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.