Students at Francis Marion University measured the mass of each M\&M candy insets ofcandies and insets ofcandies.

(a) Find the mean of the 16values on the left side of the table and the mean of the 16values on the right side of the table.

(b) Find the standard deviation of the 16mean values at the left side of the table and of the 16mean values at the right side of the table.

(c) From the standard deviation of the mean values for sets of 4candies, predict what the standard deviation is expected to be for the sets of 16candies. Compare your prediction to the measured standard deviation in (b).

• The standard deviation of a set of numbers is the distance between them and the mean value of the set of number.
• In a group of numbers, the mean is the average or most frequent value.

(a)

The denominator will be the total number of samples, and the numerator will be the sum of all sample values:

For left table

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _i}{x}_i}{n} \\ =\frac{\left(0.8799+0.9356+..........+0.8701+0.9138\right)g}{16} \\ =\frac{14.2432g}{16} \\ =0.8902g \end{gathered}$$

For right table

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _i}{x}_i}{n} \\ =\frac{\left(0.9004+0.9152+......+0.8803+0.9055\right)g}{16} \\ =\frac{14.3438g}{16} \\ =0.89649g \end{gathered}$$

(b)

The sum of the squared amounts of sample values subtracted by the mean value will be our numerator within the radical sign for clarity.

Hence, solve the equation

For left table

$$\begin{gathered} s_4=\sqrt{\frac{{\displaystyle \sum _i}{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(0.8799-0.8902\right)}^2+{\left(0.9356-0.8902\right)}^2+....+{\left(0.8701-0.8902\right)}^2+{\left(0.9138-0.8902\right)}^2}{15}}g \\ =\sqrt{7.75544\times {10}^{-4}}g \\ =2.7849\times {10}^{-2}g \\ =2.785\times {10}^{-2}g \end{gathered}$$

For right table

$$\begin{gathered} s_{16}=\sqrt{\frac{{\displaystyle \sum _i}{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(0.9004-0.89649\right)}^2+{\left(0.9152-0.89649\right)}^2+....+{\left(0.8803-0.89649\right)}^2+{\left(0.9055-0.89649\right)}^2}{15}}g \\ =\sqrt{1.25328025\times {10}^{-4}}g \\ =1.1195\times {10}^{-2}g \end{gathered}$$

(c)

The formula for predicted standard deviation${\sigma }_n$ ,

${\sigma }_n=\frac{\overline{x}}{\sqrt{n}}$

As a result, assume that the sample mean is equal to the population mean.

As a result of the replacement of values, arrive to the following equation:

Predicted Quotient:

$$\begin{gathered} \frac{{\sigma }_{16}}{{\sigma }_4}=\frac{\left({\displaystyle \frac{0.89649g}{\sqrt{16}}}\right)}{\left({\displaystyle \frac{0.89020g}{\sqrt{4}}}\right)} \\ =\frac{0.2241225}{0.4451} \\ =0.503532914 \\ =0.5 \end{gathered}$$

Observed Quotient:

$$\begin{gathered} \frac{s_{16}}{s_4}=\frac{1.1195\times {10}^{-2}g}{2.785\times {10}^{-2}g} \\ =0.429 \end{gathered}$$