Distribution of Typos The number of typographical errors per...

Name: Problem 7, Chapter 12: Calculus and Its Applications
Uploaded: 2021-06-29T18:28:46-08:00
Duration: 6 min 34 s
Channel: Barsha Rana
Description: Distribution of Typos The number of typographical errors per page of a certain newspaper has a Poisson distribution, and there is an average of 1.5 errors per page. (a) What is the probability that a randomly selected page is error-free? (b) What is the probability that a page has either two or three errors? (c) What is the probability that a page has at least four errors?

Distribution of Typos The number of typographical errors per page of a certain newspaper has a Poisson distribution, and there is an average of 1.5 errors per page. (a) What is the probability that a randomly selected page is error-free? (b) What is the probability that a page has either two or three errors? (c) What is the probability that a page has at least four errors?

Distribution of Typos The number of typographical errors per page of a certain newspaper has a Poisson distribution, and there is an average of 1.5 errors per page.
(a) What is the probability that a randomly selected page is error-free?
(b) What is the probability that a page has either two or three errors?
(c) What is the probability that a page has at least four errors?

Key Concepts

Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, assuming these events occur with a known constant mean rate and independently of the time since the last event.

Probability Mass Function (PMF)

The probability mass function for a Poisson distribution is used to calculate the probability of observing exactly k events. It is defined as P(X=k) = (?^k * e^(-?)) / k!, where ? is the mean number of events, k is the number of occurrences, and e is the base of the natural logarithm.

Cumulative and Tail Probabilities

Cumulative probability in the context of the Poisson distribution involves summing probabilities from 0 up to a certain number of events to find the likelihood of observing at most that many events. Tail probabilities are the complement of cumulative probabilities, representing the probability of observing a number of events above a given threshold, such as 'at least four errors'.

Transcript

0:00 Welcome to numerate.

00:01 In the current problem, we have a tricky situation.

00:04 Here is x, which is the number of automobiles that arrive in any given interval of one minute, any given one minute interval.

00:31 Whereas y is given to be, the inter -arrival time, inter -arrival time between any two successive arrival.

00:58 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.

01:11 Now imagine on the first minute, there came five cars, okay, five automobiles.

01:18 And then at 1 .5 minute itself, six vehicles came.

01:26 Okay.

01:26 So that means this is the interval.

01:29 So this interval thing follows y, whereas this number follows x.

01:35 Now we are given x follows a poiseau distribution and also it is given that it is expected that on an average for any one minute interval, the expected number, expectation of x is 5.

01:50 Now we know by the property of poisson distribution, the expectation of poisson distribution, it says parameter lambda.

02:00 Therefore, lambda is equals to 5.

02:03 So therefore, the first question asked over here is at least five cars.

02:12 Correct? that is.

02:14 That is.

02:14 Is probability of x greater than equal to 5.

02:19 So, we can write that as 1 minus probability x less than equal to 4.

02:35 Because at least 5 means 1 total minus cases of 0, 1, 2, 3, 4.

02:42 So how are we going to calculate? it will be this, correct? so, like, how can we evaluate this probabilities? one way is that recursive relation that we all know, that is p of x plus 1 is equal to lambda by x plus 1 into px.

03:08 So you just calculate p0 manually and then plugging in p0, you get p1.

03:15 Plugging in p1, you get p2.

03:18 Plugging in 2, we get 3 and then 4.

03:22 And then add all of them up, we get this quantity and then one minus we get this...

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