Open Math›Year 11›Mathematics Standard›Data analysis›Measures of centre and spread

Measures of centre and spread

Calculate the mean, median and mode of a dataset, compute the population standard deviation, and choose the most appropriate measure of centre for a given context.

Worked examples

Finding mean, median and mode

Straightforward

Problem

Five students scored

14, 17, 17, 19

and

23

on a quiz. Find the mean, median and mode of their scores.

Find the mean by adding all values and dividing by the number of values.

\bar{x} = \frac{14 + 17 + 17 + 19 + 23}{5} = \frac{90}{5} = 18

Find the median. The data is already in ascending order. With 5 values, the median is the 3rd value.

Ordered data:

14, 17, \mathbf{17}, 19, 23

The middle (3rd) value is

\mathbf{17}

, so the median is

17

Find the mode by identifying the value that appears most often.

The value

17

appears twice; all other values appear once. The mode is

17

Consider which measure would change most if a student scored 95 instead of 23.

New mean

= \frac{14+17+17+19+95}{5} = \frac{162}{5} = 32.4

— a large increase from

18

.

New median: ordered data is

14, 17, 17, 19, 95

— the 3rd value is still

17

, unchanged.

Mode is still

17

, unchanged.

The mean is most sensitive to the extreme value.

Answer

Mean

= 18

, median

= 17

, mode

= 17

. The mean is most affected by an extreme value.

Calculating population standard deviation

Moderate

Problem

The weights (kg) of five packages are

2, 5, 6, 7

and

10

. The mean weight is

6

kg. Calculate the population standard deviation correct to 2 decimal places and interpret the result.

Write down the population standard deviation formula.

\sigma = \sqrt{\frac{\sum(x - \bar{x})^2}{n}}

where

\bar{x} = 6

and

n = 5

Find the deviation of each weight from the mean and square it.

x

x - \bar{x}

(x - \bar{x})^2

|
|---|---|---|
|

2

-4

16

|
|

5

-1

1

|
|

6

0

0

|
|

7

1

1

|
|

10

4

16

Sum the squared deviations.

\sum(x - \bar{x})^2 = 16 + 1 + 0 + 1 + 16 = 34

Divide by

n

to find the variance, then take the square root.

\sigma = \sqrt{\frac{34}{5}} = \sqrt{6.8} \approx 2.61 \text{ kg}

Interpret the result in context.

A standard deviation of

2.61

kg means that the package weights typically differ from the mean of

6

kg by about

2.61

kg. The relatively large spread (compared to the mean) indicates the weights vary considerably.

Answer

\sigma \approx 2.61

kg. Package weights typically differ from the mean by about

2.61

kg.

Comparing datasets using mean and standard deviation

Challenging

Problem

Two mechanics record their daily repair times (minutes) over one week.

Mechanic A:

38, 44, 46, 48, 54

Mechanic B:

20, 40, 46, 52, 72

(a) Calculate the mean and population standard deviation for each mechanic correct to 2 decimal places.
(b) Compare the two mechanics and state which has more consistent repair times.
(c) State the most appropriate measure of centre for this data and justify your choice.

Calculate the mean for each mechanic.

\bar{x}_A = \frac{38 + 44 + 46 + 48 + 54}{5} = \frac{230}{5} = 46 \text{ min}

\bar{x}_B = \frac{20 + 40 + 46 + 52 + 72}{5} = \frac{230}{5} = 46 \text{ min}

Both mechanics have the same mean repair time.

Calculate the population standard deviation for Mechanic A.

Squared deviations from

\bar{x}_A = 46

(38-46)^2 = 64, \quad (44-46)^2 = 4, \quad (46-46)^2 = 0, \quad (48-46)^2 = 4, \quad (54-46)^2 = 64

\sigma_A = \sqrt{\frac{64+4+0+4+64}{5}} = \sqrt{\frac{136}{5}} = \sqrt{27.2} \approx 5.22 \text{ min}

Calculate the population standard deviation for Mechanic B.

