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 (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (A????1Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,A????1 J. Aircraft, 1983: 1028A????11032):
.736 .863 .865 .913 .915 .937 .983 1.007 1.011 1.064 1.109 1.132 1.140 1.153 1.253 1.394
Compute and compare the values of the sample mean and median.
By how much could the largest sample observation be decreased without affecting the value of the median?

Answer 1

Answer:

$Mean = 1.030$

$Median = 1.009$

$Difference = 0.141$

Step-by-step explanation:

Given

Data:

.736, .863,  .865,  .913

.915, .937, .983,  1.007

1.011, 1.064, 1.109, 1.132

1.140, 1.153, 1.253, 1.394

Solving (a): Mean and Median

Mean

Mean is calculated as thus;

Mean = Summation of observation divided by number of observations$Mean = \frac{.736 +.863 +.865 +.913 +.915 +.937 +.983 +1.007 +1.011 +1.064 +1.109 +1.132 +1.140 +1.153 +1.253 +1.394}{16}$

$Mean = \frac{16.475}{16}$

$Mean = 1.0296875$

$Mean = 1.030$ Approximated

Median

Since the number of observation is 16;

$Median = \frac{16 + 1}{2}$

$Median = \frac{17}{2}$

Median = 8.5th item

This can be determined by calculating the average of the 8th and 9th item

$Median = \frac{1}{2}(1.007 + 1.011)$

$Median = \frac{1}{2}(2.018)$

$Median = 1.009$

Solving b:

For the median to remain 1.009, the largest sample (1.394) must be greater than or equal to 1.253

Calculating the difference:

$Difference = 1.394 - 1.253$

$Difference = 0.141$

Hence, it can be reduced by 0.141