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Dora is the quality control manager of a candyproducing factory She has measured the weight of 16 boxes of candies The average ofthe weights of these 16 boxes = 450...

Question

Dora is the quality control manager of a candyproducing factory She has measured the weight of 16 boxes of candies The average ofthe weights of these 16 boxes = 450,and the standard deviation of them Is She going to providea label for the weights of boxes with 95% confidence. Whatis the range of the weights that she should report? (3 marks)

Dora is the quality control manager of a candyproducing factory She has measured the weight of 16 boxes of candies The average ofthe weights of these 16 boxes = 450,and the standard deviation of them Is She going to providea label for the weights of boxes with 95% confidence. Whatis the range of the weights that she should report? (3 marks)



Answers

A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated?

This exercise, We're going to be constructing a 99%, confidence interval for the main weight. Um garbage trucks must lift each time. Um At a given collection point no Were given a random sample of 325 containers. So the sample size there is hatred 25 and we're told that the average um collection for those 325 containers was £75.3. So the mean the sample mean 73.75.3 lbs. And this sample standard deviation is £12.8. Now let's proceed and determine the 99.8 confidence interval. Now the level of significance that corresponds to the 99.8 conflict .99.8% confidence interval alpha is 1 -0.998 and that is 0.002. So the critical value that corresponds to this given level of certificates for two tails test is three points zero 90. Now we want to use the following formula, it should get the confidence interval X mas plus or minus the margin of error, which is given by The critical value for two tailed tests, Multiplied by the sample standard deviation divided by the square of end. So let's go ahead a substitute the values. So it's going to be 75.3 class or 3.09 um times the square at the times 12.8 over the square root of 325. I wouldn't be Fit this into the calculator and solve going to obtain 75.3 plus or minus £2.19. So we are 99 .8% confident that the population means will lie between the interval given by the uh The statement that the mean lies between 75 7 75.3 plus or minus £2.19.

A shop owner wants to know how largest sample needs to be in order to construct a 95% confidence interval off the Truman cost for a large cheese pizza. So we need to find end the sample size and the owner wants to be within 15 cents. Wants to be within 15 cents were also given a standard deviation of 26 cents and again with 95% confidence level. The Z value associated with a 95% confidence level is 1.96 and with the following formula, we will determine how large that sample size needs to be. So you plug our values in 1.96 multiplied by 0.26 divided by our error, which is your 0.15 and you will square all this quantity. Simplifying the values in the bracket will have N is equal to three point for zero and which is keeping a two decimal places since our initial given information also had two decimal places and we'll square that quantity and is equal to 11.56 instance, we're dealing with finding the sample size we need to round our sample size up to the next hole number, which is 12. So our sample size of 12 pizzas are needed for is needed to Dysart to determine our 95% confidence interval for the true cost true, mean cost of a large pizza cheese pizza.

So we know that individual Candies are normally distributed. So this is for individuals. And we know that the mean weight of an individual candy is 0.1 out. And we know that the standard deviation of an individual candy is point 01 ounces, and we're going to have 16 pieces of candy placed into a passion package. And it tells us that these are independently placed in, which is important. And so our new distribution will also be normal because the original distribution was normal. So this distribution will be normal. And our first questions we want to know what will the mean B and I'm going to let that distribution be called why? And really are written a variable. Why is the sum of 16 independent Candies and there are 16 of these? And so the mean will end up being 0.1 added together 16 times or 160.1 time 16 So it will be 1.6 ounces. That will be the mean of our distribution. Now, what would the standard deviation be? Well, the important part was that these were independent because if they weren't, we wouldn't be able to find the standard deviation and so remember, for a random variable to find the variance or the standard deviation squared, we have to find the standard deviation of each of these individuals added together all the way out to the last one. And so our standard deviation of why will end up being 0.1 squared plus point Oh, one squared plus and so on. And we'll add together 16 of those so we can go time 16, and then we will need to square root that and so we'll end up having if we we can do this actually without using a calculator. The square root of this becomes 0.1 and the square root of 16 becomes four. And so our answer is 40.4 ounces. That will be our standard deviation for this distribution. So here is the mean for an individual and the standard deviation for an individual candy. And if we put 16 of those Candies in the mean weight of that package, just the candy itself will be 1.6 ounces on the average and the standard deviation weak 0.4 now. So that was part a part B. We wanted to find What's the likelihood that the weight of the package is less than 1.6 ounces and that's going to be 50%? Half of them will be less than that. So be 50% and then we want to find what number? What what? Wait has 99% of these packages are higher than that? So I go back up to our picture. Well, we'll just we'll just draw the one. And we know that the mean is 1.6, and we want 99% or this area to be 0.19 point 99 which means this area down here must be point or one. So we need to find what that Z value is because we know we're working with a normal distribution and so we can look that either up in a table or I'm going to use my inverse normal function on my calculator and type in an area of point a one keeping the main at zero and the standard deviation at one, and we'll find out what that Z value is, and that's the value is negative 2.3 to 6. So let's now convert that into knowing that a Z score is equal to the score. Well, we is why, minus the mean divided by the standard deviation. And we know that that Z score is negative 2.3 to 6, and we're going to find that y value and we know the mean is 1.6. And the standard mediation is point oh four. So we can take this number negative 2.3 to 6. Multiply it by point oh four and then add on 1.6 to solve for life. And we find out that why is equal to 1.5 Yeah, 1.507 ounces. So roughly 1.5 ounces If the pack, if you're weight of the candy 99% of the time the weight of that can be The package of 16 Candies will be higher than 1.5 ounces. And I think we've answered all the questions

All right, so I want to find the standard deviation of the sample. I know Sigma, which means the standard deviation of the population I know this has given us to Sigma is given to me. As to now I want to find the standard deviations for all the shipments. That is for the shipments that go to restaurants, the shipments that go to the grocery stores and the shipments that go to the discount outlet stores. Let's try do look at these one by one. What is the formula that we're going to use here? The formula that we're going to use here is Sigma by routine sigma by routine where N is what ended the sample size What a Sigma Sigma as to what is end in this case. And it's for to buy route force, which is nothing but too white too, which is one. So what is my standard deviation for this one? This shipment, it is one similarly over here. I will have to buy Route 16. Yeah, this is nothing but two by four. Over here I get 0.5 And over here I have to buy Route 100. There's nothing but to white 10 or this is going to turn out to be zero going toe. Now, the second question that he asks me is, would the distribution off shipping waves be better characterized by a normal model for the boxes or the pallets? The answer would be it would be ballots. Why? Because the sample size over here is large. It is 100. So, yes, even if the distribution from which it has taken is not normal, you know the sample will form an approximately normal distribution. If the sample sizes large. In this case, it is just 16. But in this case, it is 100. So the answer would be palette. And these are my ancestors.


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