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AnswerGiven,The aivals of phone calls at a telephone switching office is a Poisson random process N(t) with all artival rate 1 4 calls per second. We monitor N(t) s...

Question

AnswerGiven,The aivals of phone calls at a telephone switching office is a Poisson random process N(t) with all artival rate 1 4 calls per second. We monitor N(t) starting from t = 0 over a 10-second interval. Let Sn be the time of the airival of the nt call_What is P(N(1) =0) , the probability of no phone calls during the first second? Here, Iam denoting lambda by L in the below discussion for my convenience Poisson distribution;, P(r) = (e^L )* (L^VL! above P(r) signifies the probability of &q

Answer Given, The aivals of phone calls at a telephone switching office is a Poisson random process N(t) with all artival rate 1 4 calls per second. We monitor N(t) starting from t = 0 over a 10-second interval. Let Sn be the time of the airival of the nt call_ What is P(N(1) =0) , the probability of no phone calls during the first second? Here, Iam denoting lambda by L in the below discussion for my convenience Poisson distribution;, P(r) = (e^L )* (L^VL! above P(r) signifies the probability of "time between two successive calls i31" So, for 1 = 0 P(0) = e^-4 2 What is P(N(4)-N(3) = 4), the probability of four phone calls arrive between the third and the fourth second Here, It is nothing but P(4) P(4) = (32/3) *e^-4



Answers

The average time between incoming calls at a switchboard is 3 minutes. If a call has just come in, the probability that the next call will come within the next $t$ minutes is $P(t)=1-e^{-t / 3} .$ Find the probability of each $t$ situation. (a) A call comes in within $\frac{1}{2}$ minute. (b) A call comes in within 2 minutes. (c) A call comes in within 5 minutes.

In this problem, it is given that the number of telephone calls that arrive at a phone exchange is often modeled as a poison random variable, Assume that on the average there are 10 calls per hour. Let X denote the number of telephone calls that arrive at the phone exchange per hour. It is said on the average there are 10 calls per hour. So here Lam Brady is equal to 10 into one, Which is equal to 10. We know that probability of X equal to X for a poison random variable is He raised 2-. Lombardi. Lombardi is two x divided by X. Factorial. There are X ranges from 01, two up to infinity. So this is equal to the rest too minus can dangerous to X divided by X. Factorial. As here, Lamberti is equal to 10, so this is equal to it is to my understand 10 days to X divided by X. factorial. In the first part were asked what is the probability that there are exactly five calls in one hour? That is we have to find the probability of x equal to five Probability of x equal to five. Is the rest too minus 10, multiplied by 10, raised to five, divided by five factorial. This is equal to zero point 03 78. 0.0378. So the probability that there are exactly five calls in one hour is 0.0378. Next it is asked what is the probability that there are three or fewer calls in one hour? That is we have to find the probability of X less than or equal to three. This is equal to as the poison and and variable starts from X equal to zero. We will start from probability of X equal to zero. So this is equal to probability of X equal to zero. Blessed Probability of x equal to one. Plus Probability of x equal to two. Plus the probability of x equal to three. As we have to find probability of X less than or equal to three viva late. The probabilities from X equal to zero up to X equal to three. This is equal to The rest to -10. Then raised to zero, Divided by zero factorial bless It is 2 -10. 10. Raised to one. Now we are finding probability of x equal to one so tankers to one divided by one factorial plus The rest to -10. Then raised to Now we are finding probability of X equal to two. So 10 Rescue too Divided by two factorial plus, It is 2 -10 Dang raised to three Divided by three factorial. Now we will find these values. This is equal to zero point 00 00 sites plus zero point 00 04. Safe plus zero point 00 two seven. Bless zero point 00 seven five seven. This is equal to the sun, is equal to zero 01. See you know three 0.0103. So the probability debt there are three or fewer calls in one hour is 0.0103. Next it is asked what is the probability that there are exactly 15 calls in two hours we have Lamba is equal to 10 per hour. So for two of us, T is equal to two And so we have Lambro T. is equal to 10 into two, which is equal to 20. We have to find the probability of x equal to 15, Probability of X, equal to 15 will be the rest to minus Lamberti. But here Lamberti is 20. So he raced to -20 lamb bratty raised two X. Lamberti is 20 raised two X. X. Is 15. So 20 raised to 15 divided by X. Factorial. That is divided by 15 factorial. The rest to -20 into 20, raised to 15, divided by 15 factorial. This is equal to zero point zero faith 16. 0.0516. So the probability that they're at exactly 15 calls in two hours is 0.0516. Next disaster. What is the probability that there are exactly five calls in 30 minutes. Here We have Lambro equal to 10. We have to find the probability that there are exactly five calls in 30 minutes. So T. is equal to 30 minutes, Which is 0.5 hour As Lambro is 10 per over. So T. Will also be return in unit of our so T. is equal to 0.5 hour. Lamberti is equal to 10 into 0.5. The is equal to five, Probability that there are exactly five calls in 30 minutes Is probability of x equal to five. This is equal to he rest to minus Lamberti. That is dearest to minus five laboratories to X. That is five threes 25 divided by X factorial. That is divided by five factorial, It is 2 -5 into five, phrase to five, divided by five factorial. This is equal to zero 17 55, 0.1755. So the probability that there are exactly five calls in 30 minutes is 0.1755.

