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5. [4 marks] At a particular hospital it is estimated that each day, on average, 85% of people keep their appointments at the outpatients' clinic. Use the norm...

Question

5. [4 marks] At a particular hospital it is estimated that each day, on average, 85% of people keep their appointments at the outpatients' clinic. Use the normal distribution to approximate the probability that on a given day when 60 appointments have been booked that exactly 55 patients keep their appointment. Round to three decimal places (Hint: You can verify your answer by doing the binomial distribution calculation:

5. [4 marks] At a particular hospital it is estimated that each day, on average, 85% of people keep their appointments at the outpatients' clinic. Use the normal distribution to approximate the probability that on a given day when 60 appointments have been booked that exactly 55 patients keep their appointment. Round to three decimal places (Hint: You can verify your answer by doing the binomial distribution calculation:



Answers

The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.

What can you conclude at the 5% significance level?

Let's start this problem by writing down what we know where what is given. The average wait time in a doctor's office varies. The standard deviation of those wait times is 3.4 minutes. A random sample of 30 patients in this doctor's office has a sample standard deviation of 4.1 minutes, and one doctor believes so. We have a belief for a claim that the variants and variance is defined by Sigma Squared is greater than what was originally thought. Okay, so we're going to be running a chi square test of a single variance on this. And since we know that the standard deviation was 3.4, we then know that the variance is equal to 3.4 squared. So therefore, when we make our belief or our claim that this variance is greater than what was originally thought, it would be greater than 3.4 squared. And in order to run this test, we need a chi square test statistic. And to find that Chi Square test statistic, you will apply the formula the quantity of n minus one times the sample variance over the population variance. So our chi square test statistic will equal 30 minus one times our s squared. So we know esus 4.1. So we'll need 4.1 squared over sigma squared. And we know that Sigma squared is equal to 3.4 squared. And if you put that into your calculator and calculated out, you will get a chi square test statistic of 42.1704 So our chi square test statistic for this data is 42.1704

In problem 101. We have an urgent care facility in which patients arrive at an average rate of one patient every seven minutes, which means that it equals one in seven or one divided by seven. We will assume the duration between arrivals is exponentially distributed or it has an exponential distribution, which means we can now define a random variable X, which is the duration between patients arrival. And we can say that it follows an exponential distribution with the rate of one divided by seven. And, of course, the random variable is in minutes. Now we can continue to answer birds from a to D party. A. We want to find the probability that the time between two successive visits or deliberation is this than two minutes. Simply, we want to get the probability of the random variable to bill is then two minutes as they're under variable follows an exponential distribution. It means it has a community probability equals one minus E to the power of minus m are deployed by X. We're exile. Is this value then, to get the probability of the random variable to be less than two minutes, it equals one minus E is a part of minus M, where M is the red one, divided by seven multiplied by X X years too, which he gives or going to for a five for Bharti. We want to get the probability the time between two successive visits will be more than 15 minutes. We have here negative, then the probability of X to be greater than equals 15 minutes more than 15 minutes. It's somebody equals the compliment event one minus the compliment event or minus the probability of X to be smaller than 15 minutes. And this is a continuous distribution. We can add the equal sign. We can ignore it. Then it equals E to the bar of minus mm, multiplied by X. We have here M is one divided by seven and X here is 15. Then the answer is 4.1173 for birth C. We have an information about the loss to visit or the last arrived. The rest arrival has post since the minutes. We want to find the probability that the next arrival will be within the next five minutes. Then we want to get the probability for the random variable to be greater smaller than 15 minutes to happen before 15 minutes, given the random variable is greater than 10 minutes, this means the minutes husbands. The minutes have passed since the last visits, and we want to get the probability to happen within the next five minutes. After the 10 minutes using the complemented event, we can write it as one minus the probability for the random variable to be greater than 15, given that the random variable is greater than 10 using the Memories property, which says that we can really this as one, minus the probability of the random variable to be greater than the difference between these two values. 15 minus 10, which gives five again. This is a compliment event for the following. The probability of X to be smaller than five then equals using the cumulative probability function one minus. He was about, uh, minus. MX Emir is one divided by seven and X is five, which gives 4.51 46 Finally, for body, we want to find the probability that more than eight patients arrive during a half hour period. Eight patients arrived in half hour buried, which means 30 minutes and we put the random variable here in minutes and it's defined as the duration between patients arrival. Then we want to get it. Has the one interval happens at what time or what duration? We can answer this by getting assessor one multiplied by 30 divided by it. It's 30 divided by it, one multiplied by 30 divided by eight. Then we can get the probability for ex to be smaller than 30 divided by it as an adoration. If the duration is smaller than 30 divided by eight or three point 0.75 If the duration between arrivals is less than 3.75 this means more than eight patients will arrive during a half hour parrot, which equals one minus e to the power of minus one, divided by seven multiplied by X, which is here now 3.75 Then the probability equals 4.41 47 And this is the final answer of Bardi. Final answer. Barzee. Final answer of party and the final answer of party

