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Data handling - calculating means and standard deviations

Extracts from this document...

Introduction

ASSIGNMENT 1I

Type I

Name:

Candidate #: [               ]

School:

October 2003

Question 1

The table below shows the height for 60 students in centimeters:

Table 1

177

175

137

155

150

166

132

146

179

140

169

177

141

148

130

176

135

130

157

172

178

143

143

136

132

166

130

151

145

178

131

171

160

140

179

166

145

142

177

176

132

135

164

179

161

145

134

179

139

149

135

142

172

148

159

160

137

130

130

164

  • The mean () is calculated using the equation below:

 = image08.png

  • The standard deviation (image00.png) is calculated[1]* using the equation below:

image01.png= 17.08731661 image02.png

Question 2

  1. The table below shows the height of 60 students after adding 5 cm to each height:

Table 2

182

180

142

160

155

171

137

151

184

145

174

182

146

153

135

181

140

135

162

177

183

148

148

141

137

171

135

156

150

183

136

176

165

145

184

171

150

147

182

181

137

140

169

184

166

150

139

184

144

154

140

147

177

153

164

165

142

135

135

169

  • The mean () is calculated using the equation below:

 = image20.png

  • The standard deviation (image00.png) is calculated[2]* using the equation below:

image01.png= 17.08731661 image02.png

  1. The table below shows the height of 60 students after subtracting 12 cm from each height:

Table 3

165

163

125

143

138

154

120

134

167

128

157

165

129

136

118

164

123

118

145

160

166

131

131

124

120

154

118

139

133

166

119

159

148

128

167

154

133

130

165

164

120

123

152

167

149

133

122

167

127

137

123

130

160

136

147

148

125

118

118

152

  • The mean (image03.png) is calculated using the equation below:

 = image04.png

  • The standard deviation (image00.png) is calculated using the equation below:

image01.png= 17.08731661 image02.png

Adding 'a' to a set of data will result in an increase in the mean (image03.png) by 'a' while keeping the standard deviation (image00.png) for the data set unchangeable.

Subtracting 'a' from a set of data will result a decrease in the mean by 'a'.

Question 3

...read more.

Middle

28.0

33.8

35.4

28.2

29.6

26.0

35.2

27.0

26.0

31.4

34.4

35.6

28.6

28.6

27.2

26.4

33.2

26.0

30.2

29.0

35.6

26.2

34.2

32.0

28.0

35.8

33.2

29.0

28.4

35.4

35.2

26.4

27.0

32.8

35.8

32.2

29.0

26.8

35.8

27.8

29.8

27.0

28.4

34.4

29.6

31.8

32.0

27.4

26.0

26.0

32.8

  • The mean () is calculated using the equation below:

 = image07.png

  • The standard deviation (image00.png) is calculated[4]* using the equation below:

image01.png= 3.417463322 image09.png

Multiplying a set of data by 'a', where a>1 will result in the multiplication of the standard deviation (image00.png) by 'a'. The standard deviation will be divided by 'a' if If 0<a<1.


The mean (
image03.png) will be also multiplied by 'a'. The mean will be divided by 'a' if 0<a<1.

The changes caused by the multiplication of a data set by negative value (a<0) are investigated below:

The table below shows the same data set used in this question using a = - 3, where each value is multiplied by a;

Table 6

-531

-525

-411

-465

-450

-498

-396

-438

-537

-420

-507

-531

-423

-444

-390

-528

-405

-390

-471

-516

-534

-429

-429

-408

-396

-498

-390

-453

-435

-534

-393

-513

-480

-420

-537

-498

-435

-426

-531

-528

-396

-405

-492

-537

-483

-435

-402

-537

-417

-447

-405

-426

-516

-444

-477

-480

-411

-390

-390

-492

  • The mean (image03.png) is calculated using the equation below:

 = image10.png

  • The standard deviation (image00.png) is calculated using the equation below:

image01.png= 51.26194983 image11.png

...read more.

Conclusion

-22.9

-22.9

11.1

  • The mean () is calculated using the equation below:
  • = image19.png

Since the subtraction of 'a' from the data set values results in a subtraction of 'a' from the mean then the transformation is done by subtracting the mean which is 152.9 from each of the values of the data set.

152.9 – 152.9 = 0

(b) The table below shows the new data set which has a standard deviation=1

Table 17

10.4

10.2

8.0

9.1

8.8

9.7

7.7

8.5

10.5

8.2

9.9

10.4

8.3

8.7

7.6

10.3

7.9

7.6

9.2

10.1

10.4

8.4

8.4

8.0

7.7

9.7

7.6

8.8

8.5

10.4

7.7

10.0

9.4

8.2

10.5

9.7

8.5

8.3

10.4

10.3

7.7

7.9

9.6

10.5

9.4

8.5

7.8

10.5

8.1

8.7

7.9

8.3

10.1

8.7

9.3

9.4

8.0

7.6

7.6

9.6

  • The standard deviation (image00.png) is calculated using the equation below:

image01.png= 1

Since dividing the data set values by 'a' results the dividing of the standard deviation by 'a' then the transformation is done by dividing each of the data set values by the original standard deviation which is 17.1

17.1 / 17.1 = 1

(c) The table below shows the new data set which has a standard deviation=1

Table 18

1.4

1.3

-0.9

0.1

-0.2

0.8

-1.2

-0.4

1.5

-0.8

0.9

1.4

-0.7

-0.3

-1.3

1.4

-1.0

-1.3

0.2

1.1

1.5

-0.6

-0.6

-1.0

-1.2

0.8

-1.3

-0.1

-0.5

1.5

-1.3

1.1

0.4

-0.8

1.5

0.8

-0.5

-0.6

1.4

1.4

-1.2

-1.0

0.6

1.5

0.5

-0.5

-1.1

1.5

-0.8

-0.2

-0.1

-0.6

1.1

-0.3

0.4

0.4

-0.9

-1.3

-1.3

0.6

  • The mean () is calculated using the equation below:
  • = image19.png
  • The standard deviation (image00.png) is calculated using the equation below:

image01.png= 1

From parts (a) and (b) it is concluded that subtracting the mean then dividing over the standard deviation can complete the transformation.


[1]* Using Microsoft Excel function: STDEVP

[2]* Using Microsoft Excel function: STDEVP

[3]* Using Microsoft Excel function: STDEVP

[4]* Using Microsoft Excel function: STDEVP

[5]* Using Microsoft Excel functions: MEDIAN, QUARTILE

[6]* Using Microsoft Excel functions: MEDIAN, QUARTILE

...read more.

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