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

Question 3:

The mean of five numbers is

\sqrt{p}

. The sum of the squares of the numbers is 120 and the standard deviation is 2q. Express p in terms of q.

Solution:

\begin{array}{l} Mean \bar{x} = \frac{\sum x}{N} \\ \sqrt{p} = \frac{\sum x}{5} \\ \sum x = 5 \sqrt{p} \\ Standard deviation, σ = \sqrt{\frac{\sum x^{2}}{N} - {\bar{x}}^{2}} \\ 2 q = \sqrt{\frac{120}{5} - {(\sqrt{p})}^{2}} \\ 4 q^{2} = 24 - p \\ p = 24 - 4 q^{2} \end{array}

Question 4:

A set of positive integers consists of 1, 4 and p. The variance for this set of integers is 6. Find the value of p.

Solution:

\begin{array}{l} Variance, σ^{2} = \frac{\sum x^{2}}{N} - {\bar{x}}^{2} \\ 6 = \frac{1^{2} + 4^{2} + p^{2}}{3} - {(\frac{1 + 4 + p}{3})}^{2} \\ 6 = \frac{17 + p^{2}}{3} - {(\frac{5 + p}{3})}^{2} \\ 6 = \frac{17 + p^{2}}{3} - [\frac{25 + 10 p + p^{2}}{9}] \\ 6 = \frac{51 + 3 p^{2} - 25 - 10 p - p^{2}}{9} \\ 2 p^{2} - 10 p + 26 = 54 \\ 2 p^{2} - 10 p + 28 = 0 \\ p^{2} - 5 p + 14 = 0 \\ (p - 7) (p + 2) = 0 \\ p = - 2 (not accepted) \\ ∴ p = 7 \end{array}