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AS-Level Maths: Statistics 1for Edexcel

S1.2 Calculating means and standard deviations

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Means

Calculating means

Calculating standard deviations

Coding

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The mean is the most widely used average in statistics. It is found by adding up all the values in the data and dividing by how many values there are.

, , ,...,1 2 3 nx x x x

...1 2 3 inxx x x x

xn n

Note: The mean takes into account every piece of data, so it is affected by outliers in the data. The

median is preferred over the mean if the data contains outliers or is skewed.

Mean

Notation: If the data values are , then the mean is

This is the mean symbol

This symbol means the

total of all the x values

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If data are presented in a frequency table:

Mean

Value Frequency

… …

2x

nx

1x 1f

2f

nf

...1 1 2 2 i in n

i i

x fx f x f x fx

f f

then the mean is

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Example: The table shows the results of a survey into household size. Find the mean size.

Mean

Household size, x Frequency, f

1 20

2 28

3 25

4 19

5 16

6 6

To find the mean, we add a 3rd column to the table.

x × f

20

56

75

76

80

36

TOTAL 114 343

Mean = 343 ÷ 114 = 3.01

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Standard deviation

Calculating means

Calculating standard deviations

Coding

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There are three commonly used measures of spread (or dispersion) – the range, the inter-quartile range and the standard deviation.

( )2

variance ix x

n

( )

2

s.d. ix x

n

Standard deviation

The following formulae can be used to find the variance and s.d.

variance = (standard deviation)2variance = (standard deviation)2

The variance is related to the standard deviation:

The standard deviation is widely used in statistics to measure spread. It is based on all the values in the data, so it is sensitive to the presence of outliers in the data.

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Total: 22

Example: The mid-day temperatures (in °C) recorded for one week in June were: 21, 23, 24, 19, 19, 20, 21

( )2

variance ix x

n

Standard deviation

...21 23 21 14721

7 7x

21 0 0

23 2 4

24 3 9

19 -2 4

19 -2 4

20 -1 1

21 0 0

( )2ix xix xix

So variance = 22 ÷ 7 = 3.143

So, s.d. = 1.77°C (3 s.f.)

°CFirst we find the mean:

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There is an alternative formula which is usually a more convenient way to find the variance:

Standard deviation

( ) ( )2 2 2But, 2i i ix x x x x x 2 22i ix x x nx 2 22ix x nx nx 2 2ix nx

2

2variance ix xn

Therefore, and

2

2s.d. ix xn

( )2

variance ix x

n

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Example (continued): Looking again at the temperature data for June: 21, 23, 24, 19, 19, 20, 21

Standard deviation

14721

7x

...2 2 2 221 23 21ix

°C

Also, = 3109

.

.

2

2 23109variance 21 3 143

7s . 77.d 1

ix xn

°C

Note: Essentially the standard deviation is a measure of how close the values are to the mean value.

We know that

So,

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When the data is presented in a frequency table, the formula for finding the standard deviation needs to be adjusted slightly:

Calculating standard deviation from a table

2

2s.d. i i

i

f xx

f

Example: A class of 20 students were asked how many times they exercise in a normal week.

Find the mean and the standard deviation.

Number of times exercise taken

Frequency

0 5

1 3

2 5

3 4

4 2

5 1

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Calculating standard deviation from a table

x × f x2 × f

0 0

3 3

10 20

12 36

8 32

5 25

No. of times exercise taken, x

Frequency, f

0 5

1 3

2 5

3 4

4 2

5 1

. .2

2 2116s.d. 1 9 1 4

08

2i i

i

f xx

f

The table can be extended to help find the mean and the s.d.

TOTAL: 20 38 116

.38

201 9x

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If data is presented in a grouped frequency table, it is only possible to estimate the mean and the standard deviation. This is because the exact data values are not known.

An estimate is obtained by using the mid-point of an interval to represent each of the values in that interval.

Example: The table shows the annual mileage for the employees of an insurance company.

Estimate the mean and standard deviation.

Calculating standard deviation from a table

Annual mileage, x Frequency

0 ≤ x < 5000 6

5000 ≤ x < 10,000 17

10,000 ≤ x < 15,000 14

15,000 ≤ x < 20,000 5

20,000 ≤ x < 30,000 3

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Calculating standard deviation from a table

Mileage Frequency, f Mid-point, x f × x f × x2

0 – 5000 6 2500 15000 37,500,000

5000 – 10,000 17 7500 127,500 956,250,000

10,000 – 15,000 14 12,500 175,000 2,187,500,000

15,000 – 20,000 5 17,500 87,500 1,531,250,000

20,000 – 30,000 3 25,000 75,000 1,875,000,000

480,000

410

5,667x

TOTAL 45 480,000 6,587,500,000

26,587,500,000s.d. 10,667

47

55 11

miles

miles

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In most distributions, about 67% of the data will lie within 1 standard deviation of the mean, whilst nearly all the data values will lie within 2 standard deviations of the mean.

