From Table 15.2 we see that we must cover an interval of length of at least 306 − 96 = 210 with bins of width b = 3.49 · 68.48 · 272−1/3 = XXXXXXXXXXSince 210/36.89 = 5.69, we need at least six bins...

From Table 15.2 we see that we must cover an interval of length of at least 306 − 96 = 210 with bins of width b = 3.49 · 68.48 · 272−1/3 = 36.89. Since 210/36.89 = 5.69, we need at least six bins to cover the whole dataset.