第二章数据集的描述方法.pptVIP

This is the data from problem 3.54 in BL5ed. Give the class 10-15 minutes to compute before showing the answer. Using exact values: Midhinge = (Q1 + Q3)/2 = (2.75 + 8.25)/2 = 11/2 = 5.5 Using exact values: Midhinge = (Q1 + Q3)/2 = (2.75 + 8.25)/2 = 11/2 = 5.5 This is the data from problem 3.54 in BL5ed. Give the class 10-15 minutes to compute before showing the answer. Q1 = 1(n+1)/4 = 1(10+1)/4 = 11/4 = 2.75 Position If exact values: 75% of way Between 2 3; Value is 2.75 Q3 = 3(n+1)/4 = 3(10+1)/4 = 33/4 = 8.25 Position If exact values: 25% of way Between 8 9; Value is 8.25 Using exact values: Midhinge = (Q1 + Q3)/2 = (2.75 + 8.25)/2 = 11/2 = 5.5 The number of classes is usually between 5 and 15. Only 3 are used here for illustration purposes. 11 total employees; total salaries are $770,000. The mode is $20,000 (Union argument). The median is $30,000. The mean is $70,000 (President’s argument). Different measures are used! Throughout this chapter, we will be using the following notation, which I will introduce now. Location (Position) Concerned with where values are concentrated. Variation (Dispersion) Concerned with the extent to which values vary. Shape Concerned with extent to which values are symmetrically distributed. This is the data from problem 3.54 in BL5ed. Give the class 10-15 minutes to compute before showing the answer. Median = 6.5 Position = (n+1)/2 = (10+1)/2 = 5.5 1 2 3 5 6 7 8 8 9 11 1 2 3 4 5 6 7 8 9 10 (n = 10) (6+7)/2 = 6.5 Mode = 8 Midrange = 6 (Xsmallest + Xlargest)/2 = (1+11)/2 = 6 Shape Concerned with extent to which values are symmetrically distributed. Kurtosis The extent to which a distribution is peaked (flatter or taller). For example, a distribution could be more peaked than a normal distribution (still may be ‘bell-shaped). If values are negative, then distribution is less peaked than a normal distribution. Skew The extent to which a distribution is symmetric or has a tail. Values are 0 if normal distribution. If the value