A fatigue test is performed on rotating beam specimens where, for each rotation cycle, thespecimens experience tensile and compressive stresses of equal magnitude. The cycles-to-failureexperience with 69 specimens of 5160H steel from 1.25-in hexagonal bar stock was as follows:L 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 where L is the life in thousands of cycles, and f is the class frequency of failures.(a) Estimate the mean and standard deviation of the life for the population from which thesample was drawn.(b) Presuming the distribution is normal, how many specimens are predicted to fail at less than115 kcycles?

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oeerivona oeerivona · Answer 1 · 2020-02-25T03:27:22+01:00

Answer:

The answers to the question are

(a) Mean = 122.9 and standard deviation = 30.3

(b) Based on the available data, the number of specimens that are predicted to fail at less than 115 kcycles = 50 %

Explanation:

To solve the question, we have attached a table of the values and we use the following formula for finding the mean

(a) The mean of the frequency distribution data is

[tex]\mu =\frac{1}{N}[/tex]∑[tex]f_{i}x_{i}[/tex] where

μ = mean,

[tex]f_{i}[/tex] = frequency at point i,

[tex]x_{i}[/tex] =value of data at point i

and N = sum of the frequencies

which give, from values in the table

μ = 8480/69 = 122.9

The standard deviation is given by

[tex]S_{x} =\sqrt{\frac{Sum(f_{i} x_{i}^{2}) -N\mu^{2} }{N-1} }[/tex] which from the attached table is

[tex]\sqrt{\frac{1104600-1042180}{69-1} }[/tex] = 30.3

(b) The z score of a normal distribution is given as

[tex]z=\frac{x-\mu_{x} }{\sigma_{x} }[/tex] Therefore a value of 115 kcycles is

(115 - 122.9)/30.3 = -0.26

Therefore we have P(115) = P(z<-0.26) ≈ 34+13.6+2.2+0.2 ≈ 50%