4(3)
3, 7, 11, 5, 9, 2 dan 12

Susun set nombor ikut tertib menaik,
2, 3 (Q₁), 5, 7 (Q₂), 9, 11 (Q₃), 12

Maka julat antara kuartil = 11 – 3 = 8

(c)

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