TASK 1 GROUP 2

Task group Assignment Santiago Cacciaguerra Melich Arsen Voskanyan 1. A doctor randomly selected 40 of his 20 – 29 years old patients and obtains the following data regarding their serum HDI cholesterol. 70 56 48 48 53 52 66 48 36 49 28 35 58 62 45 60 38 73 45 51 56 51 46 39 56 32 44 60 51 44 63 5046 46 69 53 70 33 54 55 52 Class Class limit f f/n F m-point boundaries 1 28-33 3 0.075 3 30.5 27.5-33.5 2 34-39 4 0.1 7 36.5 33.5-39.5 3 40-45 4 0.1 11 42.5 39.5-45.5 4 46-51 10 0.25 21 48.5 45.5-51.5 5 52-57 9 0.225 30 54.5 51.5-57.5 6 58-63 5 0.125 35 60.5 57.5-63.5 7 64-69 2 0.05 37 66.5 63.5-69.5 8 70-75 3 0.075 40 72.5 69.5-75.5 n=40 K= 6

Lucas Bellon 3)The following data represent the miles er gallon for a 2015 Ford Fusion for six randomly selected vehicles. Compute the mean, median and mode miles per gallon 34.0 33.2 37.0 29.4 23.6 25.9 Mean: 83.1/6 = 30.5 Median: 62..6/2 = 31.3 Mode: there is no mode 23.6 + 25.9+[29.4+33.2]+34.0+37.0

Laurent Oramas 4) The following data represent the amount of time (in minutes) a random sample of eight student took to complete the online portion of an exam in Statistics course. Compute the range, sample variance , and sample standard deviation time 60.5 78.9 84.6 85.9 89.9 94.7 122.3 128.0 Range: 128 - 60.5 = 675 minutes Sample Variance: n= 8 mean =244.8/8=93.1 60.5 - 43.1 = (-32. 6)^2 = 1062.76 78.9 - 93.1 = (-14.2)^2 = 201.64 3471.94/7 = 5^2 = 495.99 minutes 84.6 - 93.1 = (-8.5 )^2 =72.25 85.9 - 93.1 = (-7.2)^2 = 51.84 89.4-93.1 = (-3.2 )^2 = 10.24 94.7 - 93.1 = (1.6)^2 = 2.56 122.3 - 93.1 = (29.2)^2 = 852.64 N = (8-1) = 7 128.0 - 93.1 = (34.4)^2 = 1218.01 ∑ = 3471.94 Sample Standard deviation: S = √ 495.99 = 22.27 minutes

S^2 = 62.09/9 =6.89 S=√6.89 =2.62 Material B N=9 mean =113.32/10=11.33 (5.78-11.33)^2 =30.80 (9.65-11.33) =2.82 (6.71-11.33)^2=21:34 (13,44-11.33) = 4,45 (6.84-11.33)^2 = 20.16 (14.71-11.33)^2= 11.42 (7.23-11.33)^2= 16.81 (16.39-1.33)^22=25.60 (8.2-11.33)^2=9.79 (24.37-11.33)^2=170.04 ∑=313.23 S^2313.23/9 = 34.80 S√34.8 = 5.89 d) Material A min = 3.17 Q1+=4.52 Q2/md = 5.78 Q3= 8.01 Max = 11.92 Material B min = 5.78 Q1= 6.84 Q2/md = 8.92 Q3 = 14.71 max= 24.37

e) f) Material A This distribution is skewed to the right Material B This distribution is skewed to the right g) No, material B does not have any outliers 6.89 - 1.5(7.98) 19.71+1.5(7.87) 6.84 - 11.81 =-4.97 14.78 + 11.81 = 26.52 Clara Ferre 6) According to the US Census Bureau, the mean of the commute time to work for a resident of Boston, Massachusetts is 27.3 minutes. Assume that the standard deviation of the commute time is 8.1 minutes to answer the following questions, using Chebyshev theorem. a) We should apply Chebyshev's theorem because we don't have information about the distribution of the commute times b) x̄ = 27.3 x̄ +- 2.5 S = 8.1 27.3 +- 2(8.1) 27.3 +- 16.2 (11.1, 43.5) At least 75% of all data values are between 11.1 and 43.5 c) x̄ +- 3.5 27.3 +-3 (8.1)