(12 total points) The hardness of a metal is determined by impressing a hardened
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Question
(12 total points) The hardness of a metal is determined by impressing a hardened point in the surface of the metal and then measuring the depth of the penetration of the point. Let X denote the hardness of a particular alloy where X-N(70,9). 9. (a) (2 points) If a specimen is acceptable only if its hardness is between 67 and 75, what is the probability that a randomly chosen specimen has an acceptable hardness? (b) (3 points) What hardness is needed so that only 5% of the alloys have a higher hardness? (c) (3 points) If the acceptable range is between 67 and 75 and the hardness of 10 randomly selected specimen are chosen, what is the expected number of acceptable specimens among then 10? Hint: This relates to one of the discrete probability distributions. (d) (4 points) What is the probability that at most 8 of the 10 selected specimens have a hardness of less than 73.84? Hin:t This relates to one of the discrete probability distributionsExplanation / Answer
a)
Let X be the hardness of the material such that X~N(70,9).
Hence, P(67<=X<=75) = P(((67-70)/9)<=z<=((75-70)/9)) = P(-0.33<=z<=0.56)
= P(z<=0.56) – P(z<=-0.33) =0.7123 – 0.3707 = 0.3416 or 34.16% Using z-table
b)
Let c be such that (70 - c, 70 + c) will contain 95% of all specimen having acceptable hardness.
Hence,
c/? = z0.05/2
since z0.025 = 1.96, c = 1.96*9 = 17.64
Hence the range would be (70-17.64,70+17.64) ie (52.36,87.64)
c)
Because the sampling is independent, it follows a Binomial distribution with p = 0.3416, therefore the mean is m = 10*0.3416 ~ 3
d)
P(X <= 73.84) = 0.89973
For a binomial distribution:
P(p = 0.9, n = 10, x = 9) = 0.38742
P(p = 0.9, n = 10, x = 10) = 0.34868
Thus, the probability that at most eight of ten specimens have a hardness of less than 73.84 is:
1 – 0.38742 – 0.34868 = 0.264
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