Values that lie more than 2 standard deviations from the mean are sometimes classed as outliers – any such values should be treated carefully.

Standard deviation is measured in the same units as the original data. Variance is measured in the same units squared.

Notes about standard deviation

Here are some notes to consider about standard deviation.

Most calculators have a built-in function which will find the standard deviation for you. Learn how to use thisfacility on your calculator.

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Examination-style question: The ages of the people in a cinema queue one Monday afternoon are shown in the stem-and-leaf diagram:

Examination-style question

2 3 means 23 years old2 3 63 1 6 64 1 2 5 6 95 0 4 76 1

a) Explain why the diagram suggests that the mean and standard deviation can be sensibly used as measures of location and spread respectively.

b) Calculate the mean and the standard deviation of the ages.

c) The mean and the standard deviation of the ages of the people in the queue on Monday evening were 29 and 6.2 respectively. Compare the ages of the peoplequeuing at the cinema in the afternoon with those in theevening.

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a) The mean and the standard deviation are appropriate, as the distribution of ages is roughly symmetrical and there are no outliers.

Examination-style question

2 3 means 23 years old2 3 63 1 6 64 1 2 5 6 95 0 4 76 1

b) . .597

597 so, 42 642861

44

2 6ix x . .2 227,131

27131 so, s.d. 42 6428614

10 9ix c) The cinemagoers in the evening had a smaller mean

age, meaning that they were, on average, younger than those in the afternoon.

The standard deviation for the ages in the evening was also smaller, suggesting that the evening audience were closer together in age.

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Sometimes in examination questions you are asked to pool two sets of data together.

Combining sets of data

Example: Six male and five female students sit an A-level examination.

The mean marks were 52% and 57% for the males and females respectively. The standard deviations were 14 and 18 respectively.

Find the combined mean and the standard deviation for the marks of all 11 students.

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Let be the marks for the 6 male students.

Let be the marks of the 5 female students.

To find the overall mean, we first need to find the total marks for all 11 students.

,...,1 6x x

,...,1 5y y

Combining sets of data

As 52x 6 52 312x As 57y 5 57 285y

312 285 597x y

.. . %. .597

54 2727 31

541

Therefore

So the combined mean is:

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To find the overall standard deviation, we need to find the total of the marks squared for all 11 students.

As s.d. 14x

Therefore,

So the combined s.d. is: (to 3 s.f.)

Combining sets of data

As s.d. 18y

2

2s.d. ix xn

( )2 2 2s.d.x n x ( )2 2 26 14 52 17,400x ( )2 2 25 18 57 17,865y

2 2 35,265x y

. . %235,26554 2 6 17

111

Notice that the formula

rearranges to give

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Calculating means

Calculating standard deviations

Coding

Coding

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Coding is a technique that can simplify the numerical effort required in finding a mean or standard deviation.

Enter some data below, and see how it changes when you add or multiply by different numbers.

Coding

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Adding

So, if a number b is added to each piece of data, the mean value is also increased by b.

The standard deviation is unchanged.

i iy ax b

y ax b s.d. s.d.y xa

Coding

More formally, if then:

Multiplying

If each piece of data is multiplied by a, the mean value is multiplied by a.

The standard deviation is also multiplied by a.

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Example: Find the mean and the standard deviation of the values in the table. Use the transformation below to help you.

15

10y x

Coding

x Frequency

50 3

60 5

70 7

80 4

90 1

y

0

1

2

3

4

Using the given transformation, add a y column to the table.

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Coding

y Frequency, f

0 3

1 5

2 7

3 4

4 1

y × f y2 × f

0 0

5 5

14 28

12 36

4 16

.35

201 75y

Total 20 35 85

. .2

2 285s.d. 1 75

21 09

0i i

i

f yy

f

To find the mean:

To find the s.d.:

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And the standard deviation of x is: 10 × 1.09 = 10.9

We can rearrange:

to get:

15

10y x

Therefore the mean of x is:

Coding

10 50x y

. .10 50 10 1 75 0 75 6 5x y

Note how the coding helped to simplify the calculations by making the numbers smaller.

You have now found the mean and standard deviation of y. To find them for the x values, you must reverse the coding.