3. Let T be exponentially distributed with mean 5 minutes....

3. Let T be exponentially distributed with mean 5 minutes. Let \(\tilde{T}\) be the same lenght of time except measured in hours. Determine the following: (a) The p.d.f. of T. (b) The probability that T is greater than 10 minutes. (c) The probability that T equals 10 minutes. (d) The variance of T (including units). (e) The squared coefficient of variation of T. (f) The probability that T is greater than 15 minutes given that T is greater than 5 minutes. (g) The Laplace-Stieltjes transform of T; i.e., \(L(s) = E[e^{-sT}]\). (The L-S transform can be used to find means and higher moments of nonnegative random variables similar to how the p.g.f. of nonnegative, integer-valued random variables can be used to compute those quantities.) (h) The distribution of \(\tilde{T}\) including any parameters. (i) The squared coefficient of variation of \(\tilde{T}\).

Transcript

0:00 Hello everyone.

00:01 So in this question as asked, let us directly move to part d.

00:06 In part d, we have to find first the probability density function, the mean and the variance for the total repair time.

00:18 So the total repair time, total repair time is given by you.

00:31 So we have fu is equal to integrating over v.

00:43 F uv ucuma v dv we are doing this to get the probability density function so this is equals to integrating from minus u to plus u 1 divided by 8 multiplied by e raised to the power minus u divided by 2 dv which gives this is equal to 1 divided by 8 e raised to the power minus u divided by 2 this is constant, integrating dv from minus u to plus u, which gives 1 divided by 8, e raised to the power minus u divided by 2.

01:22 Now integration of dv becomes v, limit being from minus u to u.

01:27 Further substituting and solving we get this to be equals to 1 divided by 4, e raised to the power minus u divided by 2 multiplied by u for all u greater than equals to 0, and for other that is u less than 0 we have the function is equal to 0 so this is our probability density function now to find the mean we'll find the expected value of u which will be integrating from 0 to infinity u f u u dv it it's d u now solving this from 0 to infinity u multiplied by 1 divided by 4, e raised to the power minus u divided by 2, u, du.

02:29 On solving this, we get this to be equals to 1 divided by 4, gamma 3 divided by 1 divided by 2 raise to the power 3, which is nothing but equals to 4, which is our mean.

02:45 Now we'll find, for variance we'll find expectation of u square also, which becomes integrating from 0 to infinity, u square f u d u proceeding in similar manner as we proceeded for expectation of u we get this to be equals to 24 substituting the values of f u then solving and integrating and putting the value limits we get this to be equal to 24 thus our variance becomes expectation of u square minus expectation of u whole square substituting the values 24 minus 4 square which is 16 so 24 minus 16 that is nothing but is equal to a which is our variance now next part e ask us to find the probability that the total repair time exceeds 4 hours that is in part e we have to find the probability that u is greater than 4 right so this is equal to integration from 4 to infinity f u u du or this will be which is equals to integration from 4 to infinity 1 divided by 4 e raised to the power minus u divided by 2 multiplied by u du now solving by integration by paths method we get 1 divided by 4 is common taking that in common, u, e raised to the power minus u divided by 2 whole divided by minus 1 divided by 2.

04:53 The limit of this becomes from 4 to infinity minus integration differentiating u, that is it gives 1 1 and then integrating e raised to the power minus u divided by 2, which gives e raised to the power minus u divided by 2, which gives e raised to the power minus u divided by 2.

05:16 D u whole divided by four multiplied by two from four to infinity right again solving this and solving first part by substituting the limits and then integrating and substituting the limits for second part we get this to be equals to one divided by four eight e raised to the power minus two minus 4, e raised to the power minus u divided by 2, limit being 0 to infinity.

05:56 It would be from 4 to infinity.

05:59 4 to infinity.

06:01 Again solving this, we get this to be equals to 2 e raised to the power minus 2 plus e raised to the power minus 2 or 3 e raised to the power minus 2, which is nothing but is equals to 0 .406 0.

06:18 Now let us come to part f of the question.

06:22 In part f, it is said that the pdf, the mean and variance of the difference between the repair time of the machine 1 and the repair time of the machine 2.

06:30 We have to find this.

06:33 So, the difference between the repair time of the machine 1 and the repair time of the machine 2 is given by the random variable v as defined previously.

06:42 So pdf of we can be defined by integrating f, by integrating fuv function over u, that is, let us write the case first where v lies between minus infinity to 0.

07:00 Then we have u lies between minus v to infinity.

07:07 So fv, u becomes integration of fu v, du from minus v to infinity...

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