SOLUTION: Email: sfmaharshi@gmail.com I'm stuck with this problem. Can you please help me understand the process of solving it? The average waiting time to be seated for dinner at a popular

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Question 847125: Email: sfmaharshi@gmail.com
I'm stuck with this problem. Can you please help me understand the process of solving it? The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with standard deviation of 3.6 minutes. Assume the variable is normally distributed. When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the following time.
a. Between 15 and 22 minutes
b. Less than 18 minutes or more than 25 minutes
c. Is it likely that a person will be seated in less than 15 minutes?
Thank you

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
'extracting' necessary problem solving info from the 'Words'
mean = 23.5 and SD = 3.6
Using TI Calculator
a. P(15≤x ≤22) = normalcdf(smaller, larger, µ, σ) = normalcdf(15 22, 23.5, 3.6)
b. P(x<18 0r x >.25) = 1 - P(18≤x ≤25)= 1 - normalcdf(18, 25, 23.5, 3.6)
c. NO, over 2 SDs from mean ( |15-23.5| = 8.5)
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