Who did you know? The general wisdom is that you're supposed to go to the dentist twice a year. So that's interesting. Well as you study this table. So this table this this contingency table. This problem of a probability distribution, probability distribution. Yes. Um ranks respondents to the survey by age group and by how often or when was the last time that they had been to the dentist? So um that here we go. Let's go and jump into our problem. So it's supposed to represent the whole U. S. Population. So now we're going to ask we can take from the survey and generalize it to the U. S. Population. Say if a american adult is selected at random, what is the probability That they fall into one of these categories? We're gonna go ahead and convert them two as we go along. So the first one asks about our first row, what's the probability that randomly selected american adult has attended? The debt has been to the dentist in within the last six months, basically is how you would translate that. Right? So that's asking for that single variable, not a joint event. So, headway on out to the margin there and get that marginal probability associated with less than six months. So that's a 0.441 out there on the table. Alright. The next one is the probability that the person, the randomly selected american adult is not in the second age category. Okay, so the easiest way to do this is to take the totality of all the people and subtract off the marginal probability for the second age group. Second column of age groups. So that's a fair point 314. So if we do that calculation uh handy Dandy calculator real quick. We end up with 0.686. So you should be interpreting these as we go along. So the first one, you know, you make a sentence of that. You're going to say like 44.1%. There is there is a 44.1 chance that a randomly selected American citizen adult has been to the dentist in within the last six months. It's not bad. That's not bad. 44 chance. So yeah, if an American adult is selected at random that there is a 44 probability that they have been to the dentist within the last six months. So this one Would go, there's a 68.6 chance that a randomly selected american adult is within the age group of 45- 64. All right. That's just saying how old they are. Really has nothing to do with the dentist. Right. All right. So let's come to the next one. So, the next one is a joint event alright. The joint event that they are over 75 and that it's been five or more years since they've been to the dentist. Okay. So people who just basically don't go to the dentist anymore and are very old. What's the probability that a random select american citizen falls in that joint event category? So you're pulling that off, you're looking for the intersection of row five there and call them for Within that black box. And that would translate as 2.2 chance. So let's say if an American adult was selected at random, there's a 2.2 chance that they are over 75 and have not been to the dentist in more than five years, five or more years, say it correctly. Alright, onward. Now we get to our nice conditional probability is really important part of the question. Right. What's the probability that they've been to the dentist sometime between two and five years. And I'm switching up my notation here given That they are an adult under 44 years old. All right. So that you've been to the dentist into in the last 2-5 years, But you're an adult under 44 given given. Right. So, we got to look only at so in these when we have given a probability distribution table and ask are conditional. Were forced to do this formula where we have to do um Joint variation divided by the given. The probability is given. So we want the probability of T4 and a one divided by The probability of the given, which was just a one. All right. So look for that joint one at row four and column one. That's 10.7 Provided by that marginal probability at the bottom of row of column one, which is .5-6. All right, pull out the handy dandy calculator for these 24 755.5 to 6. It's And we get .133 or there's a 13.3 chance who 13.3 chance that are randomly selected American adults is Under 44 and has been to the dentist in the last 2, 5 years. All right. Last but not least. They take that same conditional, but swap it around. So now it's the condition conditional probability. What's probability that they are 18 to 44 years old given That they've been to the dentist in the last 2-4, 2, 5 years. Right. So this numerator is exactly the same for this calculator calculation, but the denominator now switches to that T. four row, so My numerator still .7, but my denominator becomes the marginal probability for roti four, which is 40.1 to 2. All right, So he is. The calculator gives me 25 7 Or 57.38%. So we can say, given that a randomly selected adult Has been to the dentist in the last 2 to 5 years. The probability that they are 18-44 is 57.38%. All right, enjoy.

Hello everyone. So we're going to solve problems 151 For this fortification. We have to rest refer the hospital emergency room data that has given an example 2.8. So we have to use the base here. Um to calculate the probability that a person visit hospital for. And given that their elders do. Yes. No we have to go do In the data of 2.8. Now let's say even tell the notes that the person is a WBS and even a P C. D. Do you know the person visited hospital one hospital too. Hospital three and Hospital for respectively. Now we have to calculate the probability of be that is hospital for given that I will use the base rule here is the probably of hell be for a variety of provided vote would be off hell even a what the hell? Hmm. Somebody else. And given possibility of three bless poverty of L. Given see property of C bless City of Elk 70 probability of B. No just we have to calculate we have been used. Oh sorry for this. So we have to use the data from example 2.8 Do you answer the same property of health? Given the so in the depravity of it will be 2 42 divided by yeah 4329 priority of D. Will be 4329 headed by the total 2 to 5 to five to no property of L. Given A. When I do for you Divided piles 5-92 actually 5292 divided by the total to to fight too. He said loose Be to 70 x 6991 for the hospital. It would be to 46. We're going to be faith sales good 56,405,640 into 5640 the world by the total to 52 One more addition for hospitality that we have already calculated about 40 to develop 4329 4329 divided 5- 2- 5 to. And these values will be cancelled. No, I will solve for the remaining values. It will be 242 divided five 2 2, 2 felt too 1010. Or else we can just take the common. So it will be 2 42. We just take 250-2 common in the denominator and we'll just cancel it. So it would be 1 95 plus 2 70 Plus 2 46. That's 2 42. Mhm. And the answer of this would be 1 95 plus 2 17 plus 2 46 2:42. This would be 9:53 2 42 divided by 9 50 truth. This would be 2:42 divided by 9:53 point 254. This will be the final answer and the final remedy hopefulness through the best here. Um Thank you. Have a good day


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