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A keypunch operator has probability of 0.004 of making an error in single keystroke. Suppose program requires 500 keystrokes: Using Poisson distribution, determine ...

Question

A keypunch operator has probability of 0.004 of making an error in single keystroke. Suppose program requires 500 keystrokes: Using Poisson distribution, determine the probability that the operator makes (a) no errors (6) two or more errors_(10 pts)

A keypunch operator has probability of 0.004 of making an error in single keystroke. Suppose program requires 500 keystrokes: Using Poisson distribution, determine the probability that the operator makes (a) no errors (6) two or more errors_ (10 pts)



Answers

The number of typing errors made by a typist has a Poisson distribution with an average of four errors per page. If more than four errors appear on a given page, the typist must retype the whole page. What is the probability that a randomly selected page does not need to be retyped?

Welcome to enumerate. In the current problem. We have a tricky situation here is X. Which is the number of automobiles. Okay, that arrives that arrive in any Given interval of one minute, any given one minute intervals. Whereas why is given to this the inter arrival time, the inter arrival time between in me two successive a rivalry. That means if I draw a diagram, say imagine this is the time bar. Okay, so this is first minute, This is the second minute, this is the third minute, this is the fourth minute, this is the five minute now. Imagine on the first minute there came five cars. Okay, five automobiles and then at 1.5 million itself six uh, vagal scheme. Okay, so that means this is the interval. So this interval thing follows by whereas this number follows X. Now we are given X follows apology distribution. And also it has given that it is expected that on an aboriginal one for any one minute interval, the expected number expectation of X is fun. Now we know by by the property of poison distribution, the expectation of poison distribution. It's it's parameter lambda lambda is equals 25 So therefore The first question asked over here is at least five cars got it. That is probability of X greater than equal to five. So we can write that. Yes one minus probability X less than equal to four Because at least five minutes, one total minus Cases of 0 1, 2, 3, 4. So how are we going to calculate? Maybe this? Thanks. So like how can we can evaluate these probabilities? One where is that reconsideration that we all know that is P. Of X plus one. Physical to lambda by X plus one into PX. So you just calculate P. Zero manually and then plugging in P. Zero. You get P. One plugging in favour you get P. Two plugging in two we get three and then four and then add all of them up. We get this quantity and then one minus we get this. Okay so this is the result of the first problem. 2nd secondly they are asking that why is the inter arrival time? Okay. And also why follows exponential theater distribution for exponential got distribution. We are. No that the average interval time is one by five minutes. Okay. And exponential distribution means is one by theatre if the parameters theater which is supposed to one by five so theater will be five and also this is nothing but basically it will be won by lambda if boys oh is slammed over here. So now we get while following exponential right now what is that pdf of exponential? It is he tied to the power minus sita X. For zero less than equal to x. Flex then equals to infinity and zero. Otherwise. Now the question they have asked. Oops. Oh yeah. The question that they have asked is that oops if and zero otherwise. Okay. Now the question that they have asked is what is the probability that the in the arrival time will be less than one by five? Okay so that would be 02, one by five five. It will devour -5 eggs. The X. It is equals two. If I take five out each of the power minus five X bye minus five. From 0 to 1 x five. Now this fight this fight cancels so we will have Minour it is the power -5 into one x 5 minus minus the power minus five in 20 Therefore we have this expression okay let's do step by step. So it's the power minus one minus minus plus. It's the power zero. Anything risk. The positive is once one minus. It's the power minus one. It is equals to 0.63 21 2056. So I hope you could understand this. This is just we have calculated and subtracted from one, and if you have any conditions, let me

