Hi all,

I've solved the question but I'm not sure whether or not my answer is correct. I need your help.

Thanks in advance.

The length of time it takes to be seated at a local restaurant on Friday night is normally distributed with a mean of 15 minutes and a standard deviation of 4.75 minutes.

a. What is the probability that you have to wait more than 20 minutes to be seated?

b. What is the probability that you have to wait between 13 and 16 minutes to be seated?

a)

x= length of time

μ= 15

σ= 4.75

P(x>20)

For x=20, z=(20-15)/4.75 -> 1.052

P(x>15)= P(z> 1.052) = [total area] - [area to the left of z = 1]

b)1- 0.8536 = 0.1464

x= length of time

μ = 15

σ= 4.75

P(13<x<16) For x=13, z= (13-15)/4.75 -> -0.421

For x=16, z= 16-15/4.75 -> 0.210

P(13<x16) = P(-0.412<z< 0.210)

0.416-0.340= 0.076