Answer: 200
Step-by-step explanation:
Given : A hotel chain recently surveyed 300 new customers and asked them to rate their overall satisfaction with their recent hotel stay.
The survey estimated a mean satisfaction [tex]\mu=7[/tex] with a standard deviation [tex]\sigma=1[/tex].
Let x represents the rating given by customer.
Using formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex].
For x = 6, z= -1
For x= 8 , z=1
The probability that new customers gave the hotel a rating of 6,7, or 8 will be :-
[tex]P(6\leq x\leq8)=P(-1\leq z\leq1)=P(z\leq1)-P(z\leq-1)\\\\=P(z\leq1)-(1-P(z\leq1))\\\\=2P(z\leq1)-1\\\\=2(0.8413447)-1=0.6826894\approx0.68[/tex]
The number of new customers gave the hotel a rating of 6,7, or 8 will be :-
[tex]0.68\times300=204\approx200[/tex] [Rounded to the nearest tens]
Hence, the approximate number of new customers gave the hotel a rating of 6,7, or 8 =200
