The propagation of fatigue cracks in various aircraft parts has been the subject

Question

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 (light hours/10^4) to reach a given crack size in fastener holes intended for use in military aircraft. (a) Compute and compare the values of the sample mean x^- and median x^-. (Round your answers to four decimal places.) x^- = flight hours/10^4 x^- = flight hours/10^4 (b) By how much could the largest sample observation be decreased without affecting the value of the median? (Enter your answer to three decimal places.) 317 flight hours/10^4

Explanation / Answer

(a)

In order to find the mean, take the sum of given numbers and divide it by total numbers (i.e. 16)

sum of all numbers = 16.386

mean = 16.386/16 = 1.024

In order to find the median, arrange the given numbers in descending order and take the average of 8th and 9th numbers.

Median = (1.01 + 1.04)/2 = 1.025

(b)

Median will be changed if we introduce any number which is less than 9th number i.e. 1.04

Hence, the largest sample observation can be decreased till 1.04 i.e. 1.359 - 1.04 = 0.319

1 0.727 2 0.853 3 0.868 4 0.908 5 0.923 6 0.93 7 0.976 8 1.01 9 1.04 10 1.042 11 1.085 12 1.121 13 1.132 14 1.16 15 1.252 16 1.359

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