There are two machines available for cutting corks intended for use in wine bott
ID: 3150562 • Letter: T
Question
There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally dis- tributed with mean 3 cm and standard deviation .1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation .02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. 1. If a shipment consists 30% of corks produced by the first machine, what is the probability that a random cork is acceptable? 2. If a shipment consists 30% of corks produced by the first machine, what is the probability that a randomly picked acceptable cork comes from the second machine? 3. What should be the mean diameter of corks produced by machine 1 that optimizes this machine’s production?
Explanation / Answer
1. If a shipment consists 30% of corks produced by the first machine, what is the probability that a random cork is acceptable?
For machine 1:
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 2.9
x2 = upper bound = 3.1
u = mean = 3
s = standard deviation = 0.1
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.158655254
P(z < z2) = 0.841344746
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.682689492
For machine 2:
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 2.9
x2 = upper bound = 3.1
u = mean = 3.04
s = standard deviation = 0.02
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -7
z2 = upper z score = (x2 - u) / s = 3
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 1.27981E-12
P(z < z2) = 0.998650102
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.998650102 [ANSWER]
Let
A = acceptable
hence,
P(A) = P(1) P(A|1) + P(2)P(A|2) = 0.30*0.682689492 + (1-0.30)*0.998650102 = 0.903861919 [ANSWER]
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2. If a shipment consists 30% of corks produced by the first machine, what is the probability that a randomly picked acceptable cork comes from the second machine?
Hence,
P(2|A) = P(2) P(A|2)/P(A) = (1-0.30)*0.998650102/0.903861919 = 0.773409142 [ANSWER]
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3. What should be the mean diameter of corks produced by machine 1 that optimizes this machine’s production?
Acceptable corks are 2.9 to 3.1, so the mean should be at the middle,
u = (2.9+3.1)/2 = 3.0 cm [ANSWER]