Question 3 A Travel Weekly International Air Transport Association survey asked

Question

Question 3 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. a. What is the probability that more than 20% of the business travelers say that the reason for their most recent business trip was an internal company visit? b. What is the probability that between 18% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

Explanation / Answer

p= 0.19 , standard deviation = sqrt( p (1-p) / n )

p(p > 0.20) = p( z > 0.20 - 0.19 / sqrt(0.19*0.81/950) )

= p( z > 0.01/0.01273)

= p(z> 0.7857)

= 1 - p( z < 0.7857)

= 1 - 0.78398

= 0.21602

b)

p( 0.18<p<0.20) = p( p<0.20) - p(p<0.18)

= 0.78398 - p(z<0.18-0.19 / 0.01273) (as calculated in first part)

= 0.78398 - p(z< -0.7857)

= 0.78398 - (1-p(z< 0.7857))

= 0.78398 - 0.21602

= 0.56796

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