Final HW 2.4

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Jun 12, 2024

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2.4 2.93 (a) Range is difference of maximum data value and minimum data value. Highest amount of taxes paid per person is $7,764 and lowest amount of taxes paid per person is $2,782. So the range for the amount of taxes paid per person is $ 7764 – 2782 = 4982 (b) Range is difference of maximum data value and minimum data value. Highest percentage of personal income paid in taxes per person is 16.6% and lowest percentage of personal income paid in taxes per person is 9.1%. So the range for percentage of personal income paid in taxes per person is 16.6% - 9.1% = 7.5% 2.95 Let n is the number of data values. So = ∑ (x) - nxbar = ∑ (x) – n [∑x /n] = ∑ (x) - ∑x = 0 Hence, sum of deviation is always zero. 2.98 (a) The objective to calculate the range. The highest value is 9 and lowest value is 2. range = highest value – lowest value = 9 – 2
2.4 = 7 Therefore, the range is 7. (b) The objective is to calculate the variance . Sample data is 2, 4, 7, 8, 9. Here, the number of data values is 5. That is n = 5 The mean of the sample is calculated as follows: = 2+4+7+8 +9 / 5 = 30 /5 = 6 Thus, the sample mean is 6. X X – X̅ (X – X̅) ² 2 2 – 6 = -4 (2-6)2 = 16 4 4 – 6 = -2 (4-6)2 = 4 7 7 – 6 = 1 (7-6)2 = 1 8 8 – 6 = 2 (8-6)2 = 4 9 9 – 6 = 3 ( 9-6) 2 = 9 ∑x1 = 30 ∑( x1 – X̅) = 0 ∑( x1 – X̅)2 = 34 The variance is calculated as follows: s 2 = ∑ ( x 1 − ¯ x ) ² n − 1 = 34 / 5-1
2.4 = 34 /4 = 8.5 Therefore, the sample variance is 8.5 (c) The objective is to calculate the standard deviation. The standard deviation is calculated as follows: s = √ s 2 √ 8.5 = 2.9155 Therefore, the standard deviation is 2.9155 2.100 (a) Sample data is 7, 6, 10, 7, 5, 9, 3, 7, 5, 13. Here number of data values is 10 so. Therefore the mean of the sample is X̅ = 7+6+10+7+5+9+3+7+5+13 / 10 = 7.2 Now let us find out the variance of sample. Following table shows the calculations: X X – X̅ (X – X̅) ² 7 -0.2 0.04 6 -1.2 1.44 10 2.8 7.84 7 -0.2 0.04 5 -2.2 4.84 9 1.8 3.24 3 -4.2 17.64
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2.4 7 -0.2 0.04 5 -2.2 4.84 13 5.8 33.64 Total 73.6 So the variance of the sample is = 73.6 / 10-1 = 8.18 Hence, the sample variance is 8.18. (b) Following is the calculation for variance using formula 2.9. x x² 7 49 6 36 10 100 7 49 5 25 9 81 3 9 7 49 5 25 13 169 ∑x = 72 ∑x² = 592
2.4 (c) Standard deviation of the data will be s = √ 8.18 s = 2.86 2.102 (a) Range is difference of highest and lowest value. So range will be range = highest value – lowest value = 48 - 21 = 27 Hence, the range of the data is 27 (b) Size of random sample is 10 so n = 10 Therefore the mean of the sample is x̄ͅ > = 36+26+48+28+45+21+21+38+27+32 / 10 = 322 / 10 = 32.2 Now let us find out the variance of sample. Following table shows the calculations:
2.4 X X – X̅ (X – X̅) ² 36 3.8 14.44 26 -6.2 38.44 48 15.8 249.64 28 -4.2 17.64 45 12.8 163.84 21 -11.2 125.44 21 -11.2 125.44 38 5.8 33.64 27 -5.2 27.04 32 -0.2 0.04 Total 795.6 So the variance of the sample is = 795.6 / 10 – 1 = 795.6 / 9 = 88.4 Hence, the sample variance is 88.4. (c) Standard deviation of the data will be s = √ 88.4 s = 9.4
