8.4.4 Measures of Dispersion for Ungrouped Data, SPM Paper (Long Questions)

Question 7:
A set of data consists of twelve positive numbers.

It is given that Σ {(x - ¯ x)}^{2} = 600 and Σ x^{2} = 1032.

$It is given that Σ {(x - \bar{x})}^{2} = 600 and Σ x^{2} = 1032.$
Find
(a) the variance
(b) the mean

Solution:
(a)

\begin{matrix} Variance = \frac{Σ {(x - ¯ x)}^{2}}{N} = \frac{600}{12} = 50 \end{matrix}

$\begin{array}{l} Variance = \frac{Σ {(x - \bar{x})}^{2}}{N} \\ = \frac{600}{12} \\ = 50 \end{array}$

(b)

\begin{matrix} Variance = \frac{Σ x^{2}}{N} - {(¯ x)}^{2} 50 = \frac{1032}{12} - {(¯ x)}^{2} {(¯ x)}^{2} = 86 - 50 = 36 ¯ x = 36 \end{matrix}

$\begin{array}{l} Variance = \frac{Σ x^{2}}{N} - {(\bar{x})}^{2} \\ 50 = \frac{1032}{12} - {(\bar{x})}^{2} \\ {(\bar{x})}^{2} = 86 - 50 \\ = 36 \\ \bar{x} = 36 \end{array}$

Question 8:

A set of examination marks x₁, x₂, x₃, x₄, x₅, x₆ has a mean of 6 and a standard deviation of 2.4.

(a) Find

(i) the sum of the marks,

Σ x

$Σ x$ ,

(ii) the sum of the squares of the marks,

Σ x^{2}

$Σ x^{2}$ .

(b) Each mark is multiplied by 2 and then 3 is added to it.

Find, for the new set of marks,

(i) the mean,

(ii) the variance.

Solution:
(a)(i)

\begin{matrix} Given mean = 6 \frac{Σ x}{6} = 6 Σ x = 36 \end{matrix}

$\begin{array}{l} Given mean = 6 \\ \frac{Σ x}{6} = 6 \\ Σ x = 36 \end{array}$

(a)(ii)

\begin{matrix} Given σ = 2.4 σ^{2} = {2.4}^{2} \frac{Σ x^{2}}{n} - {¯ ¯ ¯ X}^{2} = 5.76 \frac{Σ x^{2}}{6} - 6^{2} = 5.76 \frac{Σ x^{2}}{6} = 41.76 Σ x^{2} = 250.56 \end{matrix}

$\begin{array}{l} Given σ = 2.4 \\ σ^{2} = {2.4}^{2} \\ \frac{Σ x^{2}}{n} - {\bar{X}}^{2} = 5.76 \\ \frac{Σ x^{2}}{6} - 6^{2} = 5.76 \\ \frac{Σ x^{2}}{6} = 41.76 \\ Σ x^{2} = 250.56 \end{array}$

(b)(i)

Mean of the new set of numbers

= 6(2) + 3

= 15

(b)(ii)

Variance of the original set of numbers

= 2.4²= 5.76

Variance of the new set of numbers

= 2² (5.76)

= 23.04

Leave a Comment Cancel reply