Question

There is a chance that a bit transmitted through a digital transmission channel is received in error. Let X equal the number of bits in error in the next three bits transmitted. Suppose that the performed experiment on the entire population gives the probability distribution below,

x	0	1	2	3
P(X=x)	0.656	0.292	0.049	0.003

Find

a)The cdf of X, and graph it

	x < 0	0 ? x < 1	1 ? x < 2	2 ? x < 3	3 ? x < 4
F(x)

b)The expected number of bits in error, µ_X, and the standard deviation ,?_X, in the next three bits transmitted.

c)The probability of observing at most 2 bits in error in the next three bits transmitted.

d)The probability of observing at least 1 bit in error in the next three bits transmitted.

Answer 1

(a)

The CDF is obtained by adding the values of PDF at each step.

So the CDF for this pdf is:

	x<0	0 <= x < 1	0 <= x < 2	0 <= x < 3	0 <= x < 4
F(x)	0	0.656	0.948	0.997	1

(b)

E[X] = 0*0.656 + 1*0.292 + 2*0.049 + 3*0.003 = 0.399

Var(X) = E[X^2] - (E[X])^2

E[X^2] = 0*0.656 + 1*0.292 + 4*0.049 + 9*0.003 = 0.515

So,

Var(X) = 0.515-(0.399^2) = 0.356

So,

?(X) = Var(X)^0.5 = 0.356^0.5 = 0.597

(c)

Here we are asked to calculate P(X <=2).

Looking at the CDF, we get:

P(X <=2) = 0.948

(d)

Here we are asked to calculate P(X >= 1).

Using the formula:

P(X >= 1) = 1 - P(X <1) = 1-P(X=0) = 1-0.656 = 0.344