Statistics - Estimation?

Question

When customers phone an airline to make reservations, they usually find it irritating if they are kept on hold for a long time. In an effort to determine how long phone customer are kept on hold, one airline took a sample of 167 phone calls and determined the length of time (in minutes) each caller was kept on hold. The sample standard deviation was 3.8 minutes. How many more phone customers, should be included in the sample to be 99% sure that the sample mean (X bar) of hold times is within 30 seconds of the population mean (mew) of hold times?

I Have an answer, I figured it out now :)

Standard Deviation = 3.8
Zc = 99% (2.58)
E = 30 secs (0.5) half a minute

We are trying to find N?

N = (2.58)(3.8)/ 0.5 sum of this is squared
N = 19.608 squared
N = 384.47 we always round up!
N = 385

The Q says how many MORE? so we subtract 167

N = 385 - 167
N = 218 or more

Answer 1

About another 360. The standard deviation will still be about 3.8 minutes, but the "standard error of the mean" will be 3.8 / sqrt(527) = 10 secs approx, and I think the 99% confidence interval is 3 standard deviations.

Answer 2

basically some random concepts approximately this. For the form of trouble-loose case, Pr(X=a million) is basically the reported danger, or 3 / 10. And Pr(X=4) = a million /10 to word the rule of thumb of succession, we'd desire to assume that the cube is truthful. Then we could say that we've "seen" (a prior) one 4 in 6 rolls, plus the only 4 in 10 that we easily have been given. So then the Pr(X=4) is two / sixteen. extra effective many times, Pr(X=4) = (a million + W) / (10 + 6W), the region W is 0 to infinity, and reflects how specific we are that the cube is truthful.

Answer 3

30 seconds.!