The ten numbers \({x_1},{\text{ }}{x_2},{\text{ }} \ldots ,{\text{ }}{x_{10}}\) have a mean of 10 and a standard deviation of 3.

Find the value of \(\sum\limits_{i = 1}^{10} {{{({x_i} - 12)}^2}} \).

EITHER

let \({y_i} = {x_i} - 12\)

\(\bar x = 10 \Rightarrow \bar y = - 2\)     M1A1

\({\sigma _x} = {\sigma _y} = 3\)     A1

\(\frac{{\sum\limits_{i = 1}^{10} {y_i^2} }}{{10}} - {{\bar y}^2} = 9\)     M1A1

\(\sum\limits_{i = 1}^{10} {y_i^2} = 10(9 + 4) = 130\)     A1

OR

\(\sum\limits_{i = 1}^{10} {{{({x_i} - 12)}^2} = \sum\limits_{i = 1}^{10} {x_i^2 - 24\sum\limits_{i = 1}^{10} {{x_i} + 144\sum\limits_{i = 1}^{10} 1 } } } \)     M1A1

\(\bar x = 10 \Rightarrow \sum\limits_{i = 1}^{10} {{x_i} = 100} \)     A1

\({\sigma _x} = 3,{\text{ }}\frac{{\sum\limits_{i = 1}^{10} {x_i^2} }}{{10}} - {{\bar x}^2} = 9\)     (M1)

\( \Rightarrow \sum\limits_{i = 1}^{10} {x_i^2} = 10(9 + 100)\)     A1

\(\sum\limits_{i = 1}^{10} {{{({x_i} - 12)}^2} = 1090 - 2400 + 1440 = 130} \)     A1

[6 marks]

Very few candidates answered this question well, but among those a variety of nice approaches were seen. Most candidates though revealed an inability to deal with sigma expressions, especially \(\sum\limits_{i = 1}^{i = 10} {144} \). Some tried to use expectation algebra but could not then relate those results to sigma expressions (often the factor 10 was forgotten). In a few cases candidates attempted to show the result using particular examples.