Normally distrbution with Probability

[Askhwhelp]

Assume the length of waiting at supermarket is approximately normally distributed with mean 6 minutes and standard deviation 1.5 minutes.
(1) Fund the probability that waiting time is longer than 8 minutes
My way is P(z > (8-6)/1.5) = P(z > 4/3) = .0918
(2) What length of the waiting time constitutes the 99th percentile ($Φ_.99$)?
My way is
For 99th percentile, find p(z > 2.327) = .01
find (y-6)/1.5 = 2.237 <=> y = 9.4905. For 0th percentile, y should be negative infinity. However, it does not make sense since we talk about time here. We should take 0 instead.
The waiting time length is between 0 and 9.4905
(3) Find probability that waiting time lasted longer than 6 minutes given that it lasted longer than 5 minutes
My way is P(Y>6|Y>5) = P(Y>6,Y>5)/P(Y>5) = P(Y>6)/P(Y>5) = P(z > (6-6)/1.5)/P(z > (5-6)/1.5) = P(z > 0) = P(z > -2/3) = .5/(1-.2514)=.67.

Could anyone please check (1), (2), (3) for me?

 

[mighty2000]

Yes, your calculations for (1), (2), and (3) are correct. Your method for finding the 99th percentile is also correct, but you could also use a z-score table to find the z-score corresponding to the 99th percentile, which is 2.3263. Then, you could plug this z-score into the formula (y-6)/1.5 = z to solve for y, which would give you the same result of 9.49 minutes.