Assume the length of waiting at supermarket is approximately normally distributed with mean 6 minutes and standard deviation 1.5 minutes. (1) 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
Could anyone please check (1) for me?
I assume (1) and (2) mean the same thing here.
What you have done is reasonable, except you have rounded at the wrong time: if the normal approximation was exact, the value would be closer to $9.4895$ then to $9.4905$, but there is in any case no good reason for being any more precise than $9.49$ minutes or perhaps $9$ minutes and $29$ seconds if you prefer sexagesimal.
As you say, a normal approximation cannot be exact if negative values are impossible. Your approach of putting this very small amount (about $0.00003$) at $0$ rather than negative is probably the most sensible thing to do.
In real life, queuing times are often zero and can have a right skewed distribution and so a normal approximation is probably inappropriate.