The paper “Study on the Life Distribution of Microdrills” (J. of Engr. Manufacture, 2002: 301–305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. a. Why can a frequency distribution not be based on the class intervals 0–50, 50–100, 100–150, and so on? b. Construct a frequency distribution and histogram of the data using class boundaries 0, 50, 100, . . . , and then comment on interesting characteristics. c. Construct a frequency distribution and histogram of the natural logarithms of the lifetime observations, and comment on interesting characteristics. d. What proportion of the lifetime observations in this sample are less than 100? What proportion of the observations are at least 200?

Answer: Step1: Given that, The paper “Study on the Life Distribution of Micro Drills” (J. of Engr. Manufacture, 2002: 301–305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. Step 2: a). The aim is why can a frequency distribution not be based on the class intervals 0–50, 50–100, 100–150, and so on. In the given data set, the observation is falls in the specified boundary of 50. Here the class intervals are 0-50, 50-100, 100-150 and so on. The value 50 falls in both of the intervals ‘0-50’ and ‘50-100’. So the frequency distribution is not based on the class intervals. Step 3: b). To construct a frequency distribution and histogram of the data using class boundaries 0, 50, 100, . . . , and then comment on interesting characteristics. Class Frequency Relative frequency 0-50 9 9/50 = 0.18 50-100 19 19/50 = 0.38 100-150 11 11/50 = 0.22 150-200 4 4/50 = 0.08 200-250 2 2/50 = 0.04 250-300 2 2/50 = 0.04 300-350 1 1/50 = 0.02 350-400 1 1/50 = 0.02 400-450 0 0 450-500 0 0 500-550 1 1/50 = 0.02 N = 50 Total = 1 Histogram of the data The distribution is skewed right. There is a gap in the histogram, and the outlier is ‘500-550’ interval. Step4: c). To construct a frequency distribution and histogram of the natural logarithms of the lifetime observations, and comment on interesting characteristics. Step 5: d). The aim is what proportion of the lifetime observations in this sample are less than 100 What proportion of the observations are at least 200 Proportion of the lifetime observations in this sample are less than 100 = Relative frequency for 0-50 + Relative frequency for 50-100 = 0.18 + 0.38 = 0.56 Proportion of the lifetime observations in this sample are at least 200 = Relative frequency for 200-250 + Relative frequency for 250-300 + Relative frequency for 300-350 + Relative frequency for 350-400 + Relative frequency for 450-500 + Relative frequency for 500-550. = 0.04 + 0.04 + 0.02 + 0.02 + 0 + 0 + 0.02 = 0.14. Therefore, the proportion of the lifetime observations in this sample are less than 100 is 0.56. And the proportion of the observations are at least 200 is 0.14.