Squared deviations from

\bar{x}_B = 46

(20-46)^2 = 676, \quad (40-46)^2 = 36, \quad (46-46)^2 = 0, \quad (52-46)^2 = 36, \quad (72-46)^2 = 676

\sigma_B = \sqrt{\frac{676+36+0+36+676}{5}} = \sqrt{\frac{1424}{5}} = \sqrt{284.8} \approx 16.88 \text{ min}

Compare the two mechanics.

Both mechanics average

46

min per repair, but:
- Mechanic A:

\sigma_A \approx 5.22

min (times clustered closely around

46

min)
- Mechanic B:

\sigma_B \approx 16.88

min (times vary widely — from

20

72

min)

Mechanic A has more consistent repair times.

Choose the most appropriate measure of centre and justify.

Neither dataset contains extreme outliers that would significantly distort the mean. The mean (

46

min) is the most appropriate measure of centre here because it uses all values and gives an accurate picture of the typical repair time for each mechanic.

Answer

Mean A

= 46

min,

\sigma_A \approx 5.22

min; Mean B

= 46

min,

\sigma_B \approx 16.88

min. Mechanic A is more consistent. The mean is the most appropriate measure of centre for both datasets.

Practise

Q1·Straightforward

The daily temperatures (°C) recorded over five days were

18, 21, 21, 24

and

26

. Find the mean temperature.

Explanation

\bar{x} = \frac{18 + 21 + 21 + 24 + 26}{5} = \frac{110}{5} = 22

The mean temperature is

22

°C.

Q2·Straightforward

Find the median of the dataset:

5, 8, 9, 12, 14, 17

Explanation

The dataset is already in ascending order:

5, 8, 9, 12, 14, 17

.

With

n = 6

values, the median is the average of the 3rd and 4th values:

\text{Median} = \frac{9 + 12}{2} = \frac{21}{2} = 10.5

Q3·Straightforward

A dataset contains the values

3, 5, 5, 6, 8, 52

. Which measure of centre is most affected by the extreme value

52

MeanMedianModeAll three equally

Explanation

The mean uses every value in its calculation, so a very large value like

52

pulls it upward significantly.

Without

52

\bar{x} = \frac{3+5+5+6+8}{5} = 5.4

With

52

\bar{x} = \frac{3+5+5+6+8+52}{6} \approx 13.2

The median changes from

5

5.5

— a much smaller shift. The mode (

5

) is unchanged. The mean is most affected by extreme values.

Q4·Straightforward

The number of customers served each hour over seven hours were:

4, 6, 7, 7, 9, 10, 7

. Find the mode.

Explanation

Counting the frequency of each value:
-

4

: once
-

6

: once
-

7

: three times
-

9

: once
-

10

: once

The value

7

appears most often, so the mode is

7

Q5·Moderate

The dataset

3, 5, 5, 7

has a mean of

5

. Using the population standard deviation formula

\sigma = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n}}

calculate

\sigma

correct to 2 decimal places.

Explanation

Find each squared deviation from the mean (

\bar{x} = 5

):

|

x

x - \bar{x}

(x - \bar{x})^2

|
|---|---|---|
|

3

-2

4

|
|

5

0

0

|
|

5

0

0

|
|

7

2

4

\sigma = \sqrt{\frac{4 + 0 + 0 + 4}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2} \approx 1.41

Q6·Moderate

A machine produces metal rods. The diameters (mm) of five rods are

10.0, 10.2, 10.4, 10.6

and

10.8

. The mean diameter is

10.4

mm. Calculate the population standard deviation correct to 2 decimal places.

