The masses of 20 bags, in kg, are measured.
The results are shown on the stem-and-leaf diagram.

1. List the data values and calculate the mean

The masses are
$0.9,\ 1.2,\ 2.0,\ 2.6,\ 2.6,\ 2.6,\ 3.0,\ 3.3,\ 3.3,\ 3.4,\ 3.4,\ 3.8,\ 3.8,\ 4.1,\ 4.2,\ 4.9,\ 4.9,\ 4.9,\ 5.0,\ 5.1$.

$\text{Mean}=\frac{0.9+1.2+2.0+2.6+2.6+2.6+3.0+3.3+3.3+3.4+3.4+3.8+3.8+4.1+4.2+4.9+4.9+4.9+5.0+5.1}{20}$

$=\frac{66.2}{20}=3.31$

1. Calculate the standard deviation from the listed masses

Using $\bar{x}=3.31$ and $n=20$,

$\sum x^2=266.8$

$\text{standard deviation}=\sqrt{\frac{\sum x^2}{n}-\bar{x}^2}$

$=\sqrt{\frac{266.8}{20}-(3.31)^2}$

$=\sqrt{13.34-10.9561}$

$=1.5439\ldots$

So the standard deviation is $1.54\text{ kg}$, to 3 significant figures.

1. State the effect on the mean when every mass is increased by the same amount

Each actual mass is $0.08$ kg heavier, so every data value has $0.08$ added to it.

Therefore the mean also increases by $0.08$ kg.

New mean $=3.31+0.08=3.39$ kg.

2. State the effect on the standard deviation when every mass is increased by the same amount

Adding $0.08$ kg to every mass shifts all the data values by the same amount, so the distances of the values from their mean are unchanged.

Hence the standard deviation is unchanged.

New standard deviation $=1.54$ kg.