Consider the data set \(\{ 2,{\text{ }}x,{\text{ }}y,{\text{ }}10,{\text{ }}17\} ,{\text{ }}x,{\text{ }}y \in {\mathbb{Z}^ + }\) and \(x < y\).

The mean of the data set is \(8\) and its variance is \(27.6\).

Find the value of \(x\) and the value of \(y\).

use of \(\mu  = \frac{{\sum\limits_{i = 1}^k {{f_i}{x_i}} }}{n}{\text{ }}\) to obtain \(\frac{{2 + x + y + 10 + 17}}{5} = 8\)     (M1)

\(x + y = 11\)     A1

EITHER

use of \({\sigma ^2} = \frac{{\sum\limits_{i = 1}^k {{f_i}{{({x_i} - \mu )}^2}} }}{n}\) to obtain \(\frac{{{{( - 6)}^2} + {{(x - 8)}^2} + {{(y - 8)}^2} + {2^2} + {9^2}}}{5} = 27.6\)     (M1)

\({(x - 8)^2} + {(y - 8)^2} = 17\)     A1

OR

use of \({\sigma ^2} = \frac{{\sum\limits_{i = 1}^k {{f_i}x_i^2} }}{n} - {\mu ^2}\) to obtain \(\frac{{{2^2} + {x^2} + {y^2} + {{10}^2} + {{17}^2}}}{5} - {8^2} = 27.6\)     (M1)

\({x^2} + {y^2} = 65\)     A1

THEN

attempting to solve the two equations     (M1)

\(x = 4\;\;\;\)and\(\;\;\;y = 7\;\;\;({\text{only as }}x < y)\)A1     N4

Note:     Award A0 for \(x = 7\) and \(y = 4\).

Note:     Award (M1)A1(M0)A0(M1)A1 for \(x + y = 11 \Rightarrow x = 4\) and \(y = 7\).

[6 marks]

Reasonably well done. Most candidates were able to obtain \(x + y = 11\). Most manipulation errors occurred when candidates attempted to form the variance equation in terms of \(x\) and \(y\). Some candidates did not apply the condition \(x < y\) when determining their final answer.