IB Applications and Interpretation SL · Descriptive statistics · Section 5 of 5

Variance and standard deviation

3 worked examples · By Álvaro López

Key idea

The standard deviation measures how far the data spread from their mean, and it is given in the same units as the data. It is the spread measure that goes with the mean: in Paper 1 you calculate it with a table and in Paper 2 with the GDC. You also need to know what happens to the mean and the standard deviation when a number is added to, subtracted from or multiplied by all the data.

Measuring how far the data are from the mean

If you subtract the mean from each data value, the positive and negative deviations cancel out: their sum is always zero. To measure the spread, the deviations must be made positive. That is why they are squared and averaged.

Variance is in squared units (if the data are in metres, it is in square metres). That is why the number you compare with the data and interpret is the standard deviation, which is in the same units.

Calculating with a table

With raw data grouped by value, the table has one row for each value $x_{i}$ with its frequency $f_{i}$ . Add two columns, $f_{i} x_{i}$ and $f_{i} x_{i}^{2}$ , and total all the columns.

Grouped data in intervals

With intervals you do not know the exact values. Each interval is represented by its class midpoint, and the mean and standard deviation you get are estimates.

The method for the table is the same, but using the class midpoints as $x_{i}$ . Write "estimate" in the answer: the result is not that of the original data, but the one you get by assuming that each value in the interval equals its midpoint.

Using the GDC

Interpreting the standard deviation

A large standard deviation means the data are widely spread around the mean. A small one means the data are clustered close to it. A single number says little on its own: interpret it by comparing it with another standard deviation in the same unit, or with the mean itself.

Write the interpretation as a sentence in context. For example: "the call times in the morning are more consistent than those in the afternoon, because their standard deviation is smaller." A number without a sentence does not earn the interpretation mark.

What happens when the data change

Data are often transformed: a bonus is added to each salary, euros are converted to dollars, or the scale of the marks is changed. The mean and the standard deviation change, but not in the same way.

The reason is simple. Adding $b$ shifts all the data by the same distance: the mean moves with them, but the distances between data values do not change, so the spread does not either. Multiplying by $a$ stretches or shrinks the scale: each distance is multiplied by $a$ , so the standard deviation is multiplied by $∣ a ∣$ and the variance by $a^{2}$ .

For example, if 1 point is added to every mark on an exam, the mean goes up by 1 point, but the standard deviation does not change.

Worked examples

Example 1 Basic

Goals scored in twenty matches

A team has scored a number of goals in each of its 20 matches of the season. The table shows how many matches had each result:

Goals $x_{i}$	1	2	3	4	5
Matches $f_{i}$	2	5	6	4	3

a) Find the mean number of goals per match.

b) Find the variance.

c) Find the standard deviation.

Solution

We complete the table with $f_{i} x_{i}$ and $f_{i} x_{i}^{2}$ :

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$f_{i} x_{i}^{2}$
1	2	2	2
2	5	10	20
3	6	18	54
4	4	16	64
5	3	15	75
	$n = 20$	$61$	$215$

a) $x ˉ = \frac{61}{20} = 3.05$ goals.

b) With the shortcut formula: $σ_{x}^{2} = \frac{215}{20} - 3.0 5^{2} = 10.75 - 9.3025 = 1.4475$

c) $σ_{x} = 1.4475 \approx 1.20$ goals.

Example 2 Medium

Height of some plants, with grouped data

A nursery has measured the height of 20 plants of the same species. The results are grouped:

Height (cm)	$[10, 20 [$	$[20, 30 [$	$[30, 40 [$	$[40, 50 [$
Plants	4	7	6	3

Use the class midpoints to estimate:

a) the mean;

b) the standard deviation;

c) between which two values lie the mean minus one standard deviation and the mean plus one standard deviation.

Solution

The class midpoints are $15$ , $25$ , $35$ and $45$ cm, and $n = 20$ .

a) $\sum f_{i} x_{i} = 4 \cdot 15 + 7 \cdot 25 + 6 \cdot 35 + 3 \cdot 45 = 60 + 175 + 210 + 135 = 580$ , so $x ˉ = \frac{580}{20} = 29$ cm (estimated).

b) $\sum f_{i} x_{i}^{2} = 4 \cdot 225 + 7 \cdot 625 + 6 \cdot 1225 + 3 \cdot 2025 = 900 + 4375 + 7350 + 6075 = 18700$ . Then $σ_{x}^{2} = \frac{18 700}{20} - 2 9^{2} = 935 - 841 = 94$ and the estimated standard deviation is $σ_{x} = 94 \approx 9.70$ cm.

c) The values are $29 - 94 \approx 19.30$ cm and $29 + 94 \approx 38.70$ cm. That is, between about $19.3$ and $38.7$ cm.

Example 3 Hard

Rescaling exam marks

The marks of 25 students in an exam marked out of 10 have a mean of $4.8$ and a standard deviation of $1.2$ . The teacher wants to convert each original mark $x$ into a new mark $y = a x + b$ , so that the new mean is $7$ and the new standard deviation is $3$ .

a) Find $a$ and $b$ , given that $a > 0$ .

b) A student got a $6$ on the original exam. What is their mark on the new scale?

c) The range of the original marks is $4.5$ . Find the new variance and the new range.

Solution

a) As $σ_{y} = ∣ a ∣ σ_{x}$ , we have $3 = a \cdot 1.2$ , so $a = \frac{3}{1.2} = 2.5$ .

For the mean, $y ˉ = a x ˉ + b$ : $7 = 2.5 \cdot 4.8 + b = 12 + b$ , so $b = - 5$ .

Therefore the transformation is $y = 2.5 x - 5$ .

b) $y = 2.5 \cdot 6 - 5 = 10$ . The new mark is a $10$ .

c) The new variance is $a^{2} σ_{x}^{2} = 2. 5^{2} \cdot 1. 2^{2} = 6.25 \cdot 1.44 = 9$ , which is consistent with $σ_{y} = 3$ . The new range is $a \cdot range = 2.5 \cdot 4.5 = 11.25$ .

In the exam: write the transformation $y = a x + b$ first, calculate $a$ from the standard deviation and $b$ from the mean. If you do it the other way round, the mean comes out right but the standard deviation does not.

Common mistakes

Summary

$σ_{x}^{2} = \frac{\sum f _{i} ( x _{i} - x ˉ ) ^{2}}{n} = \frac{\sum f _{i} x _{i}^{2}}{n} - x ˉ^{2}$ and $σ_{x} = σ_{x}^{2}$ , with $n = \sum f_{i}$ . The standard deviation is in the units of the data.
In grouped data, $x_{i}$ is the class midpoint and the results are estimates.
With a sample, if the question asks for it, use $s_{x}$ , which divides by $n - 1$ . Otherwise, use $σ_{x}$ .
With $y = a x + b$ : $y ˉ = a x ˉ + b$ , $σ_{y} = ∣ a ∣ σ_{x}$ and $σ_{y}^{2} = a^{2} σ_{x}^{2}$ . The $b$ only changes the mean.
Interpret the standard deviation by comparing, with a sentence in context.

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