The title of my investigation is ‘The Average student’. I am going to investigate into whether the students in my year (11) are normally distributed. I have chosen to investigate into this topic because we were required by this subject to produce a piece of coursework assignment, so we decided this would be easiest and most feasible. This leaves me with the predicament of trying measuring certain people in my year group. Are the people in my year at completely different ends of the scale or are they round about equal.

To determine a population for my course-work, I am going to go to each form and then measure people’s height and shoes size. My year compromises of 5 forms. Each form contains roughly 30 students giving a total population of 150 students. Obviously this population is far too vast, so I what I propose to do is to number each student in a form and then I will ask someone who hasn’t seen the number configuration to pick a random number. I then match up the numbers with the corresponding students and select 10 students from each form, 5 boys and 5 girls. I will therefore end up with a sample of 50 students, fulfilling the criteria of a sample of at least 50 items of single variable data. To ensure that my sample of students is as accurate as possible, I have cross-referenced my data with that of other students carrying out the same task. They have confirmed that these are the correct measurements, my cross-referencing will not affect my mathematics skills as my skills will be used on the data, no matter what the data is.

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On this page I have included results that I calculated with the help of a spreadsheet application (Microsoft Excel).

I can then later compare my own calculated results with those shown below. I have decided to pair the boys and girls off on separate charts

I have done the process of Standard deviation

Girls:

Mean Height165.4

Standard Deviation28.15296373

1st Quartile103.75

Median117.5

3rd Quartile129.25

Mean Shoe size5.6

Standard Deviation28.15296373

1st Quartile103.75

Median117.5

3rd Quartile129.25

Boys:

Mean120.1346154

Standard Deviation28.15296373

1st Quartile103.75

Median117.5

3rd Quartile129.25

To get a visual idea of the spread of my data, I decided to represent it in a stem and leaf diagram:

707

808015

90124537

100505554345

1101745448

120911023084521

1300690

14078

1506

160742

170

18016

190

200

2103

N = 52 156 6 represents 156 mins.

Stem and Leaf diagram showing the total duration of a sample of 52 films (unsorted)

To help me when constructing a cumulative frequency diagram, I have sorted the above diagram:

707

800158

90123457

100034455555

1101444578

120001112234589

1300069

14078

1506

160247

170

18016

190

200

2103

N = 52 156 6 represents 156 mins.

Stem and Leaf diagram showing the total duration of a sample of 52 films

(sorted)

As you can se from the above diagram

• The shortest film has a duration of 77 mins, and the longest is 213 mins long

• Most films are in the group 120-130 mins.

I have now decided to construct a frequency table so that I can draw a cumulative frequency graph, which will enable me to draw a box and whisker plot, and therefore visually see any or all outliers in my data. Here follows my table:

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ClassFrequencyCumulative Frequency

70=<x<8011

80=<x<9045

90=<x<100611

100=<x<110920

110=<x<120727

120=<x<1301239

130=<x<140443

140=<x<150245

150=<x<160146

160=<x<170349

170=<x<180049

180=<x<190251

190=<x<20000

200=<x<21000

210=<x<22011

Total 52

See overleaf for cumulative frequency diagram.

From Graph:

First Quartile – 129 mins

Mean – 117 mins

Third Quartile – 103 mins

From my sorted stem and leaf diagram it is possible to calculate median, first and third quartiles of my data, and so will now do so, so that I can compare these results to those obtained from my graph.

First Quartile = ¼ x 52 + ½ = position 13.5

Value = 103 mins

Median = ½ x 52 + ½ = position 26.5

Value = 117 mins

Third Quartile = ¾ x 52 + ½ = position 39.5

Value = 129 mins

As you can see, my graph values deviate from the calculated values. Therefore as the calculations will probably be more accurate than the graph results, when I draw my box and whisker plot, I will use the calculated values.

See below for box and whisker plot:

I am now going to calculate the mean and standard deviation of my data so that I can see where the outliers of my data, if any, are.

Mean

Standard Deviation

In most data sets

• About 2/3 of the values lie within 1 standard deviation of the mean

• About 95% of the values lie within 2 standard deviations of the mean

• About 99.5% lies within 3 standard deviations.

To see if this is so with my data, I will now perform the relevant calculations:

Mean = 120.13

120.13 + 27.91 = 148.04

120.13 – 27.91 = 92.22

Therefore about 2/3 of my data should lie between 92 mins and 148 mins. That is, about 34 values should be within this range. About 38 values are in this range, so this piece of information is relevant for my information.

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Mean = 120.13

120.13 + 2(27.91) = 175.95

120.13 – 2(27.91) = 64.31

Therefore about 95% of my data should lie in the range of 64 mins to 178 mins. About 49 values should be in this range. About 49 values are in this range, and so I have identified that the outliers in my data are 181, 186 and 213 mins.

SUMMARY OF WHAT I FOUND AND ACCURACY

I began this investigation with the aim of finding out the average time of a song and determining how long or short a song can be without the audience being bored or not pleased, respectively.

I found that the average duration of the sample of songs that I used was about 235seconds. The shortest film in my sample was 125.4 seconds long, and the longest was 325.2 seconds. About 2/3 of the sample of films fell between 182.seconds and 284. seconds . Therefore my media studies coursework should aim to be between about 182 seconds long and about 284seconds long. I will aim to make the song to be about 233secondslong in total length, as this is the average between these two values. I think that this investigation has been a success because I have achieved what I set out to achieve: what length my song should be for maximum audience enjoyment.

In terms of accuracy, I think that this investigation has been a success. I have constructed a table to show my graph values, where applicable, my previously calculated values and the computer’s values:

Graph ValueMy calculated valueComputer value

Mean233.5236.004

Standard Deviation51.0855.85029

Median258

mode202.2

As you can see, the only real discrepancy occurs with the computer’s standard deviation, and I can only conclude that this is because it has used a mean with more decimal places, and therefore it is more accurate. It also occurred because the calculated standard deviation was found from grouped data and therefore only an estimate. Graph discrepancies occur because of inaccurate drawing techniques. Other than that, my values are, on the whole, pretty accurate.

The quality of the investigation could be improved by using a more varied population, in an ideal world this would be an Internet site of every film ever made. As this does not exist, the site that contains the most comprehensive list of films will have to do. Which in this case is the site that I used, the HMV superstore site. My method of collection would not vary even if I used another site. If I used three randomly picked numbers for each category, then I would end up with a sample of 78 films, which I think is too produce an accurate report on.

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I think that this investigation was a success, as it helped me to gain some information that will help me in another topic, and because it was completed as accurately as I could.