Calculate measures of central tendency, including the mean, median and mode, for both raw data and data presented in a frequency table

Mode: $1$ (it occurs five times, more than any other value).

Median: with $20$ values, average the $10$th and $11$th ordered values, which are $2$ and $2$, so the median is $2$.

Mean: the total is $0(3)+1(5)+2(4)+3(3)+4(2)+5+6+9 = 0+5+8+9+8+5+6+9 = 50$, and $50 \div 20 = 2.5$.

Markers award one mark for each correct measure. A common error is reading the median off the dot plot as the tallest column ($1$, the mode) instead of locating the middle position. The mean is higher than the median here because of the values $6$ and $9$ in the right tail, a point worth a one-line comment if the question asks about skew.

Class centres: $2, 6, 10, 14$ (each is the midpoint of its interval).

Mean: $\sum fx = 2(14)+6(20)+10(11)+14(5) = 28+120+110+70 = 328$, and $\sum f = 50$, so $\bar{x} = \frac{328}{50} = 6.56 \approx 6.6$ kg.

Why an estimate: grouping discards the exact values and replaces every parcel in a class with the class centre, so the calculation assumes the data is evenly spread within each interval, which is generally not exactly true.

Markers award a mark for the class centres, a mark for a correct $\sum fx$, a mark for the division, and a mark for a clear reason that the class centre stands in for the real values. Using the upper end of each interval instead of the centre is the most common error and loses the accuracy marks.

Part (a): the data must contain one or more very expensive sales (high outliers) that lift the mean. The mean adds in the actual dollar value of every sale, so a single multi-million-dollar property pulls it well above the bulk of the data, whereas the median only depends on the middle sale's position and is barely affected.

Part (b): the median of $$680\,000$ better represents a typical house price, because it is not distorted by the extreme high sales; the mean of $$1.2$ million overstates what most buyers actually paid.

Markers reward the identification of outlier/skew as the cause and a justified choice of the median, using the idea that the mean is sensitive to extreme values while the median is resistant. A bare answer with no reference to outliers earns little.

Part (a) - mean is the sum divided by how many. There are $5$ values, so add them and divide by $5$:

so the mean is $8$.

Part (b) - median is the middle of the ordered list. The data is already in order. With $5$ values the middle one is the $3$rd: $4, 7, \mathbf{7}, 9, 13$, so the median is $7$. (Check: the median $7$ sits inside the data range $4$ to $13$, as it must.)

Part (a) - mode is the most common value. The value $5$ appears twice and every other value appears once, so the mode is $5$.

Part (b) - median with an even number of values is the average of the middle two. The list is already ordered and has $6$ values, so the two middle ones are the $3$rd and $4$th, namely $5$ and $6$:

so the median is $5.5$. (Note the median need not be one of the data values when $n$ is even.)

Build the $fx$ column. Multiply each value by its frequency:

Add the columns. The total frequency is $\sum f = 5 + 8 + 4 + 3 = 20$ and the total of the $fx$ column is

Divide. The mean of a frequency table is

so the mean is $1.25$ pets. (Check: $\sum f = 20$ matches the $20$ students given, so no row was missed.)

Part (a) - mean. Add the seven values and divide by $7$:

so the mean profit is $$77\,000$.

Part (b) - median

The list is ordered and has $7$ values, so the median is the $4$th value: $48, 52, 55, \mathbf{55}, 58, 61, 210$, giving a median of $$55\,000$.

Part (c) - mode

The only repeated value is $55$, so the mode is $$55\,000$.

Part (d) - most appropriate measure

The value $210$ is an outlier - it sits far above the others and drags the mean up to $77$, which is larger than six of the seven businesses actually earned. The median of $$55\,000$ is the most appropriate measure of the typical profit, because it is not distorted by the one unusually high value. (Check: six of the seven profits are between $48$ and $61$, so a "typical" figure near $55$ is far more representative than $77$.)

Form the $fx$ column. Multiply each goal value by its frequency:

Total the two columns. The number of games is

and the total number of goals is

Divide. The mean is

so she averages $2$ goals per game. (Check: $\sum fx = 50$ is the total goals over $25$ games, and $50 \div 25 = 2$ is a sensible game average.)

Find each class centre. The class centre is the midpoint of the interval, the average of its lower and upper ends:

Build the $fx$ column using the class centre as $x$:

Total the columns.

Divide. The estimated mean is

so the mean waiting time is about $23.5$ minutes. (This is an estimate, because grouping replaces every value in a class with its centre. Check: $23.5$ falls in the $20\text{ to }30$ class, which holds the most customers, so it is sensible.)

Part (a) - the three measures for the original data. Add the seven values:

The list is ordered with $7$ values, so the median is the $4$th value: $12, 13, 15, \mathbf{15}, 16, 17, 17$, giving a median of $15$. The most common value is... $15$ and $17$ each appear twice, so the data is bimodal, with modes $15$ and $17$.

Part (b) - replace the largest $17$ with $50$. The new data set is $12, 13, 15, 15, 16, 17, 50$. The new mean is

The new ordered list is $12, 13, 15, \mathbf{15}, 16, 17, 50$, so the median is still the $4$th value, $15$.

Part (c) - which changed, and which is better. The mean jumped from $15$ to about $19.71$, a rise of nearly $5$, while the median did not change at all (it stayed $15$). The mean changed far more because it adds in the actual size of every value, so one large outlier ($50$) pulls it upward; the median only depends on the position of the middle value, so a single extreme value barely moves it. After the correction the median of $15$ better describes a typical hour, since six of the seven hours sold between $12$ and $17$ cups. (Check: the new mean $19.71$ is larger than $6$ of the $7$ data values, the tell-tale sign that an outlier has distorted it.)