Respuesta :
Let [tex]\{x_1,x_2,\ldots,x_n\}[/tex] be the data set. The variance of the data is
[tex]s^2=\displaystyle\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}[/tex]
where [tex]\bar x[/tex] is the mean of the data. You have
[tex]\bar x=\displaystyle\sum_{i=1}^n\frac{x_i}n[/tex]
so if each data point was multiplied by 2.4, you have a "new" mean of
[tex]{\bar x}^*=\displaystyle\sum_{i=1}^n\frac{2.4x_i}n=2.4\sum_{i=1}^n\frac{x_i}n=2.4\bar x[/tex]
So in the variance formula, you have
[tex]{s^2}^*=\displaystyle\sum_{i=1}^n\frac{(2.4x_i-2.4\bar x)^2}{n-1}=2.4^2\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}=5.76s^2[/tex]
so the standard deviation of the new data set would be
[tex]s^*=\sqrt{{s^2}^*}=\sqrt{5.76s^2}=2.4s[/tex]
The original standard deviation was [tex]s=5[/tex] so the new one would be [tex]s^*=2.4\times5=12[/tex].
[tex]s^2=\displaystyle\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}[/tex]
where [tex]\bar x[/tex] is the mean of the data. You have
[tex]\bar x=\displaystyle\sum_{i=1}^n\frac{x_i}n[/tex]
so if each data point was multiplied by 2.4, you have a "new" mean of
[tex]{\bar x}^*=\displaystyle\sum_{i=1}^n\frac{2.4x_i}n=2.4\sum_{i=1}^n\frac{x_i}n=2.4\bar x[/tex]
So in the variance formula, you have
[tex]{s^2}^*=\displaystyle\sum_{i=1}^n\frac{(2.4x_i-2.4\bar x)^2}{n-1}=2.4^2\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}=5.76s^2[/tex]
so the standard deviation of the new data set would be
[tex]s^*=\sqrt{{s^2}^*}=\sqrt{5.76s^2}=2.4s[/tex]
The original standard deviation was [tex]s=5[/tex] so the new one would be [tex]s^*=2.4\times5=12[/tex].