In this question, we're going to use exponential distribution, which is defined as the probability distribution that describes the time between events which occurred continuously and in divinity at an average time. The probability density function for this distribution is if flicks equal Lambda E Power Negative Mondex or zero for longer able negative fix some banks its range from X bigger than or equal zero and zero with the zero elsewhere. Net Capital X Let X has exponential distribution with promise to London, then e X, or equal to one over Lambda and evolve actually equal to one over Lambda Square. We have one call every two minutes so we can say that average time for one called what equal to two minutes The time between colt is exponential distributed so we can say the athletics be fine between two calls. X also has exponential distribution with average too. X has also exponential distribution, with average time to on Do we need to find the parameter London so we can get it as follows to equal E X equals one over London. From this, we can say that Lambda would equal to have 1/2. So ex approximately the exponential, huh? Now we need to find the average time for five calls you need have time four five, cause cause we know that for one call we need two minutes when we finish it for a second. We also need, on average, two minutes. So for two calls, we need time. What? Equal to to by two. Equal Four minute now for five calls. We need 552 What? It could do 10 minutes for the other question be we have to find the probability then Steve function four x f of x equals toe half by e power Negative health X for X Bigger than equal 00 elsewhere. Now we need to find the probability that we need more three minutes to take the next call. So the probability the probability four more three minutes toe take the next cold will be probability of X probability for X bigger than three. We're equal integration from three to infinity. A few clicks the X for equal to integration from three to infinity for a question half e power negative hub X, the T. So, by substitution, you can see that you equal to have X so do you? We're able to help. It will be equal integration Form one and half to infinity e forward Negative. You d you After integration, we get negative. Evil or negative? You from 1.5 to infinity will equal to negative zero minus e power. Negative 1.5 Well equal to oh, point 22 3. So we can say that the probability that next call will come more than three minutes is all 0.223 Moving to the next question number C in question. See, we need to find the nineties percentile of this distribution of time between two calls. It means that with probability, a 0.9 which happened X calls what X is greater than the Europe. So you have that probability 4.9 Well, happen ex calls where X is greater than zero. If X is, If X is listened, then zero. Then we have the probability X capital less than a small what equal to integration from negative Infinity two x small off. Zero d x will equal to zero so we can have the following 0.9 equal be off X capitalist than a small well equal integration from negative infinity. Two weeks a 50 DT would equal to integration from negative infinity to zero zero deity plus integration from zero to x have e power negative 4.5 xdd By substitution, we put you equal 4.5 t and D you will be 0, 00.5 DT so it would equal to zero plus integration from Europe toe a 0.5 x for e power Negative you d you would equal to negative e forward negative you from the you to a 0.5 x well equal to one minus e power Negative 4.5 x We can minus one from opened nine. So we get ive power 0.5 x equal to 4.1 Take limb for the both sides We get that 4.5 x equal then for over 81 so X would equal to 4.6. And that's the final result for question number C moving to point D in pointy. We have that two minutes paused from last coal. Now we need to find the probability that the next call will come in next the probability Welcome. And next third minute it means that we need to find the probability that coal will happen between the second and the third minute. So probability off time. Listen, 30 minutes or x more than two minutes were equal to probability off to this, then X. This is a three over B X, more than two well equal to integration from 2 to 3 a few weeks d x an integration from two to infinity for every weeks the X for that intervention from two to ST 4.5 e power negative 0.5 X DT over division from two to infinity over and five eating it Power negative over and five x the X Why substitution you would equal to 4.5 x. Do you recall too? All 0.5 you the X. So it's equal to integration from one to 1.5 e por negative You you over integration from Wantagh infinity E power Negative! You d you After we evaluate this integration, we get that negative e power negative You from 1 to 1 toe 0.5 over negative e for negative. You, from one to infinity would equal toe negative boy e power negative 1.5 minus e for negative one over E power. Negative one equals two. All 0.393 And the final result for question number D moving to the next question e We have that one call in two minutes so we can define the variable capital. Why? Which describes the number off course per hour is the number off course per hour. Then why has postal distribution then? Why has bought so distribution? We have 60 minute in one hour. So we have 60/2 equal toe certain 30 calls per hour on average. So why I approximately b p off 30. We need to find the probability that there is less than 20 calls per hour. So probability for the there is less calls per hour. Why less than 20? It was equal to some mission from oh equals zero 19 for probability that why equal I so it would equal to some mission. I equals zero to 19 for e power Negative 30 boy, 30 I over factorial oil would equal to all point all too and that the final result Thank you

