A computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent. Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by

where c is a constant.

a. Find the value of the constant c so that p(x) is a probability mass function.

b. Find P(X = 2).

c. Find the mean number of times the packet is sent.

d. Find the variance of the number of times the packet is sent.

e. Find the standard deviation of the number of times the packet is sent.

Solution 7E

Step1 of 3:

We have A computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent.

Let X be random variable which represents the number of times the packet is sent. Assume that the probability mass function of X is given by

Where c is constant.

We need to find,

a).Find the value of the constant c so that p(x) is a probability mass function.

b).Find P(X = 2).

c).Find the mean number of times the packet is sent.

d).Find the variance of the number of times the packet is sent.

e).Find the standard deviation of the number of times the packet is sent.

Step2 of 3:

a).

We have

Substitute p(x) value in above equation we get

Where c is constant.

c

c(1+2+3+4+5) = 1

c15 = 1

c = 1/15.

Therefore, the value of c is 1/15.

b).

We have

= (1/15)2

= 2/15

Therefore, P(x = 2) = 2/15.

Step3 of 3:

c).

Mean number of times the packet is sent is given by

=

=

=

=

= 3.6666

Therefore, Mean number of times the packet is sent is 3.6666.

d).

The variance of the number of times the packet is sent is given by

=

=

=

=

= 15 - 13.4439

= 1.5560

Therefore, The variance of the number of times the packet is sent is = 1.5560.

e).

The standard deviation of the number of times the packet is sent is given by

=

=

= 1.2472

Therefore, The standard deviation of the number of times the packet is sent is 1.2472.

Conclusion:

a).The value of c is 1/15.

b).P(x = 2) = 2/15.

c).Mean number of times the packet is sent is 3.6666.

d).The variance of the number of times the packet is sent is = 1.5560.

e).The standard deviation of the number of times the packet is sent is 1.2472.