Tutors, please sign in to answer this question.
We want 85% of the calls to be answered within the new time specification, so we must first find the the z-score z0 such that P(z≤z0)=0.85. From a table or using a calculator you find z0=1.036. Now use
z = (x-µ)/σ
to find the corresponding wait time:
1.036 = (x-210)/40,
which gives x=251.45 s.
Therefore, 85% of calls should be answered within a maximum wait time of 251.45 s.
As things stand, 68.26% of the wait time values fall withing ± 1 standard deviation of the mean.
If the company wants to set a new wait time so that 85% of the calls are answered within 1 standard deviation of the mean, then I think the new average wait time standard, Tmean, is given by:
(68.26%/210 sec.) = (85%)/Tmean
Tmean = 261.5 sec.
Since, however, the questions asks the new maximum wait time, Tmax, we need to calculate thusly:
(68.26%/250 sec.) = (85%)/Tmas
Tmax = 311.3 sec.
Maralyn, I admit off the top that I'm not necessarily a good statistician. Don't take my answer as Gospel.