91Ó°ÊÓ

Referring to the 2015 North American Hotel Guest Satisfaction Index Study, Exercise 4.49, the hotel guest was delighted with their stay and would recommend the hotel is 0.12. A large hotel chain randomly samples 200 guests. More than 50 out of 200 guests were delighted with the stay and would recommend the hotel.

The probability that 50 out of 200 guests were delighted with the stay and would recommend the hotel is,

The mean is,

$\begin{array}{rcl}\mathrm{μ}& =& np\\ & =& 200×0.12\\ & =& 24\\ & & \end{array}$

The variance is,

$\begin{array}{rcl}{\mathrm{σ}}^2& =& np\left(1-p\right)\\ & =& 200×0.12×\left(1-0.12\right)\\ & =& 200×0.12×0.88\\ & =& 21.12\\ & & \end{array}$

The standard deviation is,

$\begin{array}{rcl}\mathrm{σ}& =& \sqrt{21.12}\\ & =& 4.5957\\ & & \end{array}$

For,$x=50$

The z-score is,

$\begin{array}{rcl}z& =& \frac{x-\mathrm{μ}}{\mathrm{σ}}\\ & =& \frac{50-24}{4.5957}\\ & =& 5.6575\\ & & \end{array}$

$\begin{array}{rcl}P\left(x>50\right)& =& 1-P\left(zâ\copyright ½5.6575\right)\\ & =& 1-1  \left[From S\tan dard Normal Table\right]\\ & =& 0\\ & & \end{array}$

The probability value is 0

Therefore, the guests were delighted with the stay and would recommend the hotel is 0.12. Then it will unlike to say that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel since the obtained probability value is extremely small,under this scenario is claimed likely to be false.

If the claim is likely to be true, then the probability that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel must be higher than 0.12; under this scenario is, the claim likely to be true.