The waiting time x at a fast-food restaurant during lunch time is ...

[karen]

The waiting time x at a fast-food restaurant during lunch time is approximately normally distributed with a mean of 4.5 min and a standard deviation of 1.2 min. Find the value of the 75th percentile for x.

Question 2

The waiting time x at a fast-food restaurant during lunch time is approximately normally distributed with a mean of 4.5 min and a standard deviation of 1.2 min. Find the probability that a randomly selected customer has to wait more than 6.8 min.

Question 3

The waiting time x at a fast-food restaurant during lunch time is approximately normally distributed with a mean of 4.5 min and a standard deviation of 1.2 min. Find the probability that a randomly selected customer has to wait less than 2.7 min.

[dudman123]

Answer to Question 1

75th percentile is a value such that 75 of the data is less than this value; therefore the z-score of this value is to the right of 0 such that the area between 0 and z is 0.25 . Hence,
the corresponding z value is z = +0.67 . Now, the formula z = (x  4.5) / 1.2 implies 0.67 = (x  4.5) / 1.2 . Solving for x, we get x = (0.67) (1.2) + 4.5 = 5.304 minutes.

Answer to Question 2

P(x > 6. = P(z > 1.92) = 0.5000  0.4726 = 0.0274

Answer to Question 3

P(x < 2.7) = P( z < -1.50) = 0.5000  0.4332 = 0.0668

[karen]

Wow, this really help

[31809pancho]

Thanks for the timely response, appreciate it