Statistics_and_Probability

STATISTICS AND PROBABILITY | HNC - CONSTRUCTION | | TASK ONE CONDITION NUMBER OF PERCENTAGE DEGREES RELATED CASES % ° Dermatitis 143 12.77 45.972 Muscular/Skeletal Injuries 214 19.11 68.796 Vibration White Finger 86 7.68 27.648 Mesthelioma 278 24.82 89.352 Diffuse Pleural Thickening 399 35.62 128.232 A Pie Chart Total number of cases is 143 + 214 + 86 + 278 + 399 = 1120 Therefore 100% of cases are 1120. DERMATITIS Percentage of cases that have dermatitis is (143 / 1120) x 100 = 12.77% 360 x 0.1277 = 45.972° MUSCULAR SKELETAL/INJURIES Percentage of cases that have muscular/skeletal injuries is (214/ 1120) x 100 = 19.11%. 360 x 0.1911 = 68.796° VIBRATION WHITE FINGER Percentage of cases that have vibration white finger is (86/1120) x 100 = 7.68% 360 x 0.0768 = 27.648° MESTHELIOMA Percentage of cases that have mesthelioma is (278/1120) x 100 = 24.82% 360 x 0.2482 = 89.352° DIFFUSE PLEURAL THICKENING Percentage of cases that have pleural thickening is (399/1120) x 100 = 35.62% 360 x 0.3562 = 128.232° The information in this chart would be a significant help if you were the Health and Safety Officer for a construction company. It would help you influence risk assessments in certain areas, for example, looking at which areas on site could cause Diffuse Pleural Thickening and Mesthelioma. This would allow steps to be taken to try to reduce the number of cases. This could be achieved by perhaps issuing more specialist PPE (Personal protective Equipment). It may also be a good idea to consider if there are any weaknesses in working practices identifying the need for additional training. Ensuring good hand washing facilities on site, with access to barrier creams, would possibly help to improve the number of cases of dermatitis. It is also good practice to ensure workers are following Health and Safety Guidelines and wearing appropriate gloves whilst working on site. A further survey would identify whether the muscular/skeletal injuries were a direct result of working on the site or caused by outside activities of the workers (e.g. sports etc). TASK TWO No. of power outlets in 24 houses: 10, 9, 15, 12, 12, 20, 9, 9, 11, 12, 12, 13, 14, 12, 16, 12, 11, 15, 16, 13, 13, 25, 24, 14 a) The mode (value that applies most frequently) is 12. The media (amount which separates the higher numbers from the lower numbers) is 13. The range (the difference between the highest and lowest values in the set) is 16. Number of Power Outlets | Tally | Frequency | 9 | iii | 3 | 10 | i | 1 | 11 | ii | 2 | 12 | vi | 6 | 13 | iii | 3 | 14 | ii | 2 | 15 | ii | 2 | 16 | ii | 2 | 20 | i | 1 | 24 | ii | 1 | 25 | i | 1 | b) The mean (sum of values) is 13.70833. The standard deviation (a statistical measure of variability) is 4.17528. c) A Histogram A Frequency Polygon | TASK THREE Table Q3 Year | Price £k | Year | Price £k | Year | Price £k | Year | Price £k | 1990 | 67.5 | 1995 | 62.6 | 2000 | 91.1 | 2005 | 164.5 | 1991 | 64.7 | 1996 | 66.7 | 2001 | 99.7 | 2006 | 174 | 1992 | 62.9 | 1997 | 72.8 | 2002 | 116.5 | | | 1993 | 61.5 | 1998 | 76.9 | 2003 | 135.3 | | | 1994 | 60.4 | 1999 | 82.7 | 2004 | 156.9 | | | When we look at the table below, we are able to predict the average house of a price in 2012. We have done this by using a regression line. This can be confirmed by adding the prices we are given, (67.5 + 64.7 + 62.9 + .... etc) and dividing them by the number of years (2006-1990). 