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Examgle 11 Two machines Aand B, are used to pack biscuits. A sample of 10 packets was taken from each machine and the mass of each packet; measured to the nearest g...

Question

Examgle 11 Two machines Aand B, are used to pack biscuits. A sample of 10 packets was taken from each machine and the mass of each packet; measured to the nearest gram, was noted. Find the standard Deviation of the masses ofthe packets taken in the sample from each machine Which machine, do you think, is more reliable? Explain your answerMachine A (mas: ing Machine B (mas: ing)196 198198 199200 200 201201 202205192 194195 198200 201203 204 206 207

Examgle 11 Two machines Aand B, are used to pack biscuits. A sample of 10 packets was taken from each machine and the mass of each packet; measured to the nearest gram, was noted. Find the standard Deviation of the masses ofthe packets taken in the sample from each machine Which machine, do you think, is more reliable? Explain your answer Machine A (mas: ing Machine B (mas: ing) 196 198 198 199 200 200 201 201 202 205 192 194 195 198 200 201 203 204 206 207



Answers

In the past the standard deviation of weights of certain 32.0 -oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.35 oz. Is the apparent increase in variability significant at the 0.10 level of significance? Assume package weight is normally distributed. a. Solve using the $p$ -value approach. b. Solve using the classical approach.

All right. So the question that is given to us about a normal distribution, So given that he was normal distribution to solve this problem so mean is given as ah 1 37 I want to go there on the Sigma, but is the standard division This is given as 1.6. So now, if I use the definition of standard normal distribution there is given by X minus mu upon sigma. So the first part that they have told us actually to find before that I'm going to substitute the value of mule or yours. And one sigma values 1.6. So first part, which is a bar they have asked us to find the probability that the single jar contains more than the state of contents. So the state of contents are given us 1 35 So which means I can take X greater than 1 35 This is what we're supposed to find. So to start with, I'm going to first find the value of said at X equal 1 35 so we can substitute X over here as 1 35 minus 1 37 going toe upon 1.6 to simplify, you're gonna get minus 1.37 or here after you simplify and get the value of said we can therefore now right, the probability off this particular Tom Tom's upset as Zell is greater than minus one wind, three said So since from the table we need to look this value we can write the greater than sign as one minus probability off send a lesson equal to minus 1.37 Now this condition that we have the scandal represented as one minus five off minus 1.37 So you want to take this valley from the table off the Sander normal distribution table and it won't get the blow off 1.3 and column of 0.7 I think the value and then value comes as a 0.8 So if I performed the subtraction, the answer that I'm getting here for the probability is 0.9162 So this is the first part that we've done now moving on to the second part for the second part there said that we need to find that the jars can't be more than scary contents are at least take So which means that over here the vast refused my normal distribution because n over here is given asked them their moral dangers on DDE. Therefore, I applied by normal distribution this can Britain s Stanko mother probably TV and just found you know why 9162 So the definition of by a normal distribution to find the probability and begin as n c x b days twigs and this is cuties to end minus X where we can see that Q was given by one minus b So in this particular problem, you know P is 0.96 To soak you is going to be one minus B which will be 0.8 three So no morning on the mean part probably that they told us to find is probably off They have said X should be granted an equal So say it because they say that there should be a two least eight jars So this can Britain as probability off X is equal to eight Bless probability off exit nine and next is probably off X That is equal then. So if I substituted these values or here in the formula off by a normal distribution we will have. In a way, he understands this is 10. See it be a zero wind 916 to this will be raised to eight. And he was zero pines. It'll end three years. That'll be 10 minus eight, which is to plus, if I write for X is equal to nine consistency. Nine. You know why? 9162 days to nine. Andi, don't mind. You ate 38 These 21 and the next is going to be only 10. See 10 and 0.9162 days to 10 and this time becomes a stew zero, which is actually one. So I did not write that bar the calculator. If you put all of these terms and you simplify it, an individual answers. I'm writing her. I'm getting 0.1568 Plus, the second term is giving me 0.381 plus stardom is 0.416 10. If you add all of these values on so getting here 0.9547 So we've added these probabilities because at a point of time. Only one Republican occurred. We cannot have. All the three are occurring at the same time. So it reflects or and or means audition. So we've been with the B partner going onto the sea part again for the sea parted. Told us he was a normal distribution. So normal distribution. We know Muir's 1 37.2 and ah signifies 1.6. But we write Sigma squared, which is the villians or here. So now they're told that, uh, you need to actually, So this is the first problem. They have said that the mean remains 1 37.2 But you to find the standard deviation value such that nine different person off all jars will contain more than the slated contents. So in this case, we actually have to solve this as the worst problem where they have said that 95 percenters already the number of jars which contain, you know more than the skated condense. So I can therefore write down that probability off zed. All right. Oh, is good. And then some small value off there which was given as 95% so far right that in decimal that'll be 0.95 So not from the normal table that we generally look. What we need to do is we need to find a nearby value 2.9, find the table and take that particular rule and call him and find out the value of this small. They so from the table. If I find this well, I'm getting the value off small that as minus 1.645 So no, Since I've got the value of that, we also know human sigma so I can use the formula and I can find out the value off the new Sigma. Add 0.95 probability. So I can say that they're interesting. My eyes therefore equal to X minus mu. So in this case, simplifying this part funder we could get So this is Ah, we have X is 1 35 which will be equal to the new partners 1 37.2 Bless zealot, which is minus 1.645 into Sigma. So if I simplify the subtraction and taking the terms on the other side, I get 1.645 Sigma is actually called Toto going to. So therefore, the value of signifies going to divide by 1.645 which actually gives me while I'm buying 34 or was there so this way we have done with the starving off older?