In this problem, it is given that the number of errors in a textbook follows a poison distribution with a mean of 0.01 error per page, like X denotes the number of errors in one page. So we have Lamba is equal to 0.01 error per page. So Lambright's going to 0.01. It is asked what is the probability that there are three or fewer errors in 100 pages. We have to find the probability of X less than or equal to three. But here it is asked in 100 pages. So our T is 100 And so Lamberti is equal to 0.01 in 200, which is equal to one. We have lamb bratty is equal to one probability of X is equal to X. Is here is to my muslim bratty in two laboratories to X divided by X factorial. Their X ranges from 01 up to infinity. So this is equal to the rest to -1. Yeah, one raised 2 x divided by X factorial. We know that one race to any number is again one. So this is equal to the rest to minus one in 21 that is erased, two minus one divided by X. Factorial. Using this we will find probability of X less than or equal to three probability of X less than or equal to three is equal to probability of X. Lester Are equal to three is equal to Probability of x equal to zero. As the range of X starts from zero we will find Probabilities from x equal to zero Plus probability of x equal to one Plus probability of x equal to two Plus probability of x equal to three probability of X less than or equal to three. Is probability of X equal to zero plus probability of X equal to one plus probability of X equal to two plus probability of x equal to three. This is equal to the rest to -1 divided by X factorial. That is divided by zero factorial Plus the rest to -1, divided by one factorial Plus the rest to -1, Divided by two factorial Plus the rest to -1, Divided by three factorial. This is equal to zero point 36 788, 0.36788 plus zero point 36 788 plus zero point 18 394 plus zero point 06 131. And this sound is equal to zero point 98 10. So The probability that there are three or fewer errors in 100 pages is 0.9810.

In this problem with this, given that the number of errors in the text book follows a poison distribution with a mean of 0.0 and error per page. Let X denote the number of errors in one page. So we have a number is equal to 0.01. It is asked what is the probability that there are three or less errors in 100 pages. As we have to find the probability 400 pages T is equal 200 And so Lambright is 0.01 in 200 which is equal to one. We know that for a poison random variable X probability of X equal to X is erased. Two minus limb bratty into Lambright erased two X divided by X factorial. They're x ranges from 01, two up to infinity. So this is equal to arrest 2 -1 Into one Ridge 2 x divided by X factorial. But we know that one has to X is always one. So this is equal to arrest two minus one in 21. That is nearest to minus one, divided by expect O'Ryan. We have to find Probability that there are three or less errors that is. We have to find probability of X less than Are equal to three. This is equal to Probability of x equal to zero as the poison and the variable X starts from zero. We we start computing probabilities from X equal to zero. So probability of x equal to zero plus Probability of x equal to one. Less Probability of x equal to two Plus the probability of x equal to three probability of x less than or equal to three is probability of X equal to zero probability of X equal to one. Place probability of X equal to two Plus probability of x equal to three. So this is equal to Probability of x equal to zero is erased, 2 -1, Divided by zero factorial plus Probability of x equal to one is the rest to -1 divided by one factorial Plus. The probability of x equal to two is the rest to -1, Divided by two factorial plus Probability of x equal to three is the rest to -1 Divided by three factorial. So this is equal to zero point 36 788, Bliss against zero point 36788 plus zero point 18 394 plus zero point 061 31. This is equal to zero point 98. Ban zero 0.9810. So the probability that there are three or less errors in 100 pages is 0.9810

Toe work with probability. We need to think about how many times the desired event occurs, and in this case, that would be having a defective computer and compare that to the total number of outcomes observed in this case that's looking at 1200 computers. So we know two piece of information. We know that we found nine defective computers, and we know that 1200 computers were checked. So these two pieces of information will be used to find the probability that a defective computer is found from this manufacturer. So we're going to start by setting up a ratio, our ratios, the nine defective computers compared to the 1200 computers checked. Now this is going to give us the probability of having a defective computer. I would like to look at this probability as a percentage, so I'm gonna go ahead and changes to a ratio out of 100. If I divide the top and the bottom of the ratio by 12 I will get this ratio out of 100 because 1200 divided by 12 is 100 the nine divided by 12 is this decimal 75 hundreds. It's now, this can easily be converted to a percentage. So we have this percentage 75 hundreds of a percent. So we have found that a probability of having a defective computer is 75/100 of a percent. Now, we can use this probability to figure out from an entire manufacturing lot off 15,000 computers. How many computers we can predict to be defective. Now, we found that the probability of having a defective computer is 75 hundreds of a percent. So 75/100 of a percent off. These defective compute of these computers may be defective. So 75/100 of a percent times all the computers, 15,000 Peters. This will tell us what our prediction of how many defective computers we have next to work this out. I'm gonna change this percentage to decimal. And so we get this decimal 75 10 thousands. We're gonna take this 75 10 thousands and multiply it by 15,000 and we get 112 and 5/10. So that's how many computers we could predict to be defective. But in this case, we're going around this and So we're going to say we have about 113 defective computers, so in the entire lot we can predict 113 defective computers out of all 15,000 computers.


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