Measures of Dispersion 1 Range Largest observation smallest
Measures of Dispersion 1. Range = Largest observation – smallest observation Example 1 : 1, 9, 6, 7, 12, 8, 3, 10, 11 Arrange the data in ascending order 1, 3, 6, 7, 8, 9, 10, 11, 12 Range = 12 – 1 = 11 If the range is SMALL Data are close to each other If the range is BIG Data are disperse a lot to each other Range is not a good measure of dispersion because it only depends on the highest & the smallest value Example 2 : 50, 52, 53, 54, 55, 56, 57, 90 Range = 90 – 50 = 40 If there is an extreme data ( 90 ), then the range will give a wrong picture of the data.
Range for Grouped Data Example 3 Marks 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 85 – 89 Frequency 2 3 8 13 14 6 3 Range = Largest observation – smallest observation = mid-point of the highest class – mid-point of the lowest class = 87 – 52 = 35 1
Quartile for Ungrouped Data 1. First quartile / lower quartile (Q 1) = the measure = the 25% of the total number of data 2. Third quartile / upper quartile (Q 3) = the measure = the 75% of the total number of data 3. Second quartile / median (m) = the Example 1 Find the Q 1, median & Q 3 1, 9, 6, 7, 12, 8, 3, 10 Solution Step 1 : Arrange the data 1, 3, 6, 7, 8, 9, 10, 12 = the 50% of the total number of data 4 numbers median Q 1 = 3 + 6 2 Q 3 = 9 + 10 2 measure
Example 2 Find the Q 1, median & Q 3 1, 9, 6, 7, 10, 8, 3 Solution Step 1 : Arrange the data 1, 3, 6, 7, 8, 9, 10 3 numbers median Q 1 = 3 Q 3 = 9 Example 3 Find the Q 1, median & Q 3 pg. 196 Example. 29 Number of fishes 0 1 2 3 4 5 Frequency 1 5 8 7 3 1 n = 25 Median = x 13 = 2 fishes Q 1 = x 6 + x 7 2 = 1 + 2 = 1. 5 fishes 2 Q 3 = x 19 + x 20 2 =3+3 = 3 fishes 2
Interquartile Range & Semi - interquartile Range Interquartile Range = Q 3 – Q 1 • Interquartile range gives the range for 50% of the total number of data that is located in the centre of the set. Q 1 Q 3 25% 75% 50% • more meaningful if compared to range because it avoids giving a wrong picture if extreme data appears. Semi - Interquartile Range = ½ ( Q 3 – Q 1 ) • better than interquartile range because its value is very close to the standard deviation Exercise 5. 8 pg. 196 4 Q. 1, 2, 3 &
Quartile for Grouped Data Example 1 : The table shows the marks of 50 students. Calculate the Q 1 & Q 3 mark. Marks 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 85 – 89 f 2 3 8 13 14 6 3 1 Method 1 : By Interpolation Marks f 50 – 54 2 2 55 – 59 3 5 60 – 64 8 13 65 – 69 13 26 70 – 74 14 40 75 – 79 6 46 80 – 84 3 49 85 – 89 1 50 Cumulative freq. First quartile, Q 1 = the Then, first quartile class = 60 - 64 5 59. 5 12. 5 13 Q 1 64. 5 Lower boundary By proportion (nisbah) Q 1– 59. 5 64. 5 – 59. 5 obs. = 12. 5 – 5 13 – 5 Therefore, Q 1 = 59. 5 + 4. 6875 = the 12. 5 th observation = 64. 1875 Upper boundary
Third quartile, Method 2 : By Formulae First quartile, Q 1 = the Marks f 50 – 54 2 2 55 – 59 3 5 60 – 64 8 13 65 – 69 13 26 70 – 74 14 40 75 – 79 6 46 80 – 84 3 49 85 – 89 1 50 Cumulative freq. obs. = the 12. 5 th obs. Then, first quartile class = 60 - 64 LB = 59. 5 FB = 5 f. Q 1 = 8 c = 59. 5 – 54. 5 = 64. 1875 Exercise 5. 9 pg. 203 Q. 8, 9 & 10
Example 2 : The table shows the marks of 50 students. Plot an ogive & estimate the Q 1 & Q 3 Marks 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 85 – 89 Frequency 2 3 8 13 14 6 3 1 Method 3 : From the Ogive ( Cumulative Frequency Curve ) Cumulative frequency 50 Solution Step 1 : Build a cumulative frequency table Step 2 : Plot an ogive Step 3 : Q 1 = the 12. 5 th observation Step 4 : From the ogive Upper Boundary Q 1 = 64 Q 3 40 30 20 Q 1 12. 5 10 64 0 49. 5 54. 5 59. 5 64. 5 72 69. 5 74. 5 79. 5 84. 5 “Less than” cumulative frequency < 49. 5 0 < 54. 5 2 < 59. 5 5 < 64. 5 13 < 69. 5 26 < 74. 5 40 < 79. 5 46 < 84. 5 49 < 89. 5 50 89. 5 Marks ( Upper boundary
Example 3 : The table shows the marks of 50 students. Plot an “more than” ogive & estimate the Q 1 & Q 3 Marks 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 85 – 89 Frequency 2 3 8 13 14 6 3 1 Solution 37. 5 th Q = theof If 164% the observation students the 12. 5 th observation fail Qthe estimate the 3 = paper, From the ogive passing mark. From the ogive Q 1 = 64 Q 3 = 72 Cumulative frequency 50 Upper Boundary 40 37. 5 Q 1 30 20 18 Q 3 12. 5 10 64 0 49. 5 54. 5 59. 5 64. 5 70 69. 5 72 74. 5 79. 5 84. 5 Exercise 5. 9 pg. 200 Q. 1, 2, 3, 4, 5, 6 & “More than” cumulative frequency > 49. 5 50 > 54. 5 48 > 59. 5 45 > 64. 5 37 > 69. 5 24 > 74. 5 10 > 79. 5 4 > 84. 5 1 > 89. 5 0 89. 5 Marks
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