The temperatures of eighteen girls in a class on a particular day are shown in the stem-and-leaf diagram. (Stem-and-leaf diagram shown below and reproduced in figure_description.) Key: 37 | 1 represents $37.1^\circ\text{C}$

1. Use the given range to form and solve an equation for the unknown greatest temperature.

Smallest temperature $=35.4^\circ\text{C}$.

Greatest temperature $=37.x^\circ\text{C}$.

Since the range is $1.8^\circ\text{C}$,
$37.x-35.4=1.8$
$37.x=37.2$

Hence $x=2$.

1. List all 18 temperatures in ascending order using $x=2$.

Since $x=2$, the unknown temperature is $37.2^\circ\text{C}$.

The ordered data are
$35.4,\ 35.4,\ 35.7,\ 35.7,\ 35.9,\ 36.1,\ 36.2,\ 36.4,\ 36.4,\ 36.5,\ 36.6,\ 36.7,\ 36.7,\ 36.7,\ 36.8,\ 36.9,\ 37.1,\ 37.2.$

2. Find $a$, the minimum temperature.

The minimum value is the first value in the ordered list.

$a=35.4$

3. Find $c$, the median of 18 values.

There are $18$ values, so the median is the average of the 9th and 10th values.

The 9th value is $36.4$ and the 10th value is $36.5$.

$c=\frac{36.4+36.5}{2}=36.45$

4. Find $b$ and $d$, the lower and upper quartiles.

For $18$ values, split the data into the lower 9 values and the upper 9 values.

Lower half:
$35.4,\ 35.4,\ 35.7,\ 35.7,\ 35.9,\ 36.1,\ 36.2,\ 36.4,\ 36.4$
The middle value of the lower half is the 5th value, so
$b=35.9.$

Upper half:
$36.5,\ 36.6,\ 36.7,\ 36.7,\ 36.7,\ 36.8,\ 36.9,\ 37.1,\ 37.2$
The middle value of the upper half is the 5th value, so
$d=36.7.$

1. Calculate the interquartile range using the quartiles found earlier.

$\text{Interquartile range}=Q_3-Q_1=d-b$
$=36.7-35.9=0.8$

1. Calculate the mean of the 18 temperatures.

Using $x=2$, the total of the 18 temperatures is

$35.4+35.4+35.7+35.7+35.9+36.1+36.2+36.4+36.4+36.5+36.6+36.7+36.7+36.7+36.8+36.9+37.1+37.2=654.4$

$\text{Mean}=\frac{654.4}{18}=36.355\ldots\approx 36.4$

1. Calculate the standard deviation of the 18 temperatures using the mean $36.4^\circ\text{C}$.

Using $\bar{x}=36.4$ and the 18 temperatures,
$\sigma=\sqrt{\frac{\sum (x-\bar{x})^2}{18}}$
$=\sqrt{\frac{5.22}{18}}$
$=0.5385\ldots$
$\sigma\approx 0.539^\circ\text{C}$

1. Compare the mean temperatures and state the contextual conclusion.

Girls' mean temperature $=36.4^\circ\text{C}$.

Boys' mean temperature $=36.0^\circ\text{C}$.

Since $36.4>36.0$, the girls have a higher average temperature than the boys.

1. Compare the standard deviations of the girls and boys.

Girls' standard deviation $=0.539^\circ\text{C}$

Boys' standard deviation $=0.294^\circ\text{C}$

Since $0.539>0.294$, the girls have the larger standard deviation.

2. State the contextual conclusion about the spread of temperatures.

A larger standard deviation means the temperatures are more spread out from the mean.

Therefore, the girls' temperatures are more spread out / less consistent than the boys' temperatures.