edexcel unit 04 outcome 3 t1

7/28/2019 Edexcel Unit 04 Outcome 3 t1

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EDEXCEL NATIONAL CERTIFICATE

UNIT 4 MATHEMATICS FOR TECHNICIANSOUTCOME 3

TUTORIAL 1 - STATISTICAL METHODS

Learning outcomes

On completion of this unit a learner should:1 Know how to use algebraic methods

2 Be able to use trigonometric methods and standard formula to determine areas and volumes

3 Be able to use statistical methods to display data

4 Know how to use elementary calculus techniques.

OUTCOME 3 - Be able to use statistical methods to display data

Data handling: data represented by statistical diagrams e.g. bar charts, pie charts, frequencydistributions, class boundaries and class width, frequency table; variables (discrete and

continuous); histogram (continuous and discrete variants); cumulative frequency curves

Statistical measurement: arithmetic mean; mode; discrete and grouped data.

The section on statistics has been greatly reduced in scope from the pre-2007 version but students

wanting to understand the subject at a greater depth will find plenty of material on the web site.

The following is a useful web site to visit for this tutorial.

http://www.manatee.k12.fl.us/sites/elementary/palmasola/mathlabtutstat1.htm

http://home.ched.coventry.ac.uk/Volume/vol0/ogive.htm

D.J.Dunn www.freestudy.co.uk 1
http://www.manatee.k12.fl.us/sites/elementary/palmasola/mathlabtutstat1.htmhttp://home.ched.coventry.ac.uk/Volume/vol0/ogive.htmhttp://home.ched.coventry.ac.uk/Volume/vol0/ogive.htmhttp://www.manatee.k12.fl.us/sites/elementary/palmasola/mathlabtutstat1.htm


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1. CHARTS

Statistics are used to help us analyse and understand the performance and trends in various areas of

work. These might be financial trends, things to do with the population or things to do with

manufacturing. Often we wish to present information visually with easily understood graphics and

so a variety of graphs and charts are used for this purpose. Here is an example of a PIE CHART and

a BAR CHART showing the number of parliamentary seats won by main political parties at the

2001 British general election.

The number of MPs elected must be a round or whole number so the raw data is exact. When

presenting other forms of data this is not the case as explained in the following section.

2. RAW DATA

Consider a set of statistics compiled for the height of children of age 10. First we would compile a

table of heights. This would be the raw data and it is not ranked in any order, just recorded as it

happened. Note that the larger the sample we take, the more meaningful the results will be. Supposewe measure the heights to an accuracy of 0.01 m. This is the table of raw data for 10 year old

children taken with 17 samples.

Sample No. 1 2 3 4 5 6 7 8 9 10 11 12

Height (m) 1.45 1.56 1.37 1.44 1.32 1.42 1.55 1.29 1.37 1.49 1.47 1.34

Sample No. 13 14 15 16 17

Height (m) 1.56 1.28 1.35 1.62 1.46

3. RANKED DATA

If the raw data is rearranged from shortest to tallest we have the ranked data.

Sample

number

1 2 3 4 5 6 7 8 9 10 11 12 13

Height

(m)

1.28 1.29 1.32 1.34 1.35 1.37

1.37 1.44 1.44 1.46 1.46 1.47 1.49

Sample

number

14 15 16 17 Total

Height

(m)

1.55 1.56 1.56 1.62 24.34

Data presented in this form is also called DISCRETE DATA because we jump from one value toanother in steps, in this case steps of 0.01 m.



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4. GROUPED DATA

If we used exact measurements of heights, it is unlikely we would find two children exactly the

same height so we round off the values. This causes problems as we shall see. When handling large

numbers of samples, we end up with huge lists of data so to simplify the table we group it into

bands or classes within which the rounded measurements fall. The more we round off the values,

the more likely it becomes that we will find more than one in a given class. The number of children

within each class is the frequency. Next we would have to go through the laborious task of counting

how many there are in each class. If we found a child with a height exactly on the edge of the classedge, we might decide to allocate a half to each class on either side resulting in frequency values

that are not whole numbers. It is also useful to show a band for all the values below the minimum

containing no results. The result is a FREQUENCY DISTRIBUTION TABLE.

Height 0 - 1.2 1.2 - 1.3 1.3 - 1.4 1.4 - 1.5 1.5 - 1.6 1.6 - 1.7

Mid Point or MARK 1.25 1.35 1.45 1.55 1.65

Freq. 0 2 5 6 3 1

5. GRAPHS

If we plot frequency vertically against height horizontally, we get a frequency distribution graphand this can be drawn in different ways. The values plotted are the mid point values called the

MARK. In an example like this one, we assume that all the values less than 1.25 have a frequency

of zero.

