Question #252110

In the manufacturing process, a compact disc is acceptable if its diameter is between 120.00mm and 120.20mm. A new machine produces compact discs whose diameters are normally distributed with mean 120.11mm and standard deviation 0.05mm.

(a) What is the probability that a randomly selected compact disc is acceptable?

(b) If four compact discs are randomly selected, what is the probability that at least two of them are not acceptable?

THX!



1
Expert's answer
2021-10-19T02:03:34-0400

The distribution of the diameters is N(120.11,0.052)N(120.11, 0.05^{2}) =120.11+0.05N(0,1)= 120.11 + 0.05N(0, 1)

(a) The probability that a randomly selected compact disc is acceptable is P(120<120.11+0.05N(0,1)<120.20)=P(2.2<N(0,1)<1.8)=P(120 < 120.11 + 0.05N(0,1) < 120.20) = P(-2.2 < N(0,1)< 1.8) =

=P(N(0,1)<1.8)P(N(0,1)<2.2)=0.964070.01390=0.95017=P(N(0,1)<1.8) - P(N(0,1) < -2.2) = 0.96407 - 0.01390 = 0.95017

(b) The amount acceptable discs can be described by X = Bin(n, p).

In our case: X = Bin(4, 0.95)

We have to find P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = (40)0.9500.054+(41)0.9510.053+(42)0.9520.052=0.014{4 \choose 0} *0.95^{0}*0.05^{4} + {4 \choose 1} *0.95^{1}*0.05^{3} +{4 \choose 2} *0.95^{2}*0.05^{2} = 0.014


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