Data

Categorical data sorts items into groups or labels, like colours. The other options are all measured or counted numbers, so they are numerical.

Continuous data can take any value within a range, often with decimals. Time can be 12.4 seconds, 12.41 seconds, and so on. The “number of” options are discrete, and eye colour is categorical.

The mode is the value with the highest frequency. The highest frequency is 6, which belongs to a score of 7. (6 is the frequency, not the score.)

The mean $=\dfrac{3+4+5+6+100}{5}=23.6$, which is dragged up by the outlier 100. The median is 5, sitting in the middle of the cluster, so it is the better measure of centre when there is an outlier. (Range measures spread, not centre.)

Sort the values first: 3, 4, 5, 7, 9. With 5 values the median is the middle (3rd) value, which is 5. (5.6 is the mean; you must sort before reading the middle.)

Sort first: 3, 4, 5, 7, 8, 11. With 6 values the median is the average of the two middle values: $\dfrac{5+7}{2}=6$. (Skipping the sort and averaging the original middle two gives the wrong 9.)

The mode is the most common value. 7 appears 3 times, more than any other value.

Range $=20-11=9$. (Adding the max and min gives 31; 15.5 is the median.)

Multiply each score by its frequency and add: $3\times 4 + 4\times 3 + 5\times 2 + 6\times 1 = 12+12+10+6 = 40$. Then $\dfrac{40}{10}=4$.

The total must be $5\times 12 = 60$. The four known numbers sum to $10+15+8+11 = 44$. So the fifth is $60-44 = 16$.

“70 or higher” covers the 70-79, 80-89 and 90-100 bins: $10+7+3 = 20$ students. (Forgetting the 90-100 bin gives 17.)

The mode is the value with the most dots, which is 3 (it has 7 dots). (7 is the number of dots, not the value; 17 is the total number of data points.)

There are $3+4+4+2 = 13$ values, already in order. The median is the middle (7th) value: 12, 15, 17, 20, 21, 24, 28, ... so the median is 28.

Both means are $\dfrac{42}{6}=7$. Class A range $=9-5 = 4$; Class B range $=11-3 = 8$. Same mean, but Class B is more spread out.

With an even number of values there is no single middle value, so the median is the average of the two middle ones, which are the 3rd and 4th values. (Averaging all 6 gives the mean, not the median.)

Numerical data is counted or measured with numbers, like the number of siblings. The other options sort items into groups or labels, so they are categorical.

Favourite sport is categorical data, and a column (bar) graph shows how many students chose each category, making comparisons easy. A mean or range cannot be found for categories like sports.

Mean $= \dfrac{\text{total}}{\text{how many}}$, so total $= \text{mean}\times\text{how many} = 9\times 6 = 54$.

Adding a value equal to the mean does not shift the balance point, so the mean stays the same. (Adding a value above the mean would raise it; below the mean would lower it.)

Each dot represents one data value, so 4 dots above 5 means four of the data values are equal to 5. (The height of the stack is the frequency of that value.)

Range $= \text{largest} - \text{smallest}$, so largest $= \text{range} + \text{smallest} = 15 + 8 = 23$. (Subtracting gives the incorrect 7.)

Original total $=5\times 7 = 35$. Add the new score: $35+13 = 48$. New mean $=\dfrac{48}{6}=8$. (Averaging the old mean with the new score ignores how many people each represents.)

Class A total $=20\times 75 = 1500$; Class B total $=30\times 80 = 2400$. Combined mean $=\dfrac{1500+2400}{50}=\dfrac{3900}{50}=78$. (Averaging the two means directly ignores the different class sizes.)

The range measures spread. A larger range for X means its values are more spread out, even though both sets have the same average (mean).

The mean uses every value, so one very large value pulls it noticeably higher. The median (middle position) and mode (most common value) are barely affected by a single outlier.