After being cut by a special saw, suppose the length of aluminum alloy billets,

Question

After being cut by a special saw, suppose the length of aluminum alloy billets, X, is
normally distributed with mean = 96.2 inches and standard deviation = 0.2 inches.
Suppose you randomly select 20 billets (assume independence).
(a) Determine the probability that exactly 2 are longer than 96.5 inches. Hint: First
find the probability that a single billet is longer than 96.5 inches, then use the
binomial distribution.
(b) Determine the probability that 19 or fewer are have a length less than 96.5 inches.
(c) Determine the probability that all 20 billets are between 96.0 and 96.5 inches.

Explanation / Answer

a. Assume r.v X denote length of aluminium alloy billets. P[X>96.5]=P[Z>(96.5-96.2)/(0.2)] [use formula Z=(X-mu)/sigma, where, X is the raw score, mu is population mean and sigm ais population standard devaition.

=P[Z>1.5]=1-0.9332 [for any area greater than Z, subtract the area corresponding to the Z score from 1, because entry for Z in the standard normal table is the area to the left of Z]

=0.0668

Assume r.v Y denote number of billets that are longer than 96.5 inches.

P(Y=2)=20C2(0.0668)^2(1-0.0668)^18=0.2443 [Use P(Y,r)=nCr(p)^r(1-p)^n-r, where, p is probability of success, n is number of random, independent trails, and r is speciifc number of success in n trials]

b. P(X<96.5)=P(Z<1.5)=0.9332

P(Y<=19)=0.270 [use binomial table with p=0.9 and n=20, x=19]

c. P[96<X<96.5]=P[(96-96.2)/0.2<Z<(96.5-96.2)/0.2]=P(-1<Z<1.5)=0.3413+0.4332=0.7745

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