A sequence of digital message with a total number of binary words equal to 11600

Question

A sequence of digital message with a total number of binary words equal to 11600 is divided equally (N_A = N_B = 5800) across 2 consequent segments A, and B. Each segment accommodates 10 words of possible bit-lengths {x_i}; and, the words appear randomly in follows: Determine the mean (AM) and median of the word population in each segment Discuss the stationarity in mean of the process of digital message carried by the binary words across the 2 segments Determine the Kullback-Leibler (KL) statistical divergence between the segments A and B.

Explanation / Answer

a) Determining Mean (AM) and Median

A = S / N

A = Arithmetic mean

S = Sum total of the word population

N = Number of population

ie 5800/10 = 580 , Therefore mean for both the segments A and B is 580

For finding median population is to be arranged in assending order

Segment A

Segment B

Median = n / 2 and (n / 2) + 1

n = Number of population

ie 10/2 = 5 and 5 + 1 = 6 , The elements in 5th and 6th position are for

Segment A it is 638 and 779 , Therefore Median = 638 + 779 / 2 ie 1417 / 2 = 708.5

Segment B it is 628 and 769 , Therefore Median = 628 + 769 / 2 ie 1397 / 2 = 698.5

1 111 2 151 3 162 4 527 5 638 6 779 7 796 8 816 9 862 10 958

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