The Rockwell hardness of a metal is determined by impressing ahardened point int

Question

The Rockwell hardness of a metal is determined by impressing ahardened point into the surface of the metal and then measuring thedepth of penetration of the point. Suppose the Rockwell hardness ofa particular alloy is normally distributed with mean 71 andstandard deviation 2.5.
a) If a specimen is acceptable only if its hardness is between67 and 75, what is the probability that a randomly chosen specimenhas and acceptable hardness?
b)If the acceptable range of hardness is (71- C, 71+C), forwhat value of C would 95% of all specimens have acceptablehardness?
c)If the acceptable range is as in part (a) and the hardnessof each of ten randomly selected specimens is independentlydetermined, what is the expected number of acceptable specimensamong the ten?
d)What is the probability that at most eight of tenindependently selected specimens have a hardness of less than 73.84(HINT Y=number among the ten specimens with hardness less than73.84 is a binomial variable ; what is p?)
a) If a specimen is acceptable only if its hardness is between67 and 75, what is the probability that a randomly chosen specimenhas and acceptable hardness?
b)If the acceptable range of hardness is (71- C, 71+C), forwhat value of C would 95% of all specimens have acceptablehardness?
c)If the acceptable range is as in part (a) and the hardnessof each of ten randomly selected specimens is independentlydetermined, what is the expected number of acceptable specimensamong the ten?
d)What is the probability that at most eight of tenindependently selected specimens have a hardness of less than 73.84(HINT Y=number among the ten specimens with hardness less than73.84 is a binomial variable ; what is p?)

Explanation / Answer

a)67 and 75 are each 4 away from the mean, with a standarddeviation of 2.5 this is a z-score of 1.6 area under standard normal curve from -1.6 to 1.6 is 0.8904 b) 95% C.I. corresponds to a z-score of 1.960 C = 1.960*4 = 7.84 c) 10 * .8904 = 8.9 p = normalcdf(-1E99,73.84,71,4) = 76.115 binomcdf(10,.76115,8) = 0.7299

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