(a) Experiment X is carried out nine times identically, and the value 5is obtained all nine times. Calculate the mean by definition (4-12). Then the standard deviation by definition (4-13). (b) Experiment Yis carried out nine times identically, and the integers 1 through 9 are each obtained once. Repeat the calculations of part (a) for this experiment. (c) For nine repetitions of the experimentZ.The tally is that 1, 5, and 9 are each obtained three times. Repeat the calculations. (d) Explain any differences between the results in parts (b) and (c). Is standard deviation a reasonable measure of spread?

We can solve for both quantities by directly applying the mean and standard deviation definitions given by equations (1) and (2) respectively. Here, Q refers to the outcome of each measurement, and n refers to the number of these measurements.

$\overline{\mathbf{Q}}\mathbf{=}\frac{{\displaystyle \mathbf{\sum }_{\mathbf{i}}}{\mathbf{Q}}_{\mathbf{i}}{\mathbf{n}}_{\mathbf{i}}}{{\displaystyle \mathbf{\sum }_{\mathbf{i}}}{\mathbf{n}}_{\mathbf{i}}}$………………(1)

$\mathbf{∆}\mathit{Q}\mathbf{=}\sqrt{\frac{{\displaystyle \mathbf{\sum }_{\mathbf{i}}}{\left(Q_i-\overline{Q_i}\right)}^{\mathbf{2}}{\mathbf{n}}_{\mathbf{I}}}{{\displaystyle \mathbf{\sum }_{\mathbf{i}}}{\mathbf{n}}_{\mathbf{i}}}}$……………………(2)

Use equation (1) for average calculation, and we get

$\begin{array}{rcl}\overline{\mathrm{Q}}& =& \frac{{\displaystyle \sum _{\mathrm{i}}{\mathrm{Q}}_{\mathrm{i}}{\mathrm{n}}_{\mathrm{i}}}}{{\displaystyle \sum _{\mathrm{i}}{\mathrm{n}}_{\mathrm{i}}}}\\ & =& \frac{5\times 9}{9}\\ & =& 5\end{array}$

Because all measurements yield the same result of 5 , this value of the mean is expected. In terms of standard deviation using equation (2), we get,

$\begin{array}{rcl}∆Q& =& \sqrt{\frac{{\left(5-5\right)}^2\times 9}{9}}\\ & =& 0\end{array}$

We ended up with no departure from the mean, or a standard deviation of zero because all of the measurements yielded the same result.

(b)

Use equation (1) for average calculation as before, but this time with the results 1 to 9 was collected only once.

$\begin{array}{rcl}\overline{Q}& =& \frac{1+2+3+4+5+6+7+8+9}{9}\\ & =& 5\end{array}$

In terms of standard deviation, using equation (2), we get,

$\begin{array}{rcl}∆Q& =& \sqrt{\frac{{\left(1-5\right)}^2+{(2-5)}^2+{(3-5)}^2+{(4-5)}^2+{(5-5)}^2+{(6-5)}^2+{(7-5)}^2+{(8-5)}^2+{(9-5)}^2}{9}}\\ & =& \sqrt{\frac{60}{9}}\\ & =& 2.58\end{array}$

(c)

The results of 1, 5, and 9 were obtained three times using the equation (1) we get,

$$\begin{gathered} \overline{Q}=\frac{\left(1\times 3\right)+\left(5\times 3\right)+\left(9\times 3\right)}{9} \\ \overline{Q}=5 \end{gathered}$$

Now, utilize equation (2) for the estimation of standard deviation, such that,

$\begin{array}{rcl}∆Q& =& \sqrt{\frac{{(1-5)}^2\times 3+{(5-5)}^2\times 3+{(9-5)}^2\times 3}{9}}\\ ∆Q& =& \sqrt{\frac{96}{9}}\\ & =& 3.27\end{array}$

(d)

We can draw some very important conclusions from this problem.

From this problem, we can draw some very important conclusions. The mean of all these measurements turns out to be the same, emphasizing the importance of the standard deviation, which allows us to distinguish between three different experimental results with the same average.

The difference in standard deviation between parts b) and c) stems from the fact that the spread of data from the main in part c) is more pronounced than in part b). As in part b), the difference between the mean (5) and its nearest neighbors is 1 or 2 and no more than 4. This is not the case for part c), where the given data are both on the extremes of the distribution ( 4 points apart).