A copper wire manufacturer produces conductor cables. These cables are of practi

Question

A copper wire manufacturer produces conductor cables. These cables are of practical use if their resistance lies between 0.10 and 0.13 ohms per meter. The resistance of the cables follows a normal distribution, where 50% of the cables have resistance under 0.11 ohms and 10% have resistance over 0.13 ohms. Determine the mean and the standard deviation for cable resistance. Find the probability that a randomly chosen cable can be used. Find the probability that at least 3 out of 5 randomly chosen cables can be used.

Explanation / Answer

a)

As 50% of the population is under the mean, then

Mean = u = 0.11 ohms [ANSWER, MEAN]

For a right tailed area of 0.10, the corresponding z score, using table/technology, is

z = 1.281551566

Thus, as

z = (x - u) / s

Then

1.281551566 = (0.13-0.11)/s

Thus,

s = 0.015606083 [ANSWER, STANDARD DEVIATION]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.1      
x2 = upper bound =    0.13      
u = mean =    0.11      
          
s = standard deviation =    0.015606083      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.64077578      
z2 = upper z score = (x2 - u) / s =    1.281551559      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.260834186      
P(z < z2) =    0.899999999      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.639165813   [ANSWER]

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c)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.639165813      
x = our critical value of successes =    3      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   2   ) =    0.252227114
          
Thus, the probability of at least   3   successes is  
          
P(at least   3   ) =    0.747772886 [ANSWER]
  

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