Question

Question: Hotel guest satisfaction. Refer to the results of the 2015 North American Hotel Guest Satisfaction Index Study, Exercise 4.49 (p. 239). Recall that 15% of hotel guests were “delighted” with their experience (giving a rating of 10 out of 10); of these guests, 80% stated they would “definitely” recommend the hotel. In a random sample of 100 hotel guests, find the probability that fewer than 10 were delighted with their stay and would recommend the hotel.

Short answer

The probability that fewer than 10 () delighted with their stay and would recommend the hotel is 0.2676.

Step-by-step solution

Given information

Let p denotes the proportion of guests who were delighted with their stay and would recommend the hotel.

15% of guests were delighted with their experience, and of these, 80% would recommend the hotel

Therefore,

.

A random sample of size 100 is selected.

Computing the required probability

The probability that fewer than 10 () delighted with their stay and would recommend the hotel obtain as follows

$$\begin{gathered} P\left(\hat{p}<.10\right)=P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}<\frac{0.10-p}{{\sigma }_{\hat{p}}}\right) \\ =P\left(Z<\frac{0.10-0.12}{\sqrt{\frac{0.12\times 0.88}{100}}}\right) \\ =P\left(Z<\frac{-0.02}{\sqrt{0.001056}}\right) \\ =P\left(Z<\frac{-0.02}{0.0324}\right) \\ =P\left(Z<-0.62\right) \\ =0.2676 \end{gathered}$$

In the z-table, the value at the intersection of -0.60 and 0.02 is the required probability.

Therefore, the required probability is 0.2676.