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

Question 1:

Given that the standard deviation of five numbers is 6 and the sum of the squares of these five numbers is 260. Find the mean of this set of numbers.

Solution:

\begin{array}{l} Given that σ = 6, Σ x^{2} = 260. \\ σ = 6 \\ \sqrt{\frac{Σ x^{2}}{n} - {\bar{X}}^{2}} = 6 \\ \frac{Σ x^{2}}{n} - {\bar{X}}^{2} = 36 \\ \frac{260}{5} - {\bar{X}}^{2} = 36 \\ {\bar{X}}^{2} = 16 \\ \bar{X} = \pm 4 \\ ∴ mean = \pm 4 \end{array}

Question 2:
Both of the mean and the standard deviation of 1, 3, 7, 15, m and n are 6. Find

(a) the value of m + n,

(b) the possible values of n.

Solution:

(a)

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

1 + 3 + 7 + 15 + m + n= 36

26 + m + n= 36

m + n = 10

(b)

\begin{array}{l} σ = 6 \\ σ^{2} = 36 \\ \frac{Σ x^{2}}{n} - {\bar{X}}^{2} = 36 \\ \frac{1 + 9 + 49 + 225 + m^{2} + n^{2}}{6} - 6^{2} = 36 \\ \frac{284 + m^{2} + n^{2}}{6} - 36 = 36 \\ \frac{284 + m^{2} + n^{2}}{6} = 72 \\ 284 + m^{2} + n^{2} = 432 \\ m^{2} + n^{2} = 148 \\ From (a), m = 10 - n \\ {(10 - n)}^{2} + n^{2} = 148 \\ 100 - 20 n + n^{2} + n^{2} = 148 \\ 2 n^{2} - 20 n - 48 = 0 \\ n^{2} - 10 n - 24 = 0 \\ (n - 6) (n + 4) = 0 \\ n = 6 or - 4 \end{array}