These plots simply tell us the numbers in each class

by the mark. If we want to illustrate the width of the

band we use a HISTOGRAM . Notice that the midpoint of each band is the MARK. The boxes aredrawn between the CLASS LIMITS. Because theheights were rounded off to 0.01 m the limits are

0.005 either side of the CLASS BOUNDARY andthe class boundary is the exact dividing line between

each class. The width of the band from mark to mark

or boundary to boundary is the CLASSINTERVAL. In this example the interval is 0.1.



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6. MEAN

This is one of the more common statistics you will see and it's easy to compute. All you have to do

is add up all the values in a set of data and then divide that sum by the number of values in thedataset. For our example, let the height be represented by the variable x and the frequency be f.

Sample

number

1 2 3 4 5 6 7 8 9 10 11 12 13

Height(m)

1.45 1.56 1.37 1.44 1.32 1.42 1.55 1.29 1.37 1.49 1.47 1.34 1.56

Sample

number

14 15 16 17 Total

Height

(m)

1.28 1.35 1.62 1.46 24.34

The mean value is denoted x and x = 24.34/17 = 1.432 m

We can do this a bit more simply using the frequency distribution table.

x (mark) Less than 1.2 1.25 1.35 1.45 1.55 1.65

f. 0 2 5 6 3 1 total

f x 0 2.5 6.75 8.7 4.65 1.65 24.25

The mean value is x = 24.25/17 = 1.426 m. This is not quite as accurate as the previous answer

because the values have been taken at the mid point of the band.

7. MEDIAN

Whenever you see words like, "the average person ...", or "the average income of ... you don'talways want to know the mean. Often you want to know about the one in the middle. That's the

median.

Again, this statistic is easy to determine because the median literally is the value in the middle. In

order to find it, you just line up the values in your set of data from largest to smallest. The one in the

dead-centre is your median. Our table would look like this.

Sample

number

1 2 3 4 5 6 7 8 9 10 11 12 13

Height(m) 1.28 1.29 1.32 1.34 1.35 1.37 1.37 1.44 1.44 1.46 1.46 1.47 1.49

Sample number 14 15 16 17 Total

Height(m) 1.55 1.56 1.56 1.62 24.34

The mid point in the table is point number 9 so the median value is 1.44 m. If we had an even

number of samples, say 18, then there would be two values in the middle and we should average the

two to get the median. The median and the mean are almost the same value but in some cases they

can be very different. Comparing the mean to the median for a set of data can tell you a lot about

the distribution or weighting of the sample. For example if the mean is much larger than the

median, it would mean that you have a lot of tall children.



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8. MODE

The mode is the most frequently occurring value. In the example repeated below, this will be the

class with a mark of 1.45 since there are six in this group.

x (mark) 1.25 1.35 1.45 1.55 1.65

f. 2 5 6 3 1 total

f x 2.5 6.75 8.7 4.65 1.65 24.25

It is quite possible that the mode is not unique because the same maximum figure could occur more

than once the distribution. If we don't have grouped data, there is no mode unless several occur at

the same precise height. This leads us on to the next section.

SELF ASSESSMENT EXERCISE No. 1

1. The hardness of ten steel samples was measured and the results were as follows.

Sample 1 2 3 4 5 6 7 8 9 10

Hardness 90 92 95 91 98 102 97 92 95 99

Calculate the mean. Answer 95.1

2. The thickness of 20 steel strips was measured in mm and tabulated as shown.

Sample 1 2 3 4 5 6 7 8 9 10

Thickness 19.8 19.9 19.9 20.1 20.1 19.9 20.2 19.7 19.7 19.9

Calculate the mean. Answer 19.92

3. A total daily taking in a supermarket over a period of a week is shown in the table below. The

figures are rounded off to the nearest 100. Draw a PIE chart and a BAR graph to illustrate how

the daily takings are distributed as part of the weekly total.

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

2300 2500 3100 5200 6700 8200 1900

(The solution is at the end of the tutorial)



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WORKED EXAMPLE No. 1

The graph shows the number of phone calls made at the same time to a call centre plotted over

a 24 hour period. The corresponding ogive is also given.

How many calls are received at 3 hours?

What % of the total calls is made between 24 hours and 18 hours?

SOLUTION

From the frequency graph we see that at 3 hours the number of calls being made is about 1.4 so

it would be either 1 or 2 on average.

From the ogive we see that at 18 hours the % is 7% and at 24 hours it is 93% so between these

hours 66% of the calls take place.


A company manufactures steel bars of nominal diameter 20 mm and cuts them into equal

lengths. The diameter of each length is measured at the middle for the purpose of quality

control and rounded to the nearest 0.05 mm. The results for 20 bars are given below.

