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5- In a large restaurant an average of 3 out of every 5 customer ask for water with their meal. random sample of 10 customers is selected: (a) Find the probability ...

Question

5- In a large restaurant an average of 3 out of every 5 customer ask for water with their meal. random sample of 10 customers is selected: (a) Find the probability that exactly 6 ask for water with their meal?Page 2 of 4(b) Find the probability that less than 9 people ask for water? (c) A second random sample of20 customers is selected: Find the smallest value of n such that P(X n) > 0.9 where the random variable X represents the number of customers who ask for water:

5- In a large restaurant an average of 3 out of every 5 customer ask for water with their meal. random sample of 10 customers is selected: (a) Find the probability that exactly 6 ask for water with their meal? Page 2 of 4 (b) Find the probability that less than 9 people ask for water? (c) A second random sample of20 customers is selected: Find the smallest value of n such that P(X n) > 0.9 where the random variable X represents the number of customers who ask for water:



Answers

The random variable $X$ has a binomial distribution with $n=10$ and $p=0.5 .$ Determine the following probabilities: (a) $P(X=5)$ (b) $P(X \leq 2)$ (c) $P(X \geq 9)$ (d) $P(3 \leq X<5)$

X. Is a binomial random variable. The probability of success is 0.5 and and a equals 10 number of trials. What is the probability uh that out of the 10 trials would get exactly five successes using the probability outlets set on by no meal then is 10, He is 0.5. There is the graph of our binomial distribution probability that we get exactly five successes Is .2461. Now the probability to x is less than or equal to two means the probability that we get at most two successes. So the probability of getting at most two successes, that means 01 or two is .0547. The probability that we get uh nine or more at least nine. Yeah The probability that we get at least nine successes that X is greater than or equal to nine is .0107. This last one wants us to find the probability that X is greater than or equal to three but less than five. So I circled the less than because that could be um kind of sneaky would almost miss that. So this is their way of saying um probability that X is either three or four, strictly less than five. So three or four successes. So calculate the probability that we get between three and four successes. So three or 4. Uh his .3, 2, 2, 3

In this question. The scenario is that it is lunchtime at a given restaurant and customers are arriving at the drive thru in a random manner, and we were told that they follow a Poisson process with a rate of 0.8 customers per minute, so we can say the rate is 0.8 customers per minute. Now, when something follows a Poisson process, there's two things that we can say about it. The first is that the number of arrivals that occur in non overlapping time intervals are independent, and the second is that the number of arrivals in a given time interval follows a Poisson distribution with mean lambda T. So if we're given a certain time interval, the mean of the Poisson distribution is equal to Lamberti now. For part A were asked, what is the expected number of customers in one hour and what is the corresponding standard deviation? So if we define X as the number of customers that arrive in one hour, we know that X follows a possible distribution with mean equal to Lambda Times T. And so they mean for possible distribution is simply Lambda Times T, which is 0.8 times 60 minutes in one hour because this rate is 0.8 per minute and so that we expect that there will be 48 arrivals on average in one hour now. Another property of the possible distribution is that the variance on the number of arrivals is equal to the expected number of arrivals, which tells us that's the standard deviation on the number of arrivals is equal to the square root of 48. How and this is equal to 6.93 now for Part B. Were given information that the drive thru workers cannot handle more than 10 customers in any five minute span and were asked to find the probability that too many customers arrive for the workers to handle between the time of 12:15 p.m. And 12:20 p.m. So we're talking about a time of five minutes because it's a poison process. It doesn't matter if we're talking about the time interval from 12 05 to 12 10 or 12, 10 to 12 15 or 12 15 to 12 20. All that matters is theory length of time that passes, so if we want to find the probability that the drive thru cannot handle the number of customers that occur in that five minutes. We're looking for the probability that the number of customers who arrive in that five minutes is greater than 10 because we know that they can handle up to 10 customers in five minutes. And this is equal to one, minus the probability that the number of customer arrivals is that most 10. So this is equal to one minus. The summation from X is equal to zero to 10 of E to the minus lander times T, which is equal to minus four. I got that from tee times lambda, which is 0.8 times Lambda Times t to the exponents X over x factorial. You know, just to explain this, remember, the probability of getting X arrivals is equal to eat to the negative and the tea on celebrity to the exponents X over x factorial. So here we're finding the some of the probabilities of getting zero through 10 arrivals in the next five minutes and then subtracting that from one. And so if you calculate this, it comes out to about 0.0 28 So that is the probability that too many customers come between the time of 12:15 p.m. And 12:20 p.m. For the drive through staff to handle and moving on to Part C. We're told that a customer has just arrived and were asked to calculate the probability that another customer will arrive in the next 30 seconds. So one way to think of this question is that for a Poisson process, the inter arrival times are distributed according to an exponential distribution with rate lambda. So if we say that we define a as the inter arrival time, it's the time between subsequent arrivals. It follows an exponential distribution with parameter lambda, so we want the probability that a is less than or equal to 30 seconds. This is the cumulative distribution for the exponential distribution, and we know that our rate is 0.8 per minute. So to keep this consistent, I should probably put this in terms of minutes, so that would be half a minute and remember, for an exponential distribution, a cumulative function is one minus e to the negative Lambda Key. So if we feel these numbers in, we get one minus e to the negative 0.8 times half, and that should come out to zero point 33 with your calculator. So the probability that the next customer arrives in the next 30 seconds is about 0.33 now for Part D, starting at noon. We want to determine the expected arrival time of the 1/100 customer as well as the standard deviation of that time. So recall that the inter arrival times of customers air exponentially distributed, and we want to find the average time for the 1/100 customer to arrive. So if we define the time that it takes for the 1/100 customer to arrive, it is thesis, um, of all of the inter arrival times for the 1st, 2nd, third customer and so on up to the 1/100 customer where all of the teas are distributed according to the exponential distribution with rate parameter lambda. Then why some 100 is distributed as a gamma with parameter and equals 100 and Lambda equals 0.8. Now the expected time for why 7 100? It's simply end times one over Lambda and remember one over Lambda would be the expected enter arrival time between any successive arrivals and since Lambda equals 0.8 would simply be 1.25 minutes. That is the expected time between success of any two successive arrivals. So for the 1/100 arrival, all we're doing is multiplying that by 100. So we end up with 125. So the expected time for the 1/100 arrival is 125 minutes. And the standard deviation for the 1/100 arrival is given by the square root of end times one over Lambda squared and this comes out to 12.5 seconds. So the standard deviation on the expected time for the 1/100 arrival is 12.5 seconds s. Sorry. Actually, that's minutes. So our units are minutes. So the standard deviation has the same units as the expected Valley, which are minutes

