My Notes Ask Your al is determined by impressing a hardened point into the surfa

Question

My Notes Ask Your al 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 Rockwall hardness of a particular alloy is normally distributed with mean 69 and standard deviation 3. men is acceptable only if its hardness is betwoen 63 and 75, what is the probability that a randomly chosen spedimen has an acceptable hardness? (Round your answer to four decimal places.) (b tf the acceptable range of hardness s places.) to c, 69 + c r what value or c would 95% of al specimens have acceptable harness? Round your answer to two decimal o) Ir the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimans is independently determined, what is the expected number of acceptable specimens armong the ten? (Round your answer to two decimal places.) specimens (d) what is the probability that at most eight of ten independently selacted specimens have a hardness of less then 71.527 [Hint: y- the number among tha tan specimens hardness less than 71.52 is a binomial variable: what is p?] (Round your answer to four decimal places.) with appropriste tabl

Explanation / Answer

Z= (x-u)/

a)

z1= (63-69)/3 = -2

z2= (75-69)/3 = 2

Thus, P(63<u<75) = 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 = 69 – 1.96*3 = 63.12 and x2= 69 + 1.96*3 = 74.88

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= (71.52-69)/3 = 0.84

P(u<71.52) = P(z<0.84)= 0.7995

Thus, for binomial we have

n = 10, r= 8, p=0.7995,q=1-p = 0.2005

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

= 1 – 10C9*(0.7995)^9*(0.2005)^1 – 10C10*(0.7995)^10*(0.2005)^0

= 1 – 0.2676 – 0.1067

= 0.6257

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