Question 788781
A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent business trip. 19% responded that it was for an internal company visit.
p = 0.19
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Suppose 950 business travelers are randomly selected. 
a. What is the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit? 
z(0.25) = (0.25-0.19)/sqrt[0.19*0.81/950] = 1.4907
P(p > 0.25) = P(z > 1.4907) = 0.0680
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b. What is the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit? 
Find the z-values of 20% and 15%; then find the probability between them.
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c. What is the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit?
Find 133/950 and 171/950
Find their z-values.
Find the probability between them.
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Cheers,
Stan H.
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