Question

An article in the Los Angeles Times (December 8,1991) reported that there are 40,000 travel agencies nationwide, of which 11,000 are members of the American Society of Travel Agents (booking a tour through an ASTA member increases the likelihood of a refund in the event of cancellation). a. If \(x\) is the number of ASTA members among 5000 randomly selected agencies, could you use the methods of Section \(7.8\) to approximate \(P(1200

Short answer

a) Yes, the normal approximation can be used as both np and n(1-p) are greater than 5. b) The mean value is 27.5 and the standard deviation is 4.4295. c) No, the standard deviation does not double but increases by a factor of \(\sqrt{2}\) when the sample size is doubled.

Step-by-step solution

Evaluate The Normal Approximation For Large n

In part a, with a sample size of 5000, \(x\) is the number of ASTA members among the randomly selected. The proportion of ASTA members nationwide is \(p = 11000/40000 = 0.275\). The large sample size makes it possible to use a normal approximation. However, in this scenario both \(np = 5000*0.275 = 1375\) and \(n(1-p) = 5000*0.725 = 3625\) are greater than 5, justifying the application of the normal approximation to the binomial distribution.

Calculate Mean and Standard Deviation

In part b, we use the formulas for the mean \(\mu=np\) and standard deviation \(\sigma=\sqrt{np(1-p)}\) of the binomial distribution. With n=100 and p=0.275, we have \(\mu=np=100*0.275=27.5\) ASTA members and \(\sigma = \sqrt{100*0.275*0.725} = 4.4295\). Thus, among 100 agencies, on average 27.5 are ASTA members, with a standard deviation of 4.4295.

Understand The Change In Standard Deviation When Sample Size Doubles

In part c, the question asks if the standard deviation will double with the sample size. The formula for standard deviation is \(\sigma = \sqrt{np(1 - p)}\). If n (the sample size) is doubled, the formula becomes \(\sigma = \sqrt{2np(1 - p)} = \sqrt{2} * \sqrt{np(1 - p)}\). Therefore, the standard deviation does not double but increases by a factor of \(\sqrt{2}\). Note that this is assuming that the proportion p stays the same as sample size increases.