In: Math
3.2
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.
0.54 | 1.80 | 1.52 | 2.05 | 1.03 | 1.18 | 0.80 | 1.33 | 1.29 | 1.13 |
3.34 | 1.54 | 0.08 | 0.12 | 0.60 | 0.72 | 0.92 | 1.05 | 1.43 | 3.02 |
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 | 1.63 | 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.
Consider sample data with
x = 8
and
s = 2.
(a) Compute the coefficient of variation.
(b) Compute a 75% Chebyshev interval around the sample mean.
Lower Limit | |
Upper Limit |
a. The range is the difference between the highest and lowest values in the data set.
Ordering the data from least to greatest, we get:
0.02 0.03 0.03 0.07 0.08 0.08 0.09 0.12 0.18 0.19 0.23 0.24 0.34 0.4 0.54 0.56 0.6 0.63 0.65 0.72 0.72 0.8 0.92 1.03 1.05 1.13 1.18 1.29 1.33 1.43 1.45 1.45 1.51 1.51 1.52 1.52 1.54 1.55 1.58 1.6 1.63 1.8 1.8 1.81 2.05 2.17 3.02 3.34 4.69 7.89
The lowest value is 0.02.
The highest value is 7.89.
The range = 7.89 - 0.02 = 7.87.
b.
X | X^2 | |
0.54 | 0.2916 | |
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 | |
1.13 | 1.2769 | |
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 | |
3.02 | 9.1204 | |
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 | |
0.02 | 0.0004 | |
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 | |
1.63 | 2.6569 | |
0.03 | 0.0009 | |
0.23 | 0.0529 | |
0.72 | 0.5184 | |
Total | 62.11 | 164.1805 |
c.
Create the following table.
data | data-mean | (data - mean)2 |
0.54 | -0.7022 | 0.49308484 |
1.8 | 0.5578 | 0.31114084 |
1.52 | 0.2778 | 0.07717284 |
2.05 | 0.8078 | 0.65254084 |
1.03 | -0.2122 | 0.04502884 |
1.18 | -0.0622 | 0.00386884 |
0.8 | -0.4422 | 0.19554084 |
1.33 | 0.0878 | 0.00770884 |
1.29 | 0.0478 | 0.00228484 |
1.13 | -0.1122 | 0.01258884 |
3.34 | 2.0978 | 4.40076484 |
1.54 | 0.2978 | 0.08868484 |
0.08 | -1.1622 | 1.35070884 |
0.12 | -1.1222 | 1.25933284 |
0.6 | -0.6422 | 0.41242084 |
0.72 | -0.5222 | 0.27269284 |
0.92 | -0.3222 | 0.10381284 |
1.05 | -0.1922 | 0.03694084 |
1.43 | 0.1878 | 0.03526884 |
3.02 | 1.7778 | 3.16057284 |
1.81 | 0.5678 | 0.32239684 |
2.17 | 0.9278 | 0.86081284 |
0.63 | -0.6122 | 0.37478884 |
0.56 | -0.6822 | 0.46539684 |
0.03 | -1.2122 | 1.46942884 |
0.09 | -1.1522 | 1.32756484 |
0.18 | -1.0622 | 1.12826884 |
0.34 | -0.9022 | 0.81396484 |
1.51 | 0.2678 | 0.07171684 |
1.45 | 0.2078 | 0.04318084 |
1.52 | 0.2778 | 0.07717284 |
0.19 | -1.0522 | 1.10712484 |
1.55 | 0.3078 | 0.09474084 |
0.02 | -1.2222 | 1.49377284 |
0.07 | -1.1722 | 1.37405284 |
0.65 | -0.5922 | 0.35070084 |
0.4 | -0.8422 | 0.70930084 |
0.24 | -1.0022 | 1.00440484 |
1.51 | 0.2678 | 0.07171684 |
1.45 | 0.2078 | 0.04318084 |
1.6 | 0.3578 | 0.12802084 |
1.8 | 0.5578 | 0.31114084 |
4.69 | 3.4478 | 11.88732484 |
0.08 | -1.1622 | 1.35070884 |
7.89 | 6.6478 | 44.19324484 |
1.58 | 0.3378 | 0.11410884 |
1.63 | 0.3878 | 0.15038884 |
0.03 | -1.2122 | 1.46942884 |
0.23 | -1.0122 | 1.02454884 |
0.72 | -0.5222 | 0.27269284 |
Find the sum of numbers in the last column to get.
So
So
d. coefficient of variation is %