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Published byCori Logan Modified over 9 years ago
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Measures of central tendency: Mode, median, mid-range and mean
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The mode: – the most common (i) The number of errors made by 11 students: 4 6 7 5 9 10 7 6 4 7 8 Rearranging the numbers in order 4 4 5 6 6 7 7 7 8 9 10 The mode is 7 What type of data is this ? “This is discrete and ungrouped data ”
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The mode: – the most common (ii) The number of children in 25 families: (grouped data) Number of children: 0 1 2 3 4 5 Number of families: 3 5 9 4 3 1
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The mode: – the most common (ii) The number of children in 25 families: (grouped data) Number of children: 0 1 2 3 4 5 Number of families: 3 5 9 4 3 1 What is the mode? There are 9 families with 2 students so the mode is 2 (N.B not 9!)
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The mode: – the most common (iii) Heights of 50 female students What type of data is this? (grouped continuous data) Height h cmFrequency 157 < h 159 4 159 < h 16111 161 < h 16319 163 < h 165 8 165 < h 167 5 167 < h 169 3 Total50 What is the mode??
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The mode: – the most common (iii) Heights of 50 female students (grouped continuous data) Height h cmFrequency 157 < h 159 4 159 < h 16111 161 < h 16319 163 < h 165 8 165 < h 167 5 167 < h 169 3 Total50 The modal class is 161 < h 163 with 19 students 161 and 163 are the lower and upper class boundaries
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2. The median – the middle value 1.The number of errors made by 11 students: 4 6 7 5 9 10 7 6 4 7 8
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2. The median – the middle value (i).The number of errors made by 11 students: 4 6 7 5 9 10 7 6 4 7 8 Rearranging the numbers in order 4 4 5 6 6 7 7 7 8 9 10 Number of items n = 11 median is (n+1)/2 th value = 6 th value = 7
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2. The median – the middle value (i)If another student is added who made 5 mistakes 6 7 5 9 10 7 6 4 7 8 4 5 Rearranging the numbers in order 4 4 5 5 6 6 7 7 7 8 9 10 Number of items n = 12 Median is (n+1)/2 th value = 6.5 th value = (6+7)/2 = 6.5
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2. The median – the middle value (ii) Number of children in 25 families(grouped data) Number of children : 0 1 2 3 4 5 Number of families (freq): 3 5 9 4 3 1
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2. The median – the middle value (ii) Number of children in 25 families(grouped data) Number of children : 0 1 2 3 4 5 Number of families (freq): 3 5 9 4 3 1 Cumulative frequency : 3 8 17 21 24 25 Median value is (25 + 1)/2 th = 13 th value = 9 (values 9 to 17 are all 2) Cum Freq is a running total
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2. The median – the middle value (iii) Heights of 50 students (Grouped continuous data) Height h cmFrequency f 157 < h 159 4 159 < h 161 11 161 < h 163 19 163 < h 165 8 165 < h 167 5 167 < h 169 3 totals 50
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2. The median – the middle value Median class contains ½( 50 + 1)=25.5 th item. Median class is161 < h 163 Height h cmFrequency fCumulative freq 157 < h 159 44 159 < h 161 1115 161 < h 163 1934 163 < h 165 842 165 < h 167 547 167 < h 169 350 totals50
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The mid-range -the value mid-way between the lowest and highest values (i) Student errors First re-arrange the data in numerical order 4 4 5 6 6 7 7 7 8 9 10 Mid-range is = (4 + 10)/2 =7
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The mean (or average): add up all the values and divide by the number of values A population is a collection of items (usually a large number of values); the mean of the population is A sample is a selection of the population; the mean of the sample is This is just a notation issue not really a big deal !
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The mean (or average) (i) Student errors: 4 6 7 5 9 10 6 7 4 7 8 = = = 6.64 to 3 s.f.
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or if we take a sample of just 4 students, say 7 9 6 4 Then == 6.5
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The mean (or average) Number of children in 25 families(grouped data) For grouped data where f is the frequency
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The mean (or average) ii) Number of children in 25 families(grouped data) Number of children (x): 0 1 2 3 4 5 Number of families (f) : 3 5 9 4 3 1 (xf): 0 5 18 12 12 5 = 2.08
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The mean (or average) Heights of female students (grouped continuous data) Height h cmf xi fxi f 157 < h 159 4632 159 < h 161 111760 161 < h 163 193078 163 < h 165 81312 165 < h 167 5830 167 < h 169 3504 totals508116 = 162.32 cm What value of x doe we use here? Mid-point x i 160 162 164 166 168 158
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