1041 words - 5 pages

1. What fraction of the days in 2004-2004 failed to meet the targeted daily average hold time of 110 econds? Given that the daily average hold time was normally distributed with a mean of 99.67 and a standard deviation of 24.24, what fraction of days where the call center failed to meet the targeted hold time of 110 seconds would you expect?

a) Based on the data in the exhibit 1, the number of days out of target hold time (more than 110 seconds) is 237. And total number of days is 728. Therefore, fraction of days is equal to 237/728 or 0.32.

Total number of days | 728 |

Number of days within target hold time (110 seconds or less) | 491 |

Number of days out of target hold time ...view middle of the document...

033 |

Fraction of days where the call center failed to meet targeted hold time (%) | 3% |

b) Based on the X=X+z*σ equation, we find the value of z.

X=79.50

σ=16.86

Target hold time = 110 seconds

Then, 110 = 79.50 + z * 16.68. From the equation, z is equal to 1.82. Thus, fraction of days where the call center failed to meet targeted hold time is equal to 1 - 0.9656 or 0.0344.

3. Based on the performance in April 2005, do you think that the performance of the call center has improved? Think about what the chance would be to observe such performance in April 2005 if the process was not improved.

As can be clearly seen from the data, the fraction of days out of target hold time (more than 110 seconds) reduced drastically from 0.33, that is one out of every three calls, to 0.034, that is only one out of thirty calls.

So basically if the process was not improved, then instead of just one day out of targeted hold time we would have observed 10 days out of targeted hold time in the month of April. Thus, we think that there has been a great deal of improvement in the performance after new process was implemented.

4. If we assume that call center performance during the month of September is continuing at the improved level with a mean of 79.50 and a standard deviation of 16.86, what is the probability of observing ten days that average 86.6 or more? What is the probability of observing ten days that average 74.4 or less? Use the normal distribution.

Since we are considering samples of size 10 (10 days), we need to consider the distribution of sample means. Sample means have an average of 79.50 and a standard deviation of 16.86/√10 = 5.33

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