Mean, Variance and Standard Deviation | IEB Grade 12 Maths

Q: Do I have to calculate standard deviation by hand?

No. Use your calculator's statistics mode. You must know which value to read off and how to interpret it.

Q: When should I use the median instead of the mean?

When the data is skewed or has outliers, because the median is not pulled by extreme values.

Q: Why is the mean of grouped data only an estimate?

Because the midpoint is used for every value in a class, and the real values may not be evenly spread.

Statistics carries about 20 marks in Paper 2. You need to calculate measures of centre and spread with a calculator, estimate them for grouped data, and explain what they tell you about a data set in plain words.

Mean, median and mode

Mean $\overset{x}{ˉ} = \frac{\sum x}{n}$ : the average.
Median: the middle value of the ordered data. For an even number of values, average the two middle values.
Mode: the most frequent value.

Example. Test marks of 10 learners: $45; 52; 58; 61; 63; 67; 70; 74; 81; 89$

$\overset{x}{ˉ} = \frac{660}{10} = 66 median = \frac{63 + 67}{2} = 65$

The mean is pulled towards extreme values. The median is not, so it is the better measure of centre for skewed data or data with outliers.

Variance and standard deviation by calculator

Variance is the mean of the squared deviations from the mean:

$σ^{2} = \frac{\sum ( x - x ˉ ) ^{2}}{n}$

Standard deviation $σ = σ^{2}$ measures how spread out the data is around the mean, in the same units as the data.

Use your calculator's STAT mode and read $σ_{x}$ (the population standard deviation).

Example. For the marks above, $σ \approx 12, 53$ and $σ^{2} = 157$ .

Within one standard deviation. $\overset{x}{ˉ} - σ \approx 53, 47$ and $\overset{x}{ˉ} + σ \approx 78, 53$ . Six of the ten marks lie in this interval.

Interpreting standard deviation with the normal distribution is not examinable.

Estimates for grouped data

For data grouped in class intervals, use the midpoint of each class:

$\overset{x}{ˉ} \approx \frac{\sum f \cdot x _{mid}}{\sum f}$

Example. Times (minutes) for 30 learners:

$0 \leq t < 10$ : 3; $10 \leq t < 20$ : 7; $20 \leq t < 30$ : 12; $30 \leq t < 40$ : 6; $40 \leq t < 50$ : 2

$\overset{x}{ˉ} \approx \frac{3 ( 5 ) + 7 ( 15 ) + 12 ( 25 ) + 6 ( 35 ) + 2 ( 45 )}{30} = \frac{720}{30} = 24 minutes$

The modal class is $20 \leq t < 30$ , and the median lies in the class containing the 15th and 16th values, also $20 \leq t < 30$ .

The mean of grouped data is an estimate, because we do not know where values fall inside each class.

Key terms

Mean: The sum of the values divided by the number of values.
Median: The middle value of ordered data.
Standard deviation: A measure of how far data typically lies from the mean.
Variance: The square of the standard deviation.
Modal class: The class interval with the highest frequency.

Common mistakes

Finding the median without ordering the data first.
Using the sample standard deviation ( $s_{x}$ ) instead of the population one ( $σ_{x}$ ).
Using class boundaries instead of midpoints for grouped data.
Forgetting units when describing the mean or standard deviation.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

The test marks of 10 learners are:

$45; 52; 58; 61; 63; 67; 70; 74; 81; 89$

Calculate the mean and the median.

Worked solution

Hidden - try the question first

Solution

$\overset{x}{ˉ} = \frac{660}{10}$
$= 66$
Median $= \frac{63 + 67}{2} = 65$

Question 2

5 marks total

routine

For the same marks:

(a) Calculate the standard deviation. (2)

(b) How many learners scored within one standard deviation of the mean? (3)

Worked solution

Hidden - try the question first

Solution

(a) Calculator: $σ \approx 12, 53$
(b) $\overset{x}{ˉ} - σ \approx 53, 47$
$\overset{x}{ˉ} + σ \approx 78, 53$
6 learners (58, 61, 63, 67, 70 and 74)

Question 3

6 marks total

complex

The times taken by 30 learners to finish a task are shown:

$0 \leq t < 10$ : 3
$10 \leq t < 20$ : 7
$20 \leq t < 30$ : 12
$30 \leq t < 40$ : 6
$40 \leq t < 50$ : 2

(a) Estimate the mean time. (4)

(b) Write down the modal class. (1)

Worked solution

Hidden - try the question first

Solution

(a) Midpoints: 5, 15, 25, 35, 45
$\sum f x = 15 + 105 + 300 + 210 + 90 = 720$
$= \frac{63 + 67}{2} = 65$
$= 24$ minutes
(b) $20 \leq t < 30$
(c) The 15th and 16th values lie in $20 \leq t < 30$

Question 4

8 marks total

problem-solving

Use the 10 test marks from Question 1 (mean 66, standard deviation about 12,53).

(a) The teacher adds 5 marks to every learner's score. Write down the new mean and standard deviation. Explain. (3)

(b) Instead, every mark is multiplied by 1,1. Write down the new mean and standard deviation. (2)

(c) The teacher discovers that the mark of 45 should have been 54. Calculate the correct mean, and state whether the standard deviation increases or decreases. (3)

Worked solution

Hidden - try the question first

Solution

(a) New mean $σ \approx 12, 53$
Standard deviation stays about 12,53
Adding a constant shifts every value equally, so the spread does not change
(b) New mean $\overset{x}{ˉ} - σ \approx 53, 47$
New standard deviation $\approx 13, 78$
(c) New mean $= \frac{660 - 45 + 54}{10}$
$= 66, 9$
The lowest mark moves closer to the mean, so the standard deviation decreases

FAQs

Do I have to calculate standard deviation by hand?

No. Use your calculator's statistics mode. You must know which value to read off and how to interpret it.

When should I use the median instead of the mean?

When the data is skewed or has outliers, because the median is not pulled by extreme values.

Why is the mean of grouped data only an estimate?

Because the midpoint is used for every value in a class, and the real values may not be evenly spread.

Measures of central tendency, variance and standard deviation

Mean, median and mode

Variance and standard deviation by calculator

Estimates for grouped data

Key terms

Common mistakes

Practice questions

FAQs

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