In this problem, we are going to find the mean and standard deviation of a certain random variable. So in this problem, it is said that boxes of 50 donor falls have an average weight of 16 oz, and the standard deviation is to 4.245 oz. And the empty boxes alone have an average weight of one ounce, and the standard deviation is 0.2 oz. What we need to find is the mean and standard deviation of the doughnut holes. Now, what we can do first of all, in order to find the mean is the expected value of one box, and 50 donut holes Can be written as the expected value for the weight of the box, and the expected value for the weight of $50 tools. That will be goes to the expected value for the weight of the box, and this will be 50 times the expected weight or a donut. So the expected value of the total, which is a box of 50 doughnut holes is given to be 16 oz. And empty boxes have an average weight of one ounce of this we want And plus 50 times the expected value of a donor. So that means that 16 -1, 15 is equals to 50 times the expected value of a donut hole. The weight of a donut hole. And so the expected value of the weight of a donut hole is equals to 15, divided by 50 and that is equals to 0.3. So the expected value for the weight of a donor hole is 0.3. Now we also need to find the standard deviation For that. Remember that for some the Variances act. So the variants for an empty box and 50 donut holes will be equals to the variants for the weight of the empty box, plus the variants for 50 donut holes. So that will be equals to the variance of the box, empty box and 50 times the variance of the weight of I don't know, I hope so. The variants for the total can be obtained as the square of the given standard deviation And the standard deviation for a box of 50 doughnut rules is given to the 0.245 and variants will be the square of the standard deviation. Now, the standard deviation for an empty boxes 0.2. So this will be 0.2 square Last 50 times the variance of the weight of donut holes. So that means that the variants in this case The variance for the weight of a vulnerable will be 0.245 sq -0.2 square divided by 50. And the standard deviation for the weight of a donut hole will be the square root of the variants, so that will be the square root of 0.245 square minus 0.2 square divided by 50. And the value of this is approximately equals to zero coins, 02 So in this case the Expected value, the mean will be 0.3 oz because that's what the unit the weights are given in, and the standard deviation will be 0.02 oz. So that means that the correct option in this case will be option B.

This current question at hand is the very same question that we had in 37. Problem number 37. But just a few values have been tweet on instead of lays. Now we have the readers, so the solution is going to be the exact same that we did for exercise. 37. Problem number 37. We're still Let's just go over this question and let us look at the values, the actual values that come out after the calculations. The first question says, Are the assumptions satisfied? Yes, we can say that all the conditions are satisfied if the six bags were randomly selected. This is very important. The next one says, find the mean and standard deviation of the weights. We are given six values and the mean is nothing but the average. So the evidence in this case turns out to be X bar is equal to 28.9833 28.9833 And what is the standard deviation? Standard deviation turns out to be 30.360 point 3601 This is a standard deviation for the sample. What is the formula for this. This one is submission off X by end. This gives us the mean and standard deviation. We already know the formula for standard deviation. The big one. That's right Off right it submission off X I minus X squid upon as this is a sample we're going to do n minus When we already know the values of N X, I excise the given values And what is X in this case? That's right. This is X bar. This is the mean confidence interval. What is the formula for confidence? Interval? It is going to be expired plus minus the margin of error. What is the margin of error? It is a critical or T start into s by routine and this value is by routine is known as a standard error. What is going to be t critical? In this case? We want a confidence interval for 95% confidence level and the degree of freedom is going to be an minus one, which is five. So for 95% confidence level and D f is equal to fight, the F is equal to fight. We will just take the help of a software or the tea tables, and we will find the T critical value, and this is going to give me lower and upper limits. The lower limit comes out to be 28.605 28.605 three, actually, 6053 and the upper limit is 29.3613 29 0.3613 What is the D party? Explain in context what your interval means. Well, in this case, we can say that we are 95% confident or there is a 95% probability. A chance that the true net weight of the bags of Doritos, marked with a net weight off 28.3 g, is between these two values is between these lower and upper levels. The next one says comment on the company's stated net weight off 28.3. We can see the 28.3 is actually not in this interval. It is not in this interval. In fact, 28.3 is less than this interval, so we can say that 95% confidence we have enough evidence to say that the stated net weight off 28.3 g appears to be incorrect, and the actual net weight is more than 28.3. These are going to be my answers.

All right, so this question deals with a known standard deviation, but an unknown means. And we also know that 18 is the second percentile. So we can use the standard normal distribution to find the Z score we'd expect for the second percentile using inverse norm. Point out too, with our mean of zero and our standard deviation of one. And that means that the second percentile has an expected disease score of 1054 Well, not expected. It just has a Z score of that. So then we can use the formula Z score equals X minus mu over sigma, and we can plug in what we know. So then we can solve this from you and simplifying. We find that mu equals 19.23 ounces.


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