A call center in Perth, Australia receives an average of 1.3...

A call center in Perth, Australia receives an average of 1.3 calls per minute. By looking at the date, a Poisson discrete distribution is assumed for this variable. Calculate each of the following. a. The probability of receiving no calls in the first minute of its office hours. b. The probability of receiving 1 call in the first minute. c. The probability of receiving 3 calls in the first minute.

A call center in Perth, Australia receives an average of 1.3 calls per minute. By looking at the date, a Poisson discrete distribution is assumed for this variable. Calculate each of the following.
a. The probability of receiving no calls in the first minute of its office hours.
b. The probability of receiving 1 call in the first minute.
c. The probability of receiving 3 calls in the first minute.

Key Concepts

Poisson Distribution

The Poisson distribution is a discrete probability distribution used to model the number of events occurring within a fixed interval of time or space when these events happen with a constant average rate and independently of the previous events. It is particularly useful for rare events and scenarios where the event count is discrete and can potentially occur at any moment within the specified period.

Rate Parameter (?)

The rate parameter, often denoted by ? (lambda), represents the average number of events occurring over the fixed interval in the Poisson distribution. It determines the distribution's shape and is critical for calculating the probability of a given number of events. In any application of the Poisson distribution, accurately identifying or estimating ? is essential for predicting outcomes.

Probability Mass Function (PMF) for the Poisson Distribution

The probability mass function for the Poisson distribution provides the probability of obtaining exactly k events in a given interval. The formula is expressed as P(X = k) = (?^k * e^(-?)) / k!, where k is the number of occurrences. This formula is fundamental for solving problems involving discrete event counts, enabling the computation of probabilities for various specific outcomes.

Transcript

00:01 In this question, we are going to use the exponential distribution, which is defined as the probability distribution that describes the time between events which occur continuously and independently at an average time.

00:15 The probability density function for this distribution is f of x equal lambda e power negative lambda x or 0.

00:29 For lambda e power negative x, lambda x, its range from x bigger than or equal 0, and 0 will be 0 elsewhere.

00:47 Let capital x, let x has exponential distribution with parameter lambda.

00:55 Then ex will equal to 1 over lambda and varx will equal to 1 over lambda square.

01:03 We have one call every two minutes, so we can say that average time for one call will equal to two minutes.

01:13 The time between calls is exponential distributed.

01:17 So we can say that let x be time between two calls.

01:35 X also has exponential distribution with average two.

01:43 X has also exponential distribution with average time 2 and we need to find the parameter lambda so we can get it as follows 2 equal x equal 1 over lambda from this we can say that lambda would equal to half, 1 over 2.

02:17 So, x approximately be exponential half.

02:30 Now we need to find the average time for 5 calls.

02:35 We need average time for 5 calls.

02:48 We know that for 1 call, we know that for 1 call, we need to.

02:50 2 minute.

02:52 When we finish it for a second, we also need on average 2 minutes.

02:57 So for 2 calls, we need time 2 by 2 equal 4 minutes.

03:18 Now for 5 calls we need 5 by 2, 2, 2 equal to 2 2 2 2 2 2 2 2 2 2...