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.
0.52 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 3.05
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 0.02 0.07 0.65 0.40 0.24 1.51 1.45
1.60 1.80 4.69 0.08 7.89 1.58 1.64 0.03 0.23 0.72
(a) Find the range.
(b) Use a calculator to calculate ?x and ?x2.
(c) Use the results of part (b) to compute the sample mean, variance, and standard deviation for the time to failure.
(d) Use the results of part (c) to compute the coefficient of variation.
What does this number say about time to failure?
1) The standard deviation of the time to failure is just slightly smaller than the average time.
2) The coefficient of variation says nothing about time to failure.
3) The standard deviation of the time to failure is just slightly larger than the average time.
4) 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.
1) A small CV indicates more consistent data because the value of s in the numerator is smaller.
2) A small CV indicates more consistent data because the value of s in the numerator is larger.

Answer 1

Answer:

range = 7.87

X = 62.12, X²= 164.3516

mean = 1.2424

variance = 1.779

standard deviation = 1.33379

coefficient of variation = 107.3559%

Step-by-step explanation:

a.) The range =

highest number - lowest number

= 7.89 - 0.02

= 7.87

b.) from the attachment I solved for X and X²

∑X = 62.12

∑X² = 164.3516

c.)

the mean = ∑x/n

n = 50

= 62.12/50

= 1.2424

variance =

σ² = [∑x²-(∑x)²/n] / n-1

= [164.3516 - (62.12)²/50] / 50-1

= 164.3516 - 77.17788/49

= 87.173712/49

= 1.779

standard deviation = √σ

= √1.779

= 1.33379

d. coefficient of variation

= σ/mean * 100

= 1.33379/1.2424 * 100

= 107.3559

This number tells us that the the standard deviation of the time to failure is larger than the average time. option 3 is the correc4 answer here