Assume that flaws along a magnetic tape follow a Poisson dis

Assume that flaws along a magnetic tape follow a Poisson distribution with a mean of 0.1 flaws per meter. Let X denote the distance between two successive flaws. a) What is the mean of X? b) What is the probability that there are no flaws in 10 consecutive meters of tape? c) What is the probability that there are at least 3 flaws in 10 consecutive meters of tape? d) How many meters of tape need to be inspected so that the probability of at least one flaw is 90%? e) Suppose an inspector starts at the beginning of the tape and measures a distance between the first flaw and the second flaw, the distance between the second and third flaw, and so forth until two consecutive flaws are found that are at least 8 meters apart. What is the probability that the first time the distance between consecutive flaws exceeds 8 meters is between the 4th and 5th flaw?

Solution

flaws along a magnetic tape follow a Poisson distribution with a mean of 0.1 flaws per meter.

let Y be the random variable denoting the number of flaws in 10 consequtive meters of tape.

then Y follows a Poisson distribution with mean=0.1*10=1

Let X denote the distance between two successive flaws.

so P[X>t]=P[distance between two successive flaws >t]=P[no occurance of flaws in t meters of the tape]=P[Z=0]

where Z~Poisson(0.1t)

so P[X>t]=P[Z=0]=e^-0.1t

or P[X<t]=1-e^-0.1t t>0

but this is the CDF of an exponential distribution with mean 1/0.1

so X follows an exponential distribution with pdf f(x)=0.1e^-0.1x      x>0

                                                                          =0 otherwise

a) so mean of X is E[X]=1/0.1=10   [answer]

b) probability that there are no flaws in 10 consequtive meters of tape is

P[Y=0] where Y~Poisson(1)

so P[Y=0]=e^-11⁰/0!=e^-1=0.36788 [answer]

c) probability that there are at least 3 flaws in 10 consecutive meters of tape is

P[Y>=3]=1-P[Y=0]-P[Y=1]-P[Y=2]=1-e^-1-e^-11¹/1!-e^-11²/2!=1-0.919699=0.080301 [answer]

d) let n meters of tape need to be inspected.

let S denote the number of flaws in n meters of the tape.

so S~Poisson(0.1n)

given that P[S>=1]=0.9 or 1-P[S=0]=0.9

or, P[S=0]=0.1 or, e^-0.1n=0.1

taking log on both sides

-0.1n=log0.1=-2.30

or, n=2.30/0.1=23

so 23 meters of tape need to be inspected. [answer]

e) we have X denote the distance between two successive flaws. and we have X follows an exponential distribution with cdf P[X<t]=1-e^-0.1t

so the distance between two consequtive flaws exceeds 8 meters is

P[X>8]=1-P[X<8]=e^-0.1*8=e^-0.8=0.449

so the probability that the first time the distance between consecutive flaws exceeds 8 meters is between the 4th and 5th flaw= P[distance between 1st and 2nd flaw<8 meters]*P[distance between 2nd and 3rd flaw<8 meters]*P[distance between 3rd and 4th flaw<8 meters]*P[distance between 4th and 5th flaw>8 meters]

=(1-0.449)*(1-0.449)*(1-0.449)*0.449=0.0751 [answer]