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A sample of size n is \({x_1},{x_2},...,{x_n}\).

\({\bar x_n}\) and \(s_n^2\) denote the sample mean and variance.

\({\bar x_{n + 1}}\) and \(s_{n + 1}^2\) represents these quantities when an additional observation \({x_{n + 1}}\) is added to the sample.

a.

We know,

\(\begin{array}{c}\sum\limits_{i = 1}^{n + 1} {{x_i}} &=& \sum\limits_{i = 1}^n {{x_i}} + {x_{n + 1}}\\ &=& n{{\bar x}_n} + {x_{n + 1}}\\\frac{{\sum\limits_{i = 1}^{n + 1} {{x_i}} }}{{n + 1}} &=& \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}\;\;\;\;\;\;\left( {{\rm{Dividing}}\;{\rm{by}}\;n + 1} \right)\\{{\bar x}_{n + 1}} &=& \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}\end{array}\)

Hence, showed.

b.

It is known that,

\(\begin{array}{c}ns_{n + 1}^2 &=& \sum\limits_{i = 1}^{n + 1} {{{\left( {{x_i} - {{\bar x}_{n + 1}}} \right)}^2}} \\ &=& \sum\limits_{i = 1}^{n + 1} {{{\left( {{x_i} - \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}} \right)}^2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{Referring}}\;{\rm{to}}\;{\rm{part}}\;{\rm{a}}} \right)\\ &=& \sum\limits_{i = 1}^{n + 1} {x_i^2 - } {\left( {n + 1} \right)^{ - 2}}{x_{n + 1}}\\ &=& \sum\limits_{i = 1}^{n + 1} {x_i^2 - } nx_n^{ - 2} + x_{n + 1}^2 + nx_n^{ - 2} - \left( {n + 1} \right)x_{n + 1}^{ - 2}\\ &=& \left( {n - 1} \right)s_n^2 + \left\{ {x_{n + 1}^2 + nx_n^{ - 2} - \left( {n + 1} \right)x_{n + 1}^{ - 2}} \right\}\\ &=& \left( {n - 1} \right)s_n^2 + \left\{ {x_{n + 1}^2 + nx_n^{ - 2} - nx_{n + 1}^{ - 2} - x_{n + 1}^{ - 2}} \right\}\\ &=& \left( {n - 1} \right)s_n^2 + {\left( {{x_{n + 1}} + nx_n^{ - 1} - nx_{n + 1}^{ - 1} - x_{n + 1}^{ - 1}} \right)^2}\\ &=& \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - \bar x} \right)^2}\end{array}\)

Therefore,

\(ns_{n + 1}^2 = \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - \bar x} \right)^2}\).

c.

The sample mean thread elongation of 15 strands is 12.58 mm.

The sample standard deviation of 15 strands is 0.512 mm.

The value of 16^th strand elongation value is 11.8.

Referring to parts (a) and (b),

The sample mean for all 16 elongation observations is computed as,

\(\begin{array}{c}{{\bar x}_{15 + 1}} &=& \frac{{15\left( {12.58} \right) + 11.8}}{{16}}\\ &=& \frac{{200.5}}{{16}}\\ &=& 12.53\end{array}\)

Thus, the sample mean for all 16 elongation observations is 12.53 mm.

The sample variance for all 16 elongation observations is computed as,

\(\begin{array}{c}s_{n + 1}^2 &=& \frac{{14}}{{15}}{\left( {0.512} \right)^2} + \frac{{{{\left( {11.8 - 12.58} \right)}^2}}}{{16}}\\ &=& 0.245 + 0.88\\ &=& 0.238\end{array}\)

The sample standard deviation for all 16 elongation observations is computed as,

\(\begin{array}{c}{s_{n + 1}} &=& \sqrt {s_{n + 1}^2} \\ &=& \sqrt {0.238} \\ &=& 0.532\end{array}\)

Therefore, the sample standard deviation for all 16 elongation observations is 0.532mm.