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Question 2 (a) Mode = 4 hours Mean = 6.9 hours (to 1 dp) Ordered array: 3, 4, 4, 5, 8, 9, 10, 12 There are 8 values so the median = 41/2th value in order ie, (5 + 8)/2 = 6.5 hours Median = 6.5 hours (b) Range: Max.=12 Min.=3 Range = 12 - 3 = 9 Lower quartile = n/4th value in order ie, 8/4 (2nd) value = 4 Upper quartile = 3n/4th value in order ie, (3*8)/4 (6th) value = 9 Inter quartile range = 9 - 4 = 5 x = 55 x2 = 455 Population variance: 2 = x2 - x n n 2 = 455 - 55 8 8 2 = 56.875 - 6.8752 = 9.609375 The data are from a sample, hence the correction factor is required: s2 = n/(n-1)*2 = 8/7*9.609375 = 10.982143 s = 10.982143 = 3.314 (to 3 dp) (c) Mode = 4 hours Mean = (55+30)/9 = 85/9 = 9.4 hours (to 1 dp) Median = 5th value in order = 8 hours Range: Max.=30 Min.=3 Range = 30 - 3 = 27 Inter quartile range: Lower quartile = n+1/4th value in order ie, 10/4 (2.5th) value = 4 Upper quartile = 3(n+1)/4th value in order ie, (3*10)/4 (7.5th) value = 11 Inter quartile range = 11 - 4 = 7 Standard deviation: x = 85 x2 = 1355 Population variance: 2 = x2 - x n n 2 = 1355 - 85 9 9 2 = 61.358025 Again, the correction factor is required: s2 = n/(n-1)*2 = 9/8*61.358025 = 69.027778 s = 69.027778 = 8.308 (to 3 dp) (d) In the revised data set the unusual value of 30 hours would affect the mean, therefore the median value of 8 hours is the most appropriate measure to use. This would be partnered with the inter-quartile range. Question 4 Maximum = 9 Minimum = 0 so Range = 9 – 0 =9 Mode = 2 or 3 errors (both have the same frequency) Hence the distribution is bi-modal. There are 100 values so median = 501/2th ordered value Lower quartile = 25th in order (n/4 approximation OK here) Upper quartile = 75th in order (3n/4 approximation OK here) x 0 1 2 3 4 5 6 7 8 9 f 15 18 19 19 10 8 7 2 1 1 Position 1 to 15 16 to 33 34 to 52 53 to 71 72 to 81 82 to 89 90 to 96 97 to 98 98 to 99 99 to 100 Median = 2 errors Lower Quartile = 1 error Upper Quartile = 4 errors Hence inter-quartile range = 4 – 1 = 3 errors To find mean and standard deviation a table of calculations is required: x = number of errors, f = number of days x 0 1 2 3 4 5 6 7 f 15 18 19 19 10 8 7 2 fx 0*15 = 0 1*18 = 18 2*19 = 38 57 40 40 42 14 fx2 02*15 = 0 12*18 = 18 22*19 = 76 171 160 200 252 98 8 9 1 1 f= 100 8 9 fx = 266 64 81 2 fx = Mean = fx/f = 266/100 = 2.66 errors 2 = fx2 - fx n n 2 = 1120 - 266 100 100 2 = 11.2 – 7.0756 = 4.1244 The data are from a sample so the correction factor must be applied: s2 = n (n - 1) s2 = 100 * 2 /99 * 4.1244 = 4.1660606 standard deviation (s) = 2.041 (to 3 dp) The data provided are discrete –so to display the distribution a bar chart can be used: Figure 3.1 Distribution of errors made No errors made per day by computer system over 100 days in auditing dept. Frequency 20 15 10 5 0 0 1 2 3 4 5 6 7 8 9 Errors made per day From the graph it is apparent that the data provided are quite skewed, so the the most appropriate descriptive measures to use are the median and inter-quartile range. Question 6 (a) Firstly, the data provided indicate that no one has been with the company for more than 50 years. At present the last category is open ended therefore 50 years can be used as the upper limit. The categories are also uneven in size so in order to construct a histogram, frequency densities are required. These are shown below: Years service 0-<5 5 - < 15 15 – < 25 25 - < 35 35 - < 50 Frequency 105 231 173 85 31 Class width 5 10 10 10 15 Frequency density 105 /5 = 21 231 /10 = 23.1 173 /10 = 17.3 85 /10 = 8.5 31 /15 = 2.07 The resulting histogram is: Figure 3.4 Histogram of years of service Number of years employees of JFS chemicals have been with the company. 25 Frequency Density 20 15 10 5 10 20 50 30 40 Years of service (b) The histogram indicates that the modal group is 5 to less than 15 years. (c) Required calculations are: Years service 0-<5 5 - < 15 15 – < 25 Frequency (f) 105 231 173 Mid-pt (x) 2.5 10 20 fx fx2 105* 2.5= 262.5 2310 3460 105*2.52 =656.25 23100 69200 25 - < 35 35 - < 50 85 31 625 30 42.5 2550 1317.5 9900 76500 55993.75 225450 Mean = fx/f = 9900/625 = 15.84 years (to 2dp) Population variance: 2 = fx2 - fx n n 2 = 225450 625 9900 625 2 = 360.72 – 250.9056 = 109.8144 Data relates to all employees of the company so they form a population. Standard deviation = 109.8144 = 10.48 (to 2dp) (d) In order to estimate the mean and inter-quartile range for the data an ogive is required: Figure 3.5 Ogive of years with company Cumulative Frequency Ogive showing amount spent by 81 shoppers on luxury goods 90 80 70 60 50 40 30 20 10 0 0 20 40 60 80 100 120 Amount (£) As there are 625 employees in the company the median position is 625+1/2 = 313, the lower quartile position is (625+1)/4 = 156½ and the upper quartile position is 3(625+1)/4 = 469½ From the graph these positions relate to approximately 14 years, 7 years and 23 years respectively. Hence median = 14 years and inter-quartile range = 23 – 7 = 16 years (e) From the histogram created in part (a) the data is slightly skewed in shape therefore the median and inter quartile range are the most appropriate measures of central tendency and dispersion to use.