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Chapter 18: A. Basic Statistics and Applications 229
EXAMPLE 18.23
The following data give the survival time (in hours) of 50 parts involved in a field test
under extreme operating conditions.
60 100 130 100 115 30 60 145 75 80 89 57 64 92 87 110 180
195 175 179 159 155 146 157 167 174 87 67 73 109 123 135 129 141
154 166 179 37 49 68 74 89 87 109 119 125 56 39 49 190
a. Construct a frequency distribution table for the above data.
b. Construct frequency and relative frequency histograms for the above data.
Solution:
1. Find the range of the data
2. Determine the number of classes
R = 195 – 30 = 165
m = 1 + 3.3 log 50 = 6.57
By rounding it we consider the number of classes to be equal to seven.
3. Compute the class width
Class width = R/m = 165/7 = 23.57
By rounding up this number we have class width equal to 24. As noted earlier, we always
round up the class width to a whole number or to any other convenient number that
may be easy to work with. Note that if we round down the class width, then some of the
observations may be left out of our count and not belong to any class. Consequently,
the total frequency will be less than n. The frequency distribution table for the data in
this example is shown as Table 18.9.
Table 18.9 Frequency distribution table for the survival time of parts.
Class Tally Frequency Relative Cumulative
frequency frequency
[30–54) ///// 5 5/50 5
[54–78) ///// ///// 10 11/50 16
[78–102) ///// //// 9 8/50 24
[102–126) ///// // 7 7/50 31
[126–150) ///// / 6 6/50 37
[150–174) ///// / 6 6/50 43
[174–198] ///// // 7 7/50 50
Total 50 1
Part IV.A.4
Continued