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 69 and standard deviation 3.

(a) If a specimen is acceptable only if its hardness is between 64 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 (69 − c, 69 + c), for what value of c would 95% of all specimens have acceptable hardness? (Round your answer to two decimal places.)


(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 72.84? [Hint: Y = the number among the ten specimens with hardness less than 72.84 is a binomial variable; what is p?] (Round your answer to four decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question

Answer 1

Solution:

(a) Let X be the hardness of the metal, then X ~ N(69, 3), and therefore:
P(64 ≤ X ≤ 74) = P[64-69/3 ≤ Z ≤ 74-69/3] = P(-1.67 ≤ Z ≤ 1.67)
= P ( Z<1.67 )−P (Z<−1.67 )
= 0.9525 - 0.0475 = 0.905
(b) |c|/σ = z_0.05/2, since z_0.025 = 1.96, c = (1.96)(3) = 5.88
(c) Because the sampling is independent, its distribution is Binomial with p = 0.905, therefore the mean is μ = n*p = (10)(0.905) = 9.05

(d) First P(X < 72.84) = 0.89973, then, use the 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 72.84 is:
       1 – 0.38742 – 0.34868 = 0.264