Produce a frequency distribution table using bands of 0.1 mm. Draw a histogram. Draw the cumulative frequency diagram. Determine the mean, median and mode of the samples.

Use the tables and plots to show your solutions

Sample 1 2 3 4 5 6 7 8 9 10

diameter 19.9 19.8 20.1 19.9 19.7 20.1 20.0 19.6 19.7 20.1

Sample 11 12 13 14 15 16 17 18 19 20

diameter 20.2 20.0 19.9 19.8 20.1 20.0 19.7 19.6 19.9 20.2



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SOLUTION

Total = 398.3 Total samples n = 20 mean = 398.3/20 =19.915

RANKED ORDER

Sample 1 2 3 4 5 6 7 8 9 10

diameter 19.6 19.6 19.7 19.7 19.7 19.8 19.8 19.9 19.9 19.9

Sample 11 12 13 14 15 16 17 18 19 20

diameter 19.9 20.0 20.0 20.0 20.1 20.1 20.1 20.1 20.2 20.2

The median is the middle value of the samples when they are ranked in order and this is the 9 th

sample so the median is 19.9 mm.

The most frequently occurring values are 19.9 and 20.1 so there is no unique mode.

FREQUENCY DISTRIBUTION TABLE

d 19.55- 19.65- 19.75- 19.85- 19.95- 20.05- 20.15- 20.25 TotalsMark 19.6 19.7 19.8 19.9 20 20.1 20.2

f 2 3 2 4 3 4 2 20f d 39.2 59.1 39.6 79.6 60 80.4 40.4 398.3cum.f 2 5 7 11 14 18 20

n = 20 Mean = 398.3/20=19.915

The median divides the ogive into two vertically and this appears to be just 19.92. Again thefigure taken from the ranked table is the more accurate. Remember the ogive is plotted using

the upper limits of bands.



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A number of resistors were measured and rounded off to the nearest 0.5 . The frequency table

resulting is shown. Plot a frequency polygon and bar graph and determine the mean, median

and mode.

f 1 2 4 8 15 10 5 2 1R 119.25 119.75 120.25 120.75 121.25 121.75 122.25 122.75 123.25

SOLUTION

Complete the table by calculating the accumulative frequency

If we rank the data in order we get

f 1 2 4 8 15 10 5 2 1 Total = 48R 119.25 119.75 120.25 120.75 121.25 121.75 122.25 122.75 123.25fR 119.25 239.5 481 966 1818.75 1217.5 611.25 245.5 123.25 Total = 5822

Mean = 5822/48 = 121.23

Plot the frequency distribution and we get the following.

The median is the middle class and this is 121.25

The most frequent value is 121.25 and this is the mode.



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SELF ASSESSMENT EXERCISE No. 2

1. The accuracy of 100 instruments was measured as a percentage and the results were grouped

into bands of 1% as shown. Find the mean, median and mode.

Range Mid freq

61.5-62.5 62 162.5- 63 2

63.5- 64 3

64.5- 65 4

65.5- 66 8

66.5- 67 12

67.5- 68 13

68.5- 69 18

69.5- 70 14

70.5- 71 10

71.5- 72 5

72.5- 73 473.5- 74 3

74.5- 75 2

75.5-76.5 76 1

Answers 68.88, 69 and 69

2. The breaking strengths of 150 spot welds was measured in Newton and grouped into bands of20 N as shown. Find the mean, median and mode.

Range f

160-10 2

180-200 6

200-220 10

220-240 28

240-260 50

260-280 31

280-300 15

300-320 8

Answers 251.47 N, 230 N and 250 N



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3. The diameters of a number of components are measured to the nearest 0.1 mm. The distribution

is shown.

Diameter mm 9.6 9.7 9.8 9.9 10.0 10.1 10.2 10.3 10.4

Number 3 9 36 88 122 90 44 7 1

Draw the histogram and ogive and deduce the mean, the median and the mode.

(mean = 10.001 mm, median = 10 mode = 10 (the most frequent value)

4. The graph shows the number of hours of daylight at a particular latitude plotted against the timeof the year. The grid lines correspond to the middle of the month. How many hours of daylight

would be expected in mid August? What would be the total hours of daylight received as a %

between mid May and mid July?

(Answers 14 and approximately 70%)



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SOLUTIONS TO SELF ASSESSMENT EXERCISE No.1

Monday

Tuesday

Wed

Thursday

Friday

Saturday

Sunday

PIE CHART

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

9,000

Monday Tuesday Wed Thursday F riday Saturday Sunday

Series1

BAR GRAPH


edexcel unit 04 outcome 3 t1

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