The stem-and-leaf diagram shows the test scores of the boys and girls from a particular class. [Stem-and-leaf diagram printed with two columns labelled "Boys" (left) and "Girls" (right) sharing a central column of stems. Key (Boys) 8 | 3 means 38. Key (Girls) 4 | 1 means 41.]

1. Write the boys' scores in ascending order from the stem-and-leaf diagram.

From the boys' side of the stem-and-leaf diagram, the scores are
$30,\ 32,\ 34,\ 34,\ 38,\ 47,\ 52,\ 54,\ 65,\ 68,\ 85.$

2. Find the median value, which is labelled $b$ on the box-and-whisker plot.

There are $11$ boys' scores.

The median is the $\frac{11+1}{2}=6$th value.

$b=47$

3. Find the lower quartile $Q_1$ and upper quartile $Q_3$, labelled $a$ and $c$.

The lower half is
$30,\ 32,\ 34,\ 34,\ 38,$
so the lower quartile is the middle value:
$a=Q_1=34.$

The upper half is
$52,\ 54,\ 65,\ 68,\ 85,$
so the upper quartile is the middle value:
$c=Q_3=65.$

1. Use the median condition after one more score is added.

The original boys' scores in ascending order are

$30,32,34,34,38,47,52,54,65,68,85.$

After one more boy's score is added, there will be $12$ scores, so the median will be

$\frac{\text{6th score}+\text{7th score}}{2}.$

If the added score is $47$, the ordered scores include two $47$s, so the 6th and 7th scores are both $47$.

$\text{new median}=\frac{47+47}{2}=47$

Hence the median remains unchanged.

1. Justify why standard deviation is more accurate for measuring consistency.

Standard deviation takes into account every score in the data set when measuring spread from the mean. The interquartile range only measures the spread of the middle $50\%$ of the scores, ignoring the lower and upper quarters. Hence standard deviation gives a more accurate measure of the consistency of the class's test scores.