A Travel Weekly International Air Transport Association survey asked business tr

Question

A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent business trip. Nineteen percent responded that it was for an internal company visit. Suppose 950 business travelers are randomly selected. What is the probability that more than 25 percentage of the business travelers say that the reason for their most recent business trip was an internal company visit. What is the probability that between 15 percentage and 20 percentage of the business travelers say that the reason for their most recent business trip was an internal company visit? 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?

Explanation / Answer

p = .19
n = 950
Mean = np = .19*950 = 180.5
Stdev = sqrt(.19*950*.81) = 12.09

a.P(p>.25) = ?

Z = (.25-.19)/(sqrt(.19*.81/950)) = 4.714 which is offcourse more than Z=1.96
indicating P(p>.25) is almost 0

b. P(.15<X<.20) = P(-3.1<Z<.79) = .7852-.001 = .7842

c. P( 133<X<171) = P( -3.93<Z<-.79)
we have normalized using mean and stdev
= .2148

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