1616.70/16 = 101.04375 This gives the average price over the 16 year period. We can then multiply 101.04375 by 22 (2012-1990) = 2222.9625 or £222.3K TASK FOUR The annual profits made by a construction company are found to be distributed normally with a mean of £22m and a standard deviation of £17m. Using tables of the standard normal distribution find: a) The probability of making a profit in excess of £31m. Formula Z=x-Mσ Mean = £22m Standard deviation = £17m P (x= ≥31m) Z=31-2217 Z=917 Z = 0.529 tabulated value of 0.53 = 0.7019 Therefore 1 – 0.7019 = 0.2981 The probability of making a profit in excess of £31m is 0.2981. b) The probability of making a loss P (x= <0m) Z=0-2217 Z=-2217 Z = -1.294 tabulated value of 1.294 = 0.9015 The probability of making a loss is 0.9015 a) The probability of making a profit between £20m and £26m. P (x= ≥20m) Z=20-2217 Z=-217 Z = -0.1176 tabulated value of 0.12 = 0.5478 Therefore 1 – 0.5478 = 0.4522 P (x= ≥26m) Z=26-2217 Z=417 Z = 0.2353 tabulated value of 0.23 = 0.5909 Therefore 1 – 0.5909 = 0.4091 0.4522 + 0.4091 = 0.8613 The probability of making a profit between £20m and £26m is 0.8613. TASK FIVE On average labourers are absent without authorisation 8 days per week. The construction site requires 45 labourers. If 45 labourers are employed, what is the probability of 45 labourers being present' How many labourers have to give a probability of at least 95% being present' Formula e-λ λrr! P (x = 45) = e-λ λrr! Therefore @ 45 labourers λ = 45 × 852 × 5- (8 + 20) = 360232 = 1.5517 (52 weeks of 5 working days minus 8 Bank Holidays and 20 days leave) P (x = 0) = e-λ = e-1.5517 = 0.2118 = 21.18% The Probability of 45 labourers being present when 45 labourers are employed is 21.18%. This would not be enough, if we require a 95% attendance. Therefore @ 46 labourers λ = 46 × 852 × 5- (8 + 20) = 368232 = 1.5862 P (x = 0) = e-λ = e-1.5862 = 0.2047 = 20.47% P (x = 1) = e-1.5862 1.586211! = 0.32471 = 0.3247 = 32.47% Thus 20.47 + 32.47 = 52.94%. This still does not meet our requirements. Therefore @ 47 labourers λ = 47 × 852 × 5- (8 + 20) = 376232 = 1.6207 P (x = 0) = e-λ = e-1.6207 = 0.1978 = 19.78% P (x = 1) = e-1.6207 1.620711! = 0.32051 = 0.3205 = 32.05% P (x = 2) = e-1.6207 1.620722! = 0.51942 = 0.2597 = 25.97% Thus 19.78 + 32.05 + 25.97 = 77.8% This still does not meet our requirements. Therefore @ 48 labourers λ = 48 × 852 × 5- (8 + 20) = 384232 = 1.6552 P (x = 0) = e-λ = e-1.6552 = 0.1910 = 19.1% P (x = 1) = e-1.6552 1.655211! = 0.31621 = 0.3162 = 31.62% P (x = 2) = e-1.6552 1.655222! = 0.52342 = 0.2617 = 26.17% P (x = 3) = e-1.6552 1.655233! = 0.86646 = 0.1444 = 14.44% Thus 19.1 + 31.62 + 26.17 + 14.44 = 91.33% This still does not meet our requirements but it is close. Therefore @ 49 labourers λ = 49 × 852 × 5- (8 + 20) = 392232 = 1.6897 P (x = 0) = e-λ = e-1.6897 = 0.1845 = 18.45% P (x = 1) = e-1.6897 1.689711! = 0.31191 = 0.3119 = 31.19% P (x = 2) = e-1.6897 1.689722! = 0.52702 = 0.2635 = 26.35% P (x = 3) = e-1.6897 1.689733! = 0.89046 = 0.1484 = 14.84% P (x = 4) = e-1.6897 1.689744! = 1.504624 = 0.0627 = 6.27% Thus 18.45 + 31.19 + 26.35 + 14.84 + 6.27 = 97.1% Again this is not the exact amount. As labourers only come in whole parts, this would be the most accurate number available to us. 49 labourers will have to be employed if you wish to ensure 45 labourers for at least 95% of the time.