Measures of Dispersion

Complete 1 page APA formatted article: Measures of Dispersion. Measure of Dispersion &nbsp.Question Source: Harris & Taylor 2009]&nbsp. Scenario: A group of patients enrolling for a trial had a normal distribution for weight. The mean weight of the patients was 80 kg. For this group, the standard deviation (SD) was calculated to be 5 kg.&nbsp.a) Calculate the value for 1 SD below the mean weight of 80 kg.&nbsp.Mean – 1(SD) = 80 – 5 = 75 kgb) Calculate the value for 1 SD above the mean weight of 80 kg.Mean + 1(SD) = 80 + 5 = 85 kgc) Since 68.2% of the subjects will be included in ±1 SD, state the weight range for 68.2% of the patients.The weight range for 68.2% of the patients is between 75 kg and 85 kg.d) Calculate the weight range for 95.4% of the patients (i.e. ±2 SD).Mean – 2(SD) = 80 – 2(5) = 80 – 10 = 70 kgMean + 2(SD) = 80 + 2(5) = 80 + 10 = 90 kgThe weight range for 95.4% of the patients is between 70 kg and 90 kg.e) Calculate the weight range for 99.7% of the patients (i.e. ±3 SD).Mean – 3(SD) = 80 – 3(5) = 80 – 15 = 65 kgMean + 3(SD) = 80 + 3(5) = 80 + 15 = 95 kgThe weight range for 99.7% of the patients is between 65 kg and 95 kg.f) Based on scenario described above, draw a normal distribution shaped graph, and illustrate the region, which would relate to ±2 SD [mean weight 80 kg. standard deviation 5 kg]. Explain your graph.&nbsp.A normal distribution graph based on the given scenario is given below. The shaded region would relate to ±2 SD [mean weight 80 kg. standard deviation 5 kg].The areas below and above mean value will be equal that is 50%. The area between mean and 1 SD on both sides will be equal to 34.1% and the area between 1 SD and 2 SD on both sides will be equal to 13.6%.Question 2 [Source: Salkind 2008]&nbsp.Identify whether these distributions are negatively skewed, positively skewed, or not skewed at all, and why. Sketch a graph to show what the distribution curve might look like.i) This talented group of athletes scored very high on the multi-stage fitness test.Positively skewed: This is because very few of the athletes (talented group of athletes) will score very high on the multi-stage fitness test.ii) On a particular biology test, all students achieved the same score.&nbsp.Not Skewed At All: This is because the standard deviation in this case will be zero. The distribution curve will be a vertical line as all students achieved the same score.Question 3 [Source: Harris & Taylor 2009]&nbsp.What percentages of subjects are included in ±1 SD, ±2 SDs, or ±3 SDs from the mean?&nbsp.About 68.2% of subjects are included in ±1 SD from the mean.About 95.4% of subjects are included in ±2 SDs from the mean. About 99.7% of subjects are included in ±3 SDs from the mean.Question 4 [Source: Harris & Taylor 2009]&nbsp.SD should only be used when the data have a normal distribution. However, means and SDs are often wrongly used for data that are not normally distributed. A simple check for a normal distribution is to see if 2 SDs away from the mean are still within the possible range for the variable.&nbsp.For example, if we have some length of hospital stay data with a mean stay of 10 days and a SD of 8 days then calculate ±2 SD for this data, i.e.&nbsp.The length of stay will be between -6 days and 26 days.&nbsp.Are these values for ‘length of stay’ plausible? Why?&nbsp.No, these values for ‘length of stay’ are not plausible. This is because ‘length of stay’ cannot be a negative number. It should be at least one day (length of stay ? 1 day).What does this tell you about the distribution of the data?&nbsp.This tells us that the distribution of the data is not normally distributed (positively skewed). In other words, there are very few patients with very long length of hospital stay.Question 5&nbsp.Use the following set of 16 scores that consists of income levels ranging from about $50,000 to about $200,000. What is the best measure of central tendency and why?$199,999$98,789$90,878$87,678$87,245$83,675$77,876$77,743$76,564$76,465$75,643$66,768$65,654$58,768$54,678$51,354The best measure of central tendency is median value. Because there is one score with very high-income level of $199,999 (Outlier) and all other scores consists of income levels ranging from about $50,000 to about $100,000. Thus, the data is positively skewed.&nbsp.