Travel Weekly International Air Transport Association survey asked business trav

Question

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. responded that it was for an internal company visit. Suppose randomly selected. What is the probability that between 15percentage and 20 percentage of the business travelers say that the reason for their most recent business trip was an internal company visit?

Explanation / Answer

percentage of travellers who said that they was for an internal company visit = 19%

proprtion of travellers who said that they was for an internal company visit = 0.19

sample size i.e n = 950

standard deviation of prpoportion of tarvellers that was for an internal company visit

=[p*(1-p)/n]0.5 = [0.19*(1-0.19)/950]0.5 = 0.0127

z value for p=0.15 is (0.15-0.19)/0.0127 = -3.1496 , corresponding p value using z table is 0.0008

P(proportion<0.15) = 0.0008

z value for p=0.20 is (0.20-0.19)/0.0127 = 0.7874, corresponding p value using z table is 0.7845

P(proportion<0.20) = 0.7845

P(0.15<proportion<0.20) = 0.7845-0.0008 = 0.7837

P(15%<ppercentage of traveelers for an internal company visit<20%) = 0.7837

probabilty that bertween 15% and 20% of the business travellers say that the reason for their most recnt bsiness trip was ans internal company visit = 0.7837

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