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

1. Interpret the stem-and-leaf diagram and find the total mass.

Using the key, $0|9$ means $0.9$ kg, so the data values are
$0.9,\ 1.2,\ 2.0,\ 2.6,\ 2.6,\ 2.6,\ 3.0,\ 3.3,\ 3.3,\ 3.4,\ 3.4,\ 3.8,\ 3.8,\ 4.1,\ 4.2,\ 4.9,\ 4.9,\ 4.9,\ 5.0,\ 5.1$.

Total mass:
$0.9+1.2+2.0+2.6+2.6+2.6+3.0+3.3+3.3+3.4+3.4+3.8+3.8+4.1+4.2+4.9+4.9+4.9+5.0+5.1=69$

2. Calculate the mean mass.

There are $20$ bags.

$\text{Mean mass}=\frac{69}{20}=3.45$

1. Use the standard deviation formula with the given data and the mean from part (a).

From part (a), $\bar{x}=3.45$.

For the 20 masses,
$\sum x^2=266.8.$

So
$\text{standard deviation}=\sqrt{\frac{\sum x^2}{n}-\bar{x}^2}$
$=\sqrt{\frac{266.8}{20}-3.45^2}.$

2. Calculate and round the standard deviation.

$$\sqrt{\frac{266.8}{20}-3.45^2}= \sqrt{13.34-11.9025}$$

$=\sqrt{1.4375}=1.1989\ldots$

Hence, to 3 significant figures,
$\text{standard deviation}=1.20\text{ kg}.$

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

Each bag is actually $0.08\text{ kg}$ heavier, so every data value has $0.08$ added to it.

Hence the mean also increases by $0.08\text{ kg}$.

Using the mean from part (a), the new mean is
$3.45+0.08=3.53\text{ kg}$.

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

Adding $0.08\text{ kg}$ to every mass shifts all the data values by the same amount. The distances of the masses from their mean are unchanged.

Therefore, the standard deviation remains the same.