The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J ? . Aircraft? 983: 1028–1032): a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?

Problem 39E Answer: Step1: The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J. Aircraft, 1983: 1028–1032): 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 We need to find, a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median Step2: a). 1).A mean is an average of given data, It is the sum of individual scores divided by the number of individuals and it is given by n xi x = i=1 n Consider, i x 1 0.736 2 0.863 3 0.865 4 0.913 5 0.915 6 0.937 7 0.983 8 1.007 9 1.011 10 1.064 11 1.109 12 1.132 13 1.14 14 1.153 15 1.253 16 1.394 n x = 16.475 n = 16 i=1 i n xi x = i=1 n = 16.475 16 = 1.0296. Therefore,mean of the given data is 1.0296. 2).median is the middle value in a given set of data. To find the median, we arrange the observations in order from smallest to largest value. If there is an odd number of observations, the median is the middle value. If x +xe is an even number of observations, the median is the average of the two middle values. i.e,12 2 Consider the data in Ascending order, 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 Since we have n = 16 hence the given data is even.so, as per the definition of median If there is an even number of observations, the median is the average of the two middle values. i.e,Median = x1+x2 2 = 1.007+1.011 2.018 = 2 = 1.009 Therefore,the median of the given data is 1.009. b). From the given data we can say that 1.394 can be decreased until it reaches 1.011 i.e,by largest value - middle value = 1.394 – 1.011 = 0.383 The largest of the 2 middle values. If it is decreased by more than 0.383, the median will change.