# Question

To construct a histogram for these data, we select one of the classes as a base. It is often convenient to choose the shortest class as the base (although it is not necessary to do so). Using this choice, the 0 to $ 10K class is the base. This means that we will draw a rectangle over the 0 to $ 10K class having a height equal to 162 (the frequency given for this class in the published data). Because the other classes are longer than the base, the heights of the rectangles above these classes will be adjusted. To do this we employ a rule that says that the area of a rectangle positioned over a particular class should represent the relative proportion of measurements in the class. Therefore, we proceed as follows. The length of the $ 10K to 25K class differs from the base class by a factor of (25 10) (10 0) 3 2, and, therefore, we make the height of the rectangle over the $ 10K to 25K class equal to (2 3) (62) 41.333. Similarly, the length of the $ 25K to 50K class differs from the length of the base class by a factor of (50 25) (10 0) 5 2, and, therefore, we make the height of the rectangle over the $ 25K to 50K class equal to (25) (53) 21.2.

a. Use the procedure just outlined to find the heights of the rectangles drawn over all the other classes (with the exception of the open- ended class, $ 500K).

b. Draw the appropriate rectangles over the classes (except for $ 500K).

a. Use the procedure just outlined to find the heights of the rectangles drawn over all the other classes (with the exception of the open- ended class, $ 500K).

b. Draw the appropriate rectangles over the classes (except for $ 500K).

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