You know the problem? We were told that we have a binomial random very we're in this tent and P This .01. Now we would like to find them following probability. Now if we have venice 10 mps 100.1, this means that the probability of acts is equal to 10, choose X Times Point on 1 to the X Times .99 To the 10 -40s. And then just comes from following our binomial distribution properties. Now here on part A we would like to find the probability That X is five. So the probability that access five, This means plug in 54 X. And so this is 10, choose five Times Point on 1 to the 5th Times .99 to the 5th Evaluating this because it's a very small number of about two 396 times 10 To the -8 power. That is the probability that X is equal to find. Yeah. No. And be we want the probability that x is less than or equal to two. So this just means we want the sum of all the probabilities from extra and from zero the two of P of X. Which is the sum as expose from 0 to 2 of 10 shoes x times point on one of the arts Times .99 To the 10 -X. It just means we're going to take all of those values Plug 0 1 and two and not evaluated. I don't know. We do. We got our probability of 0.9 nine 99 on C. We want the probability X is greater than or equal tonight. So that just means it's the sum As extras from 9 to 10 of p vivax. And so we do the exact same thing. They don't on the last problem. Except we're going to pull you in nine and 10 and add those together instead. And when we do this gives us a small number of 9.9, 1 times 10 To the -18. So it's practically zero On the we want the probability that acts is between three and 5. Mhm. And this is the sum As that goes from 3 to 5 P of X. So again, the same thing as we've done in the problems above, except we put in 34 and five and then add them all together. And this will give us 1.138 Times 10 to the -4.

We're told that the tip percentage of the restaurant has a mean value of 18% the standard deviation of 6%. In part, A were asked to find the approximate probability that the sample mean tip percentage for a random sample of 40 bills is between 16 and 19%. Well, we're going to let X bar a the simple, mean tip percentage for he's 40 bills now by the Central Limit Theorem. We have that X bar is normally distributed approximately because the number of samples is pretty three large, and we have that the expected value of X far is the mean which we're told is 0.18 18%. Well, 18% actually 18. Mm and the standard deviation. Well, this is same as the standard deviation of the population sigma over the square root of the number of bills and were given that the staring deviation of the population is 6% and N is 40. So we have the standard deviation of X bar is approximately 640 and therefore probability that X bar lies between 16 and 18. Sorry, 16 and 19 uh, percent. Well, this by Central Limit theorem is approximately five of 19 minus 18 mean over the standard deviation, which is 6/40 minus phi of 16 minus 18 over six over square into 40. This is five one point of five minus five of negative 2.11 approximately, and this is approximately equal to using either a table of values or computer 0.8357 Next in Part B, we're told that the sample size is now 15 rather than 40 and we're asking the probability I was requested. In part. A can be calculated from be given information. So again we have that. Well, there's an issue here, so by convention and should be greater than 30 to apply the Central Limit theorem, so using the same procedure for n equals 15 would not be appropriate.


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