We will round up to two and make each bar or class interval two units wide. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Next, calculate the width of each bar or class interval. The largest value is 74, so 74 + 0.05 = 74.05 is the ending value. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.Ħ0 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data. Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. Choose a starting point for the first interval to be less than the smallest data value. Many histograms consist of five to 15 bars or classes for clarity. To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Ahab's English class of 40 students received from 90% to 100%, then, f = 3, n = 40, and RF = f n f n = 3 40 3 40 = 0.075.
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