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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]