91影视

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

a.

Let\({y_i} = {x_i} - \bar x\)for\(i = 1,2,...,n\).

The sample mean of y is given as,

\(\begin{array}{c}\bar y &=& \frac{{\sum {{y_i}} }}{n}\\ &=& \frac{1}{n}\sum {{y_i}} \\ &=& \frac{1}{n}\sum {\left( {{x_i} - \bar x} \right)} \\ &=& \frac{1}{n}\sum {{x_i}} - \bar x\\ &=& \bar x - \bar x\\ &=& 0\end{array}\)

Therefore, the sample mean of y is 0.

The sample variance of y is computed as,

\(\begin{array}{c}s_y^2 &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{y_i} - \bar y} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{y_i} - 0} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{y_i}} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} \\ &=& s_x^2\end{array}\)

Thus, the sample variance of y is\(s_x^2\).

The sample standard deviation of y is given as,

\(\begin{array}{c}{s_y} &=& \sqrt {s_y^2} \\ &=& \sqrt {s_x^2} \\ &=& {s_x}\end{array}\)

Thus, the sample standard deviation of y is \({s_x}\).

b.

Let\({z_i} = \frac{{{x_i} - \bar x}}{s}\)for\(i = 1,2,...,n\).

The sample mean of z is given as,

\(\begin{array}{c}\bar z &=& \frac{{\sum {{z_i}} }}{n}\\ &=& \frac{1}{n}\sum {{z_i}} \\ &=& \frac{1}{n}\sum {\frac{{\left( {{x_i} - \bar x} \right)}}{{{s_x}}}} \\ &=& \frac{1}{{{s_x}}}\frac{{\sum {{x_i}} }}{n} - \bar x\\ &=& \frac{1}{{{s_x}}}\bar x - \bar x\\ &=& 0\end{array}\)

Therefore, the sample mean of z is 0.

The sample variance of z is computed as,

\(\begin{array}{c}s_z^2 &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{z_i} - \bar z} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{z_i} - 0} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{z_i}} \right)}^2}} \\ &=& \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {\frac{{{x_i} - \bar x}}{{{s_x}}}} \right)}^2}} \\ &=& \frac{1}{{s_x^2}}\left( {\frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} } \right)\\ &=& \frac{1}{{s_x^2}}*s_x^2\\ &=& 1\end{array}\)

Thus, the sample variance of z is 1.

The sample standard deviation of z is given as,

\(\begin{array}{c}{s_z} &=& \sqrt {s_z^2} \\ &=& \sqrt 1 \\ &=& 1\end{array}\)

Thus, the sample standard deviation of z is 1.