Flaws occur at random along the length of thin copper wire. Suppose that the num

Question

Flaws occur at random along the length of thin copper wire. Suppose that the number of flaws follows Poissons distribution with a mean of 2.3 flaws per millimetre. Let D demote the distance between any two consecutive flaws alon the thin copper wire

A) what distribution does D follow? state its PDF and CDF

B) Suppose that you are detecting flaws along the thin copper wire. You have inspected 3 millimetres of wire and none has been detected. What is the probability that you will detect a flaw within the next 0.5 millimetre of wire?

Explanation / Answer

A.)

Probability (flaw in each part) ~ poisson(lambda) = poisson(2.3)

Hence P(no flaws in first (d-1) parts) = [ P(no flaws in one part) ]d-1

= exp(-2.3)d-1 = exp(-2.3(d-1))

P( flaws in the dth part) = 1-P( no flaw in the dth part) = [1-exp(-2.3) ]

D is the distance between two consecutive flaws

This means there are no flaws in each first (D-1) parts and there is a flaw in Dth part

Hence pdf f(d) = P(no flaws in each first (d-1) parts)P( flaw in dth part)= [1-exp(-2.3)]exp(-2.3(d-1))

Thus,

f(d) = { [1-exp(-2.3)]exp(-2.3(d-1))   d>0

                      0,   otherwise }

Let the cdf be F(d) =P(D<d+1) = f(d=1) + f(d=2) + f(d=3) .... +f(d=d)

= [1-exp(-2.3)][ exp(-2.3(0)) + exp(-2.3(1)) +exp(-2.3(2)) +.... +exp(-2.3(d-1))

= [1-exp(-2.3)][ 1 + exp(-2.3(1)) +exp(-2.3(2)) +.... +exp(-2.3(d-1))

The above expression in GP with ratio = exp(-2.3)

Hence,

[1-exp(-2.3)][ 1 + exp(-2.3(1)) +exp(-2.3(2)) +.... +exp(-2.3(d-1))

= [1-exp(-2.3)] (1- exp(-2.3)d)/(1- exp(-2.3)) = [ 1- exp((-2.3)d) ]

Thus cdf F(d) = { (1- exp((-2.3)d) , for d>0

                           0 , otherwise }

B.)

To find P(3<d<3.5 | d>3) = P(3<d<3.5 , d>3) / P(d>3)

= P(3<d<3.5) / P(d>3)   [Because intersection of 3<d<3.5 & d>3 is 3<d<3.5]

P(d>3) = 1 - F(3) =1- (1- exp(-2.3*3) ) = exp((-2.3)3) = exp(-6.9)

P(3<d<3.5) = F(3.5) - F(3) = (1- exp(-2.3*3.5) ) - (1- exp(-2.3*3) ) = exp(-6.9) - exp(-8.05)

Hence,

P(3<d<3.5 | d>3) = P(3<d<3.5) / P(d>3) = [exp(-6.9) - exp(-8.05) ] / exp(-6.9)

= 1 - { exp(-8.05)/exp(-6.9) }

= 1 - 0.316637 = 0.683

Let me know if anything is unclear

Cheers :)

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