There is a chance that a bit transmitted through a digital transmission channel

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, µX, 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.

x

0

1

2

3

P(X=x)

0.656

0.292

0.049

0.003

Explanation / Answer

(a)

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

So the CDF for this pdf is:

(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

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

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