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2.4 Hence, the standard deviation of data is 9.4 2.105 (a) Number of towns randomly selected is 12 so X̄ͅ = 559+815+767+668+651+895+1106+1375+861+1559+888+1106 / 12 = 11250 / 12 = 937.5 Hence, the mean will be 937.5. (b) finding out the variance of sample X X – X̅ (X – X̅) ² 559 -378.5 143262.3 815 -122.5 15006.25 767 -170.5 29070.25 668 -269.5 72630.25 651 -286.5 82082.25 895 -42.5 1806.25 559 -378.5 143262.3 1106 168.5 28392.25 1375 437.5 191406.3 861 -76.5 5852.25 1559 621.5 386262.3 888 -49.5 2450.25 1106 168.5 28392.25 1106 168.5 28392.25 Total 986613 So the variance of the sample is = 986613 / 11 = 89692.1
2.4 So the standard deviation will be s = √ 89692.1 = 299.5 Hence, the standard deviation is 299.5. 2.10 9 (a) Following is the dot plot of the data: 24 26 28 30 32 34 0 1 2 3 4 5 6 Recruits for a police academy Exercise capacity (in minutes) Frequency (b) There are 20 recruits so. n = 20 Therefore the mean of the sample is X̅ = 25+27+30+33+30+32+30+34+30+27+26+25+29+31+31+32+34+32+33+30 / 20 = 601 / 20 = 30.05 (c) Range is difference of highest and lowest value. So range will be
2.4 range = highest value – lowest value = 34 – 25 = 9 Hence, the range of the data is 9. (d) finding out the variance of sample. X X – X̅ (X – X̅) ² 25 -5.05 25.5025 27 -3.05 9.3025 30 -0.05 0.0025 33 2.95 8.7025 30 -0.05 0.0025 32 1.95 3.8025 30 -0.05 0.0025 34 3.95 15.6025 30 -0.05 0.0025 27 -3.05 9.3025 26 -4.05 16.4025 25 -5.05 25.5025 29 -1.05 1.1025 31 0.95 0.9025 31 0.95 0.9025 32 1.95 3.8025 34 3.95 15.6025 32 1.95 3.8025 33 2.95 8.7025 30 -0.05 0.0025 Total 148.95 So the variance of the sample is = 148.95 / 20 – 1
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2.4 = 7.84 Hence, the sample variance is 7.84. (e) Standard deviation of the data will be s = √ 7.84 = 2.8 Hence, the standard deviation of data is 2.8 (f) Following is the required graph: 24 26 28 30 32 34 0 1 2 3 4 5 6 Recruits for a police academy Exercise capacity (in minutes) Frequency Red line started at mean and has length 2.8 (equal to standard deviation). Blue line shows the range (9) of the data. (g) Range covers the distribution horizontally. Standard deviation shows the variation of data values from the mean. Excluding data value 30, distribution seems to be rectangular.
2.4 X X – X̅ (X – X̅) ² 46 -4 16 55 5 25 50 0 0 47 -3 9 52 2 4 ∑x1 = ∑( x1 – X̅) = 0 ∑( x1 – X̅)2 = 54 Range of set 1: Range is difference of highest and lowest value. So range will be range = highest – lowest = 55 – 46 = 9 Hence, the range of the data of set 1 is 9. X X – X̅ (X – X̅) ² 30 -20 400 55 5 25 65 15 225 47 -3 9 53 3 9 ∑x1 = ∑( x1 – X̅) = 0 ∑( x1 – X̅)2 = 668 Range of set 2: Range is difference of highest and lowest value. So range will be range = highest – lowest = 65- 30 = 35
2.4 Hence, the range of the data of set 2 is 35. Above measures show, the set 2 has more variations in data values than set 1 X̅ ( x 0304, 0305 – alt + x)
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