A set of 15 observations has mean 11.5 and variance 9.3. One observation of 22.1 is considered unreliable and is removed. Find the mean and variance of the remaining 14 observations.

\(\frac{{\sum\limits_{i = 1}^{15} {{x_i}} }}{{15}} = 11.5 \Rightarrow \sum\limits_{i = 1}^{15} {{x_i} = 172.5} \)     (A1)

new mean \(= \frac{{172.5 - 22.1}}{{14}}\)     (M1)

= 10.7428… = 10.7 (3sf)     A1

\(\frac{{\sum\limits_{i = 1}^{15} {x_i^2} }}{{15}} - {11.5^2} = 9.3\)     (M1)

\( \Rightarrow \sum\limits_{i = 1}^{15} {x_i^2 = 2123.25} \)

new variance \(= \frac{{2123.25 - {{22.1}^2}}}{{14}} - {(10.7428...)^2}\)     (M1)

= 1.37 (3sf)     A1

[6 marks]

Most candidates were successful in finding the correct value of the mean; however, the variance caused many difficulties. Many candidates affirmed that there were no differences in the variance as it remained constant; some others got wrong results due to premature rounding of figures. Many candidates lost the final mark because they rounded their answers prematurely, resulting in a very inaccurate answer to this question.