In this problem it is given that the number of telephone calls that arrive at a phone exchange is often modeled as a poison random variable, Assume that on the average there are 10 calls per hour. Let X denote the number of telephone calls that arrive at the phone exchange per hour. It is said on the average there are 10 calls per hour. So here Lam Brady is equal to 10 into one Which is equal to 10. We know that probability of X equal to X for a poison random variable. Is He raised 2- Lombardi. Lombardi is two x divided by X. Factorial. There are X ranges from 01, two up to infinity. So this is equal to the rest too minus 10. Dangerous to X divided by X. Factorial. As here Lamberti is equal to 10. So this is equal to it is to my understand 10 days to X divided by x factorial. In the first part were asked what is the probability that there are exactly five calls in one hour? That is we have to find the probability of X equal to five, Probability of x equal to five. Is the rest too minus 10, Multiplied by 10. raised to five divided by five factorial. This is equal to zero point 03 78. 0.0378. So the probability that there are exactly five calls in one hour is 0.0378. Next. It is asked What is the probability that there are three or fewer calls in one hour. That is we have to find the probability of x less than or equal to three. This is equal to as the poison and and variable starts from X equal to zero. We will start from probability of X equal to zero. So this is equal to probability of X equal to zero, blessed Probability of x equal to one plus Probability of x equal to two Plus the probability of x equal to three. As we have to find probability of X less than or equal to three viva late. The probabilities from X equal to zero up to X equal to three. This is equal to The rest to -10 Then raised to zero Divided by zero factorial bless It is 2 -10, 10 raised to one. Now we are finding probability of X equal to one. So tankers to one divided by one factorial plus The rest to -10 then raised to now we are finding probability of X equal to two so 10. Rescue too Divided by two factorial plus It is 2 -10 Dang raised to three Divided by three factorial. Now we will find these values. This is equal to zero point 00 00 sites plus zero point 00 04 safe plus zero point 00 two seven. Bless zero point 00 seven five seven. This is equal to the sun is equal to zero 01. See you know three 0.0103 So the probability debt there are three or fewer calls in one hour is 0.0103. Next. It is asked what is the probability that there are exactly 15 calls in two hours we have Lamba is equal to 10 per hour. So for two of us T is equal to two And so we have Lambro. T. is equal to 10 into two which is equal to 20. We have to find the probability of x equal to 15, Probability of x equal to 15 will be the rest to minus Lamberti. But here Lamberti is 20. So he raced to -20 lamb bratty raised two X. Lamberti is 20 raised two X. X. Is 15. So 20 raised to 15 divided by X factorial. That is divided by 15 factorial. The rest to -20 into 20, raised to 15 divided by 15 factorial. This is equal to zero point zero faith 16, 0.0516. So the probability that they're at exactly 15 calls in two hours is 0.0516. Next disaster, what is the probability that there are exactly five calls in 30 minutes. Here We have Lambro equal to 10. We have to find the probability that there are exactly five calls in 30 minutes. So T is equal to 30 minutes Which is 0.5 hour As Lambro is 10 per over. So T will also be return in unit of our so T is equal to 0.5 hour. Lamberti is equal to 10 into 0.5, Which is equal to five, Probability that there are exactly five calls in 30 minutes Is probability of x equal to five. This is equal to he rest to minus Lamberti. That is dearest to minus five laboratories to X. That is five threes 25 divided by X factorial. That is divided by five factorial, It is 2 -5 into five, phrase to five, divided by five factorial. This is equal to zero 17 55, 0.1755. So the probability that there are exactly five calls in 30 minutes is 0.1755.

All right, this question asks us, but a Passat industry shin with an average of 48 calls per hour. So party asks, what is the probability of getting three calls in five minutes? So let's first compute the expected amount of calls in that time interval. So they're 48 calls in 60 minutes, and we're dealing with five minutes and working that Oh, you'll see that the minute that the minutes cancel and we're left with calls, so we expect four calls. So now we can plug are expected value before in for the mean and three in for observed so probability that X equals three equals four to the third power times he to the negative forth all over three factorial. And that is equal to point 1954 All right, and then Part B asks for 10 calls in 15 minutes. So once again we expect 48 calls in 60 minutes times 15 minutes, minutes cancel. And in this case, we're left with 12 as our expected value. So we want to find the probability that there are 10 calls in this interval which is are expected value raised, Thio the observed times e to the negative of are expected all over our observed factorial, and that works out to be 0.10 48 Part See asks us for the expected amount of calls if the operator takes a five minute break. So once again, 48 calls for 60 minutes times five minutes, and that equals four waiting. And that is our expected. And it wants the probability that there are no calls so probability that X equals zero and that is our main. Two are observed times e to the negative of are mean all over our observe factorial and remember that zero factorial is one, and we end up with point zero 18 32 All right, and then Part D asks a similar question. But this case, if the operator takes a three minute break, what's the probability is no calls again? 48 calls in 60 minutes, times a three minute break. That's our new expected value, which becomes 2.4 calls, and it wants the probability that X equals zero and that is just equal to are mean to the observed times e to the negative of are mean all over our observed factorial. And that turns out to be 0.9 07


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