Chapter 10, Problem 5PT

To determine

To calculate mean and standard deviation

Answer to Problem 5PT

Mean is 3 and standard deviation is 3.68

Explanation of Solution

Given information:

Data Nate collected is,

1, 1, 2, 3, 4, 5, 12

Formula used:

Mean is,

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Standard deviation is,

  $sd=\sqrt{\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n}}$

Calculation:

For mean, substituting the value,

  $\overline{x}=\frac{1+1+2+3+4+5+12}{7}$

So,

  $\overline{x}=\frac{28}{7}$

On solving,

  $\overline{x}=4$

Hence, mean is 4

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{{(1-3)}^2+{(1-3)}^2+{(2-3)}^2+{(3-3)}^2+{(4-3)}^2+{(5-3)}^2+{(12-3)}^2}{7}}$

So,

  $sd=\sqrt{\frac{4+4+1+0+1+4+81}{7}}$

On solving,

  $sd=\sqrt{13.57}$

On solving,

  $sd=6.68$

Hence, standard deviation is 6.68