ANSWERS should be typed please! answer all parts of question please! any wrong/i

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Question # 2 A digitized video transmission carries a randomly mixed set of video packets. The occurrence statistics of the packets are designated in 10 groups. The video-transmission involves streaming of packets wherein the statistics of occurrence of bit rate may depend strongly on bit-rate values in consequent packets streamed and the RV is considered as of variable bit rate (VBR) type The number of occurrences of each group and the corresponding random value of the associated random bit rates in the packets (RVs) are as follows Events enumerated Bit rate (R) Video in Mbps: RV 6.4 5.9 0.6 6.2 5.76.1 6.35.7 5.6 0.2 packets Number of 80 | 60 | 50 | 40 | 50 | 60 | 75 40 30 20 occurrenceS Determine the following of the RV (Bit rate -R) Find the upper and lower bounds of the average performance of the RV State with reason the most appropriate mean value (AM HIM or GM) that describes the average performance of the bit rate I. ?. III. Find the expected mean, median, mode, variance and standard deviation IV. Plot the cumulative probability distribution (CDF) of the RV. Answer-hints: Median 5.8: ? = 1.9431

Explanation / Answer

Sol :-

A digitized video transmission conveys a haphazardly blended arrangement of video bundles. The event measurements of the parcels are planned in 10 gatherings.

A) the upper and lower bounds of the average performance of the RV:

Now, R - 48.7

ni = 505

Rini = 2237.5

The normal execution of the arbitrary variable is probably going to be around of 4.87mbps.
?The upper limit is 5mbps.
The lower limit is 4.5mbps.

B) From the above piece we had seen that the normal execution of arbitrary variable is 4.87mbps.

What's more, the normal the vast majority of bit rate will false need to it.
The information is the greater part of the qualities symmetrically closer to the 4.87mbps.

C)  the mean, median, mode:

   Mean:

           Mean = Rini/ni

           We know that the values of Rini and ni

Rini = 2237.5

ni = 505

So, mean = 2237.5/505

Mean = 4.43

Median:

Median can be given as, now n = 505

The median will be average of 242th or 253th observation.

                       So, Median = (6.4 + 6.4) / 2

                                               = 128/2

                                   Median = 64

Mode:

  

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