Hi and good day ,

Based on the secondary ( ungrouped data ) dataset given , please help me find :

1. . Measures of dispersion

• Range
• Interquartile range
• Semi-interquartile range / quartile deviation
• Mean deviation / average absolute deviation
• Variance
• Standard deviation

1. . Measure of skewness
2. Pearson coefficient of skewness

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• THANK YOU

2 distribution (A and B) may have the same central locations (mean, median or mode) but different dispersions or spreads lfthe data are widely dispersed, the central location is said to be less representative of the whole data The more spread out or dispersed the data, the larger is the measures ofdispersion If the data is more concentrated, the measures of dispersion is smaller If the data are the same, the measures of dispersion will be zero Therefore, these measures of dispersion are always give positive values The range is the difference between the largest and the smallest value in a set of data ' Range =UB LB Range = largest value smallest value A simple measure of dispersion : 35.5 _ 23'5 =12 It measure the total spread of the data a Range is really inuenced by the extreme valuesin'z ! J Eg: Find the range for the following data. Range for Grouped om 12.5, 23.2, 51.1, 14.6, 33.3, 42.5 The range is the difference between the UB of Solution: Range = largest value 'smallest value = 51.1 12.5 \".jf' =3as the highest class and the LB of the lowest class Range: UB- LB Interquartile Range Semi-interquartile Range / Quartile Deviation The difference between the third and first quartiles in a set of data Semi-interquartile range = Interquartile range / 2 Interquartile range = Q3 - Q1 Eg : Find the semi-interquartile range This measure the spread in the middle 50% of the Semi-interquartile range data = Interquartile range / 2 =10.5 /2 It is not influenced by the extreme values =5.25 Eg : The weight (kg) for 20 students are 66, 54, 48, 52, 67, 42, 44, 56, 63, 60, 51, 49, 50, 54, 53, 60, 48, 59, 54 and 50. Find the interquartile range. Q1 =49.25, Q3 = 59.75 Interquartile range = Q3 - Q1 =59.75 - 49.25 = 10.5Mean Deviation / Average Absolute Deviation Eg: Given 45, 66, 53, 21, 66. Find the mean deviation. 45+ 66+ 53+ 21+66 Mean= = 50.2 Ex- mean 5 Mean deviation = n Mean deviation Ex- mean where x= the observation n 45- 50.2 + 66 - 50.2 + 53-50.2+ 21-50.2 + 66-50.2 n = the number of observation = 5 = 12.56Variance Standard Deviation A primary measure of variation. The Measure the fluctuations of data value of this measurement is in the values above and below its mean. original unit of the data Its computation results in squared Large standard deviation mean large units variability within the data set Variance and Standard Deviation Variance and Standard Deviation for Ungrouped Data (Eg: Given 45, 66, 53, 21, 66. Find the variance and 17 the standard deviation Variance: n- = n n- n 1 = (452 + 662 +532 +212 +662)_(251)2 5-1 5 = 346.7 Standard deviation: S= VS = V346.7 =18.6199 S = VS2 SDetermine the difference. between the mean and the mode of the distribution Pearson Coefficient of Skewness x- mode 3( x - median Skewness = or Skewness = S S skewness = 0, symmetrical skewness = +ve value, skewed to the right or positively skewed skewness = -ve value, skewed to the left or negatively skewedHeart failure prediction dataset Age , X Resting blood pressure , Y 40 140 49 160 37 130 48 138 54 150 39 120 45 130 54 110 37 140 58 120 39 130 49 136 42 120 54 140 38 115 43 120 60 110 36 120 43 100 44 120 49 100 44 120 40 124 36 150 53 130 51 130 53 124 56 120 54 113 41 125