Suppose that the weights of airline passenger bags are norma

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.

Solution

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]

b)

Here, c is the margin of error,

c = z * sigma

As alpha/2 = (1-0.87)/2 = 0.065, the correspondingz is

z = 1.514101888

Thus,

c = 1.514101888*3.09 = 4.678574833 [answer]

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c)

Expected number = 16*0.246330025 = 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]