Question

Kevlar epoxy is a material used on the NASA space shuttles. Strands of this epoxy were tested at the 90% breaking strength. The following data represent time to failure (in hours) for a random sample of 50 epoxy strands. Let x be a random variable representing time to failure (in hours) at 90% breaking strength. 0.51 1.80 1.52 2.05 1.03 1.18 0.80 1.33 1.29 1.12 3.34 1.54 0.08 0.12 0.60 0.72 0.92 1.05 1.43 3.04 1.81 2.17 0.63 0.56 0.03 0.09 0.18 0.34 1.51 1.45 1.52 0.19 1.55 0.01 0.07 0.65 0.40 0.24 1.51 1.45 1.60 1.80 4.69 0.08 7.89 1.58 1.65 0.03 0.23 0.72 (a) Find the range. (b) Use a calculator to calculate Σx and Σx2. (Round your answers to two decimal places.) Σx = Σx2 = (c) Use the results of part (b) to compute the sample mean, variance, and standard deviation for the time to failure. (Round your answers to two decimal places.) x = s2 = s = (d) Use the results of part (c) to compute the coefficient of variation. (Round your answer to the nearest whole number.) % What does this number say about time to failure? The standard deviation of the time to failure is just slightly smaller than the average time. The coefficient of variation says nothing about time to failure. The standard deviation of the time to failure is just slightly larger than the average time. The standard deviation is equal to the average. Why does a small CV indicate more consistent data, whereas a larger CV indicates less consistent data? Explain. A small CV indicates more consistent data because the value of s in the numerator is smaller. A small CV indicates more consistent data because the value of s in the numerator is larger.

Answer 1

(a)

Minimum value: 0.01

Maximum value: 7.89

Range = maximum - minimum = 7.89 -0.01 = 7.88

(b)

Following table shows the calculations done by excel:

	X	X^2
	0.51	0.2601
	1.8	3.24
	1.52	2.3104
	2.05	4.2025
	1.03	1.0609
	1.18	1.3924
	0.8	0.64
	1.33	1.7689
	1.29	1.6641
	1.12	1.2544
	3.34	11.1556
	1.54	2.3716
	0.08	0.0064
	0.12	0.0144
	0.6	0.36
	0.72	0.5184
	0.92	0.8464
	1.05	1.1025
	1.43	2.0449
	3.04	9.2416
	1.81	3.2761
	2.17	4.7089
	0.63	0.3969
	0.56	0.3136
	0.03	0.0009
	0.09	0.0081
	0.18	0.0324
	0.34	0.1156
	1.51	2.2801
	1.45	2.1025
	1.52	2.3104
	0.19	0.0361
	1.55	2.4025
	0.01	0.0001
	0.07	0.0049
	0.65	0.4225
	0.4	0.16
	0.24	0.0576
	1.51	2.2801
	1.45	2.1025
	1.6	2.56
	1.8	3.24
	4.69	21.9961
	0.08	0.0064
	7.89	62.2521
	1.58	2.4964
	1.65	2.7225
	0.03	0.0009
	0.23	0.0529
	0.72	0.5184
Total	62.1	164.313

So,

(c)

The mean is

The variance is

The standard deviation is

(d)

The standard deviation of the time to failure is just slightly larger than the average time.

A small CV indicates more consistent data because the value of s in the numerator is smaller.