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.
Time taken for hare to catch up with the tortoise will be found as follows: time taken=distance/(relative speed) relative speed=(speed of hare-speed of tortoise) speed of hare=500/20=25 cm/sec speed of tortoise=250/50=5 cm/sec relative speed will be: (25-5)=20 cm/ sec thus the total time taken for hare to catch up with the tortoise will be: 1000/20 =50 sec
