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.

0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.001 1.011 1.064 1.109 1.132 1.140 1.153 1.253 1.397

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?

A) Mean = 1.0295

Median = 1.006

B) the largest sample Observation can be reduced to 1.3735

Step-by-step explanation:

A) I) The mean Value is the addition of all the values, then dividing by the number of values occured.

0.736 + 0.863 + 0.865 + 0.913 + 0.915 + 0.937 + 0.983 + 1.001 + 1.011 + 1.064 + 1.109 + 1.132 + 1.140 + 1.153 + 1.253 + 1.397 = 16.472

16.472 ÷ 16 = 1.0295

1.0295 is the mean value

II) The median value is the value at the center of the point. We have two values at the middle point.

(1.001 + 1.011) ÷ 2 = 1.006

1.006 is the median value.

Therefore the mean value which shows the actual middle value of the numbers if the numbers where evenly distributed is greater than the median value which show middle value of the numbers in the result.

The difference between them is

1.0295 - 1.006 = 0.0235

B) The largest sample Observation is 1.397. it can be decrease by subtracting the difference between the mean and median from the number.

Wherefore; the difference between mean and median is 0.0235

Therefore;

1.397 - 0.0235 = 1.3735

If the largest sample Observation is reduced to 1.3735, the median value will not be affected.

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i