Three ways to say how spread out the data is.
The data
2, 4, 4, 4, 5, 5, 7, 9
Find the variance.
The mean is 40 ÷ 8 = 5, and the squared deviations total 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32. Divide by the 8 values: variance = 4. D is the standard deviation — the variance is the step before the square root.
range = highest − lowest
the definition
= 9 − 2 = 7
one subtraction
Change the 9 to 90 and the range jumps to 88 while everything else is untouched. It says nothing about the six values in the middle.
mean = 40 ÷ 8 = 5
the centre
| Value | Distance from 5 | Squared |
|---|---|---|
| 2 | −3 | 9 |
| 4, 4, 4 | −1 each | 1 each |
| 5, 5 | 0 | 0 |
| 7 | 2 | 4 |
| 9 | 4 | 16 |
| total | 0 | 32 |
The raw distances always total zero — the positives and negatives cancel exactly. Squaring removes the signs so the spread can be measured at all.
variance = 32 ÷ 8
the mean squared distance
= 4
in squared units
SD = √4
back to the original units
= 2
the standard deviation
The square root is what makes the answer comparable to the data itself. Variance is an intermediate step.
values huddle near the mean
values scattered wide
An SD of 2 says a typical value sits about 2 away from 5. Six of our eight values fall inside 3 to 7.
| Set | Values | Mean | SD |
|---|---|---|---|
| A | 49, 50, 50, 51 | 50 | ≈ 0.7 |
| B | 20, 40, 60, 80 | 50 | ≈ 22.4 |
Same mean, completely different stories. Averages alone never describe a data set.
It comes from a square root of squared numbers. An SD of zero means every value is identical; a negative one is impossible.
Negative values. Find the standard deviation of −6, −2, 0, 2, 6.
mean = 0 ÷ 5 = 0
the values cancel in pairs
squares: 36 + 4 + 0 + 4 + 36 = 80
(−6)² = +36
variance = 80 ÷ 5 = 16, SD = 4
the 0 still counts in n
What is different: a mean of 0 is fine, and squaring makes every distance positive. A spread of values around 0 is just as spread as one around 50.
A surd answer, two routes. Find the standard deviation of 1, 3, 5, 7.
mean = 16 ÷ 4 = 4; squares 9 + 1 + 1 + 9 = 20
distances −3, −1, 1, 3
variance = 20 ÷ 4 = 5, SD = √5
about 2.24
quicker: (1 + 9 + 25 + 49) ÷ 4 − 4² = 21 − 16 = 5
mean of the squares minus the square of the mean
What is different: 5 is not a perfect square, so the SD stays as √5. The second route skips the distances entirely, which helps when the mean is awkward.
5 is the variance. The question asked for the standard deviation, so take the square root.
| Change to every value | Mean | SD | Variance |
|---|---|---|---|
| add c | + c | unchanged | unchanged |
| multiply by k (k > 0) | × k | × k | × k² |
+ 7: SD stays 3, variance 9
every value and the mean move together
× 4: SD = 3 × 4 = 12
every distance from the mean is 4 times as long
variance = 12² = 144
or 9 × 4² — never 9 × 4
Both together, × 2 and then − 5: SD = 2 × 3 = 6 and variance 36. The − 5 changes nothing.
Adding one new value works through the totals instead. Eight values with variance 9 have squared distances totalling 8 × 9 = 72. A ninth value equal to the mean adds a distance of 0, so the mean stays put, the total stays 72, and the variance becomes 72 ÷ 9 = 8: smaller, because n grew.
3 + 7 = 10 or (3 + 7)² = 100 treat the shift as spread. Adding the same number to every value moves the data; it does not widen it.
In these lessons we divide by n, the number of values, when we average the squared distances (Step 3), and our questions do the same. Some calculators divide by n − 1 for a sample; if a choice matches that, it is not the answer here.
Expect a small set to compute (the mean comes out whole), two sets with the same mean to compare, or a change to every value: raised, doubled, or both. Shifts leave the SD alone; multiplying by k multiplies the SD by k and the variance by k². Ten-second check: the SD is never negative and never bigger than the range.
The data
6, 8, 10, 12, 14
Find the standard deviation.
The mean is 10 and the squared deviations add to 40, so the variance is 40 ÷ 5 = 8 and the standard deviation is √8 ≈ 2.83. A stops at the variance — the two are always one square root apart.
