# Deviation Essay Examples

## Statistics

0
(0)

Deviation 5.785 .714 .490 Variance 33.466 .510 .240 Skewness.480 -1.265 .433 Std. Error of Skewness.172 .172 .172 Kurtosis -.568 .122 -1.831 Std. Error of Kurtosis .342 .342 .342 Range 27 2 1 Minimum 9 1 1 Maximum 36 3 2 Sum 3923 510 279 1. For the gender variable, what is the appropriate measure of central tendency and what is the value for it? The mode is appropriate. This identifies males as the most common gender with a frequency of 121 while the females are the less common gender with a frequency of 79. 2. For the time variable, what are the two appropriate measures of central tendency and what are the values? For the time variable, the two appropriate measures of central...

• Words: 1375
• Pages: 5

## Conflict Resolution

0
(0)

deviation among the groups involved in the communication. Conflict Resolution The best ways in which conflict can be resolved is by encouraging the communication among the people involved in the conflict. Whenever there is good communication, the people involved in the conflict will have the opportunity to detect where they normally go wrong that results in the conflict and reconcile effectively (Myatt, 2012). The other way for effective conflict resolution is to avoid taking sides. The person who reconciles those who are involved in a conflict should not take sides. And also the individual involved in conflict resolution should learn the techniques of resolving disputes so that s/he can become a...

• Words: 275
• Pages: 1

## Data Project

0
(0)

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...

• Words: 550
• Pages: 2

## Penicillin antibiotic has more effectiveness on gram-positive bacteria

0
(0)

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...

• Words: 550
• Pages: 1

## Protein Assay

0
(0)

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...

• Words: 1100
• Pages: 4

## Response to two discussion board entries

0
(0)

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....

• Words: 275
• Pages: 1

## Manova Project

0
(0)

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...

• Words: 550
• Pages: 2

## lab report

0
(0)

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...

• Words: 825
• Pages: 3

## Sampling Distributions

0
(0)

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...

• Words: 275
• Pages: 1