4) Suppose that highly sensitive computer data received on an infra-red link mus

Question

4) Suppose that highly sensitive computer data received on an infra-red link must be received successfully 3 times (that is, in triplicate) before it is used to update a control process. The link is over a considerable distance and a) (4) On average, how many times must a piece of data be transmitted before it is used to update the control b) (6) What is the (approximate) probability that, of the next 100 data transmissions, less than half are success- data is only transmitted without error about 60% of the time. process? ly received?

Explanation / Answer

let X be the number of times a piece of data be transmitted before it is used to update the control process.

let data transmitted successfully without error is termed as success.

and the data transmitted with error is termed as failure.

let p be the probability os success and q=1-p be the probability of failure.

by question p=0.6 hence q=0.4

the computer must receive 3 successfully data transmission before it is used to update a control process.

so there must be 3 successes.

let Y be the random variable denoting the number of failures precedding the 3rd success.

so X=number of times a piece of data be transmitted before it is used to update the control process

      =number of failures+number of successes

      =Y+3

and Y follows a negative Binomial distribution with parameters r=3 and p=0.6

so pmf of Y is P[Y=y]=3+y-1Cy0.630.4y     y=0,1,2,............

now average number of times must a piece of data be transmitted before it is used to update the control process is

E[X]=E[Y]+3

now E[Y]=r*q/p=3*0.4/0.6=2

so E[X]=2+3=5

so on average 5 times must a piece of data be transmitted before it is used to update the control process. [answer]

b) let S be the number of transmissions of the next 100 transmissions received successfully.

so S~Bin(100,0.6)

so E[S]=100*0.6=60 and standard deviation of S is sqrt(100*0.6*0.4)=24

since here number of trials=100 is very large so the distribution of S can be approximated by Normal distribution with mean=60 and standard deviation=24

so S~N(60,242) approximately

so approximate probability that of the next 100 data transmissions, less than half are successfully received is

P[S<50]=P[(S-60)/24<(50-60)/24]=P[Z<-0.4167]   where Z~N(0,1)

            =0.338449 [using minitab] [answer]

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