Question

The data points given were:

With the true mean calculated as : 24,387.28

f. Given the distribution type and parameter what is the probability of seeing rates within ± 5000 of the true mean? Does the result correlate to behavior associated with the standard normal distribution? Show work that supports your answer. Data Packets Per Hr (Bin) Packets Per HR 22506 10000 12500 28707 15000 21567 17500 23168 20000 24782 22500 27809 26081 25000 31527 27500 30000 25964 32500 20922 26206 35000 20305 37500 40000 27316 14585 25908 25318 29754 26286 11908 24783 24184 20286 21376 29561 28873

Answer 1

Answer:

Given data:

10000,12500,15000,17500,20000,22500,25000,27500,30000,32500,35000,3500,40000,22506,28707,21567,23168,24782,27809,26081,31527,25964,20922,26206,20305,27316,14585,25908,25318,29754,26286,11908,24783,24184,20286,21376,29561,28873

Now to find the mean($\mu$):

Mean($\mu$)=sum of observations/number of observations

=10000+12500+15000+17500+..................................................+20286+21376+29561+28873/38

=90062/38

=23702.16

Therefore the population mean($\mu$) is 23702.16.

Therefore the probability of seeing rates within $\pm$5000 of the true mean is 24,387.28

Yes,the result correlate to behaviour associated with the standard normal distribution.