Explanation

Squared deviations from the mean (

\bar{x} = 10.4

):

|

x

x - \bar{x}

(x - \bar{x})^2

|
|---|---|---|
|

10.0

-0.4

0.16

|
|

10.2

-0.2

0.04

|
|

10.4

0

0

|
|

10.6

0.2

0.04

|
|

10.8

0.4

0.16

\sigma = \sqrt{\frac{0.16+0.04+0+0.04+0.16}{5}} = \sqrt{\frac{0.40}{5}} = \sqrt{0.08} \approx 0.28 \text{ mm}

Q7·Moderate

For a dataset with

n = 6

values, the sum of squared deviations from the mean is

54

. Calculate the population standard deviation.

Explanation

\text{Variance} = \frac{\sum(x - \bar{x})^2}{n} = \frac{54}{6} = 9

\sigma = \sqrt{9} = 3

The population standard deviation is

3

Q8·Moderate

Two classes sit the same test. Class A has mean

72

and standard deviation

4

. Class B has mean

72

and standard deviation

15

. Which statement is correct?

Class A's scores are more consistent than Class B's.Class B's scores are more consistent than Class A's.Class A's scores are more spread out than Class B's.Both classes performed equally in all respects.

Explanation

Both classes have the same mean (

72

), so their average performance is equal. However, the standard deviation measures spread around the mean.

Class A:

\sigma = 4

(scores clustered tightly around

72

)
Class B:

\sigma = 15

(scores much more spread out)

Class A's scores are more consistent because they vary less from the mean.

Q9·Challenging

Dataset A is

4, 6, 8, 10, 12

. The mean is

8

. Calculate the population standard deviation of Dataset A correct to 2 decimal places.

Explanation

Squared deviations from the mean (

\bar{x} = 8

(4-8)^2 = 16, \quad (6-8)^2 = 4, \quad (8-8)^2 = 0, \quad (10-8)^2 = 4, \quad (12-8)^2 = 16

\sigma_A = \sqrt{\frac{16+4+0+4+16}{5}} = \sqrt{\frac{40}{5}} = \sqrt{8} \approx 2.83

Q10·Challenging

Dataset B is

1, 5, 8, 11, 15

. The mean is

8

. Calculate the population standard deviation of Dataset B correct to 2 decimal places.

Explanation

Squared deviations from the mean (

\bar{x} = 8

(1-8)^2 = 49, \quad (5-8)^2 = 9, \quad (8-8)^2 = 0, \quad (11-8)^2 = 9, \quad (15-8)^2 = 49

\sigma_B = \sqrt{\frac{49+9+0+9+49}{5}} = \sqrt{\frac{116}{5}} = \sqrt{23.2} \approx 4.82

Q11·Challenging

Dataset A is

4, 6, 8, 10, 12

(mean

= 8

\sigma \approx 2.83

) and Dataset B is

1, 5, 8, 11, 15

(mean

= 8

\sigma \approx 4.82

). Which conclusion is correct?

Dataset B has greater spread than Dataset A.Dataset A has greater spread than Dataset B.Both datasets have the same spread.Dataset A has a higher mean than Dataset B.

Explanation

Both datasets have the same mean (

8

), so they are centred at the same value.

However,

\sigma_B \approx 4.82 > \sigma_A \approx 2.83

, which means the values in Dataset B are more spread out from the mean than those in Dataset A.

Dataset B has greater spread.

Q12·Challenging

A real estate agent records house sale prices (

\$000

s) in a suburb:

480, 510, 525, 540, 1200

. The mean is

\$651\,000

and the median is

\$525\,000

. Which measure of centre is most appropriate for representing the typical sale price, and why?

Median, because it is less affected by the extreme sale price of $\$1\,200\,000$.Mean, because it uses all five values in its calculation.Mode, because it is the most frequently occurring value.Mean, because it is always larger than the median.

Explanation

The sale price of

\$1\,200\,000

is an extreme value (outlier) that pulls the mean up to

\$651\,000

, which is higher than four of the five actual prices.

The median (

\$525\,000

) is the middle value in the ordered dataset and is not distorted by the outlier. It better represents the price a typical buyer would pay.

The mode cannot be calculated here because all values are distinct. The median is the most appropriate measure.