Measure of Dispersion Question Source: Harris amp. Taylor 2009] 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.a) Calculate the value for 1 SD below the mean weight of 80 kg.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.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]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.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 amp. Taylor 2009]What percentages of subjects are included in ±1 SD, ±2 SDs, or ±3 SDs from the mean?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 amp. Taylor 2009]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.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.The length of stay will be between -6 days and 26 days.Are these values for ‘length of stay’ plausible? Why?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?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 5Use 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.