Question

The Zimmerman Agency conducted a study for Residence Inn by Marriott of business travelers who take trips of five nights or more. According to this study, 37% of these travelers enjoy sightseeing more than any other activity that they do not get to do as much at home. Suppose 120 randomly selected business travelers who take trips of five nights or more are contacted. What is the probability that fewer than 40 enjoy sightseeing more than any other activity that they do not get to do as much at home?

Answer 1

Given: Proportion of travelers who enjoy sightseeing more than any other activity that they do not get to do as much at home

= 0.37

Let X be the event of travelers enjoying sightseeing more than any other activity that they do not get to do as much at home.

Then, the proportion is given by the formula,

We have to find:

Since, we are interested in success or failure of an event, we may go for a binomial distribution, with pmf

Hence,

Since, the above computation might be tedious, using an online calculator,

P(X<40) = 0.1774  

Also, another approach is using the normal approximation,

Here, Mean = n*p = 120*0.37= 44.4 > 10

Also, Variance = n*p*(1-p) = 120*0.37*(1-0.37)=27.92 > 10

We may now convert p to Z,

  

  

Hence,

  

  

From standard normal table,

We get

The probability that fewer than 40 enjoy sightseeing more than any other activity that they do not get to do as much at home is approximately equal to 0.2 .