Deviation Essay Samples and Topic Ideas
deviation =9.01, variance=81.21, range=65, interquartile range = 8.95, first quartile = 8.30, size (n) = 500, and the third quartile = 17.25. Figure SEQ Figure * ARABIC 4. An image displaying summary statistics for the Taxi05 data set A box plot of fare was also displayed using the ggplot2 package (see figure 5.). It indicates that fare is positively skewed has significant outliers for any cost above 30. Figure SEQ Figure * ARABIC 5. A box plot showing the distribution of fare from the Taxi05 data set A side-by-side boxplot of fare separated by call was also created (see figure 6.). From the figure, it is clear that the average cost of fare for dispatch calls is higher than the cost of...
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deviation was 0.8164966. Discussion After incubation, the growth of bacteria was supposed to cover the whole of each bacterium sample, only leaving out the region around the penicillin-coated discs. This region is also known as the zone of inhibition (Harley, 2004). The longer or, the larger this region is an indication that penicillin was effective in inhibiting bacterial growth. This was the case with S. aureus which had a significantly larger region as compared to the other plates, including that with gram-positive E-coli. Hence, confirming our hypothesis that indeed gram positive are more susceptible to penicillin. This susceptibility is further explained by morphological differences between...
deviation is given by the relation; =xi-m2n-1It is obtained as; (1.327/5)= 0.515Therefore the concentration of the unknown according to Lowry assay is given as; 0.598 + or -0.515 ug/ml Part II. Bradford Assay The amount(uL) of BSA needed from the 2.0mg/L BSA standard needed to prepare 760 uL of 0.1mg/L is determined using the dilution equation as follows; M1V1=M2V2M1=760uL*0.1mgL-12.0mgL-1=38uLThe amount of buffer (NaH2PO4) needed to make up the new solution is given as; 716-38=722uLTable 2.0. Bradford Assay (1-6) Tube# 1 2 3 4 5 6 Vol. of 0.1 mg/mL BSA(µL) 0 10 20 40 60 80 Vol. of H20 (µL) 800 790 780 760 740 720 Vol. of Bio-rad Dye Reagent (µL) 200. 200. 200. 200. 200. 200. Amount of...
deviation used in the graph to determine the length and number of treats are small and this affect the general presentation of the graph, which shows a narrow and tall curve. In my observation, the data was not well tabulated due to accuracy and the standard material used causing an imperfect curve. To improve the curve data must be concentrated at the center and avoid extreme values that would affect the shape of the curve. Discussion Response 2 The data presentation conducted on the grades is less making it easy to tabulate in a graph. Considering the fewer data involved the standard deviation, in this case, will be small, and this explains the wave type of curve obtained from the graph....
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Deviation N Usefulness online website 21.45 3.804 11 nurse practitioner 23.17 5.750 12 videotape 21.10 4.095 10 Total 21.97 4.633 33 Difficulty online website 9.18 3.125 11 nurse practitioner 9.83 3.881 12 videotape 9.40 1.838 10 Total 9.48 3.043 33 Importance online website 15.64 6.786 11 nurse practitioner 8.67 7.402 12 videotape 8.70 6.430 10 Total 11.00 7.479 33 A substantial statistical difference was not identified between information from the online website, nurse practitioners, and videotapes of nurse practitioners, F (6, 56) = 1.495, p > .0005; Wilk's Λ = 0.743, partial η2 = .138, see REF _Ref506380119 h * MERGEFORMAT Table 2. Therefore, we accept the null hypothesis...
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Deviation = √ (Variance) Variance = (20.16-20.314)2 + (20.16 – 20.314) 2 + (20 – 20.314) 2 + (20.56 – 20.314) 2 + (20.69 – 20.314) 2 Variance = ((-0.154)2 + (-0.154)2 + (-0.314) 2 + (0.246) 2 + (0.376) 2) / 5 = 0.34792 / 5 = 0.069584 √ 0.069584 = 0.02638 Data Exercise 2 x- x+ xs Diameter of image 2 28.2 100 9.5 12 34.2 100 5.5 27.3 52.9 100 2.1 10.8 36.8 90 4.1 20.3 45.25 95 2.5 Results Converging Lens Diverging Lens X- X+ Xs di do f di do f 2.0000 28.2000 100.0000 71.8000 27.7220 20.0000 -1.5220 2.0000 -6.3683 12.0000 34.2000 100.0000 65.8000 28.7336 20.0000 -6.5336 12.0000 -14.3429 27.3000 52.9000 100.0000 47.1000 34.7601 20.0000 -9.1601 27.3000...
deviation of 29lb. If a woman is randomly selected, the probability that her weight is between 140lb-211lb will be as given below; We calculate the z- score for a row value between 140 and 211.To decide on the value, we can get the average of the two values as shown below; Average of 140 and 211= 140+211=351/2=175.5=176. Z-SCORE FOR 176= 176-143=33/29= 1.13793=1.14 Z-score = Row score – Mean Standard Deviation The z-score obtained has a positive value indicating that it is above the mean...
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deviation. Which forecasting method would you recommend? 3-week moving average Week Actual Rentals Predicted Value Difference 1 1202 2 1503 3 1444 4 1254 1383 129 5 1609 1400 209 6 1499 1435 64 7 1689 1454 235 8 1555 1599 44 Sum of Difference 681 Average Difference 136.2 4-week moving average Week Actual Rentals Predicted Value Difference 1 1202 2 1503 3 1444 4 1254 5 1609 1350.75 258 6 1499 1452.5 47 7 1689 1451.5 238 8 1555 1512.75 42 Sum of Difference 585 Average Difference 146 Since the 3 week moving average has a smaller MAD, then it is better for forecasting this data. Question 9 page 436 A ski repair...
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Deviation Hand Length 50 7.0806 7 3.25 0.424022082 0.651169779 Hand Width 50 7.8895 8 6 1.232196173 1.110043321 Height 50 67.296 67 17 17.21875918 4.149549275 From the summary statistics, the sample population is 50, the mean hand length is 7.0806, hand width is 7.8895, and the mean height is 67.296. The median is the value at the middle of the observed data, and in this output, we have hand length to be 7, hand width is 8 and height 67. The difference between the maximum and the minimum data values give the Range, and in the data, we have Hand length 3.25, hand Width 6 and Height 17. The variance indicates how much the data is spread out from the mean, and from the output hand length has a value of...
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deviation = 1.9 The Z-Score for the data is given as Z=Sample mean-Population meanStandard deviationSubstituting we obtain, Z=63-64.61.9=-1.61.9=-0.8421Problem 3 The mean of the sample distribution is the same as that of the population. Since there is only a single sample, the best approach to estimate the population proportion is to use the sample proportion. The percentage without doctorates becomes p=(110-66)110=410=0.4 or 40%Given the population N=349, to obtain the number of teachers who do not have doctorates we multiply N*p=349×4100=139.6~140 Teachers have no doctorates. Problem 4 The formula for finding the marginal error is given as E=2S√n . Assuming a confidence interval of 95%,...
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