Problem 5 (15 points) A fast-food restaurant is interested...

Problem 5 (15 points) A fast-food restaurant is interested in studying its arrival of customers. During the busy lunch period they have observed an average of 20 customers arriving per hour. We assume the interarrival time are exponential random variables. (1) If a customer has just entered the store, what is the probability of another arrival in the next 10 minutes? (2) What is the expected number of customers arriving in a five minute window?

Transcript

0:00 Hello everyone.

00:01 So in this question there are so many parts.

00:04 Let us start by the given information.

00:07 The given data is that a fast food restaurant has customers arriving at the checkpoint corner at the rate 25 customers per hour.

00:18 That is, rate is equal to denoted by lambda is equal to 25.

00:24 Next, the average service time at the till is 75 seconds.

00:30 That is we have mu is equals to 3 ,600 divided by 75, 3 ,600.

00:43 Here denotes that in one hour there are 3 ,600 seconds.

00:47 So this will be equals to 48.

00:52 The first question is, what is utilization? so utilization is given as lambda divided by mu or 25 divided by 48, substituting the values in solvent we get this to be equals to 0 .520 -833 or 52 .0833 % approximately.

01:24 So this is our utilization.

01:27 Now, it is given that assume that utilization is 0 .56.

01:34 Let utilization is equals 0 .5 .6.

01:39 Let utilization is equals 0 .0 .5.

01:43 5, 6.

01:45 So we have to find the probability that every zero customers at the customer.

01:58 So we have to find the probability that there are zero customers at the counter either in a line or being served.

02:07 Okay? probability of zero customers becomes one minus utilization or 1 minus 0 .5.

02:28 6 which will be equals to 0 .44.

02:31 Next in this same part we have to find the probability of two or more customers, 2 or more customers.

02:51 Right.

02:52 So this will be same as 1 minus probability of 0 customer plus 1 customer right.

03:14 So this becomes 1 minus probability of 0 customer.

03:17 So this becomes 1 minus probability of 0 customer minus probability of 1 customer 1 minus 0 .44 minus the probability of 1 4 4 minus the probability of 1 customer becomes utilization was 0 .5 6 raised to the power k plus 1 here k is equals to 1 so 1 plus 1 so in solving this we get this to be equals to 1 minus 0 .44 minus 0 .3136, which is same as on solving 0 .2464.

04:04 Next, we have to find the probability that there are 1 or 2 customers at the counter.

04:12 1 or 2 customers.

04:19 Customers.

04:22 Right.

04:23 So this is equals to probability of 1 customer plus probability of 1 customer plus probability of 2 customers.

04:35 Which is same as 0 .56 raised to the power 1 plus 1 is 2 plus 0 .56 raised to the power 2 plus 1 is 3.

04:48 On solving this we get this to be equals to 0 .336 plus 0 .175616 which is same as 0 .489216...

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