**Example data set:** 6, 3, 6, 13, 7, 7 in a table:

**Mean** the average value,

**Mode** the value that occurs most often (highest frequency) e.g. The example data set has 2 modes: 6 and 7

**Median** the middle value when the data set is ordered low to high. Even number of values: the median is the average of the two middle values. Find for larger values as n + ½ .

e.g. data set from low to high: 3, 6, 6, 7, 7, 13

**Range** largest x-value − smallest x-value

e.g. range=13−3=10

**Variance**

**Standard deviation**

**Grouped data** data presented as an interval, e.g.10 < x ≤ 20 where:

- lower boundary = 10

- upper boundary = 20

- interval width = 20 − 10 = 10

- mid-interval value (midpoint) =
^{(20 + 10)}/_{2}= 15

Use the midpoint as the x-value in all calculations with grouped data.

Adding a constant to all the values in a data set or multiplying the entire data set by a constant influences the mean and standard deviation values in the following way:

Q_{1} the value for x so that 25% of all the data values are ≤ to it first quartile = 25^{th} percentile

Q_{2} median = 50^{th} percentile

Q_{3} third quartile = 75^{th} percentile

Q_{3} − Q_{1} interquartile range (IQR) = middle 50 percent

Example: Snow depth is measured in centimetres: 30, 75, 125, 55, 60, 75, 65, 65, 45, 120, 70, 110. Find the range, the median, the lower quartile, the upper quartile and the interquartile range.

First always rearrange data into ascending order: 30, 45, 55, 60, 65, 65, 70, 75, 75, 110, 120, 125

- The range
125−30=95cm

2. The median: there are 12 values so the median is between the 6

^{th}and 7^{th}value.^{ (65 + 70)}/_{2}= 67.5 cm3. The lower quartile: there are 12 values so the lower quartile is between the 3

^{rd}and 4^{th}value.^{ (55 + 60)}/_{2}= 57.5 cm4. The upper quartile: there are 12 values so the lower quartile is between the 9

^{th}and 10^{th}value.^{ (75 + 110)}/_{2}= 92.5 cm5. The IQR

92.5 − 57.5 = 35cm