The Rockwell hardness of a metal is determined by impressing a hardened point in

Question

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 68 and standard deviation 3 (a) If a specimen is acceptable only if its hardness is between 62 and 74, what is the probability that a randomly chosen specimen has an acceptable hardness? (Round your answer to four decimal places.) b) If the acceptable range of hardness is 68 places c, 68 + c ,for what value of c would 95% of all specimen have acceptable hardness? Round your answer to od ci ma (c) If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten? (Round your answer to two decimal places.) specimens (d) what is the probability that at most eight of ten independently selected specimens have a hardness of less than 71.84? [Hint: Y= the number among the ten specimens with hardness less than 71.84 is a binomial variable; what is p?] (Round your answer to four decimal places.)

Explanation / Answer

Z= (x-u)/

a)

z1= (62-68)/3 = -2

z2= (74-68)/3 = 2

Thus, P(62<u<74) = P(-2<Z<2)

= 0.9772 - 0.0228=0.9544

b)

z score for 95% confidence interval = +- 1.96

thus, x1 = u - z* and x2 = u + z*

x1 = 68 – 1.96*3 = 62.12 and x2= 68 + 1.96*3 = 73.88

C = 1.96*3 = 5.88

c)

Expected number = Probability * total count

= 0.9544*10 = 9.544

Rounding off to 2 decimals = 9.54

d)

z= (71.84-68)/3 = 1.28

P(u<71.52) = P(z<1.28)= 0.8980

Thus, for binomial we have

n = 10, r= 8, p=0.898,q=1-p = 0.102

P(atmost 8) = 1 – P(exactly 9)- P(exactly 10)

= 1 – 10C9*(0.898)^9*(0.102)^1 – 10C10*(0.898)^10*(0.102)^0

= 1 – 0.3873 – 0.3410

= 0.2717

Note: You need to know z score probability to solve this one

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