Suppose that the weights of airline passenger bags are normally distributed with

Question

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds.

a) What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds? Give your answer to four decimal places.  

b) Let X represent the weight of a randomly selected bag. For what value of c is P(E(X) - c < X < E(X) + c)=0.87? Give your answer to four decimal places.  

c) Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 16 that weigh greater than 50 lbs? Give your answer to four decimal places.  

d) Assuming the weights of individual bags are independent, what is the probability that 4 or fewer bags weigh greater than 50 pounds in a sample of size 16? Give your answer to four decimal places.

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    50      
u = mean =    47.88      
          
s = standard deviation =    3.09      
          
Thus,          
          
z = (x - u) / s =    0.686084142      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.686084142   ) =    0.246330025 [ANSWER]

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b)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.87      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.065      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.514101888      
By symmetry,          
z2 =    1.514101888      
          
As          
          
u = mean =    47.88      
s = standard deviation =    3.09      
          
Then

c = z*sigma = 1.514101888*3.09 = 4.678574834 [ANSWER]

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c)

E(x) = 0.246330025*16 = 3.9412804 [ANSWER]

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D)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    16      
p = the probability of a success =    0.246330025      
x = the maximum number of successes =    4      
          
Then the cumulative probability is          
          
P(at most   4   ) =    0.64337499 [ANSWER]

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