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Boltled water disinbutor wants ostimate Ihe amount waler contained 1-gallon bctlles purchased Irom nationally known waler boitling company The wate botling corpany&...

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Boltled water disinbutor wants ostimate Ihe amount waler contained 1-gallon bctlles purchased Irom nationally known waler boitling company The wate botling corpany'$ specifications state that the= standard devialion the amoun ol watcr equal to 0. 01 gallon_ random sampie 50 bottles selected and (nc sample mean amount waler pur Aqllon bottle 998 qallon. Comptole pana through (d) Consinuct 95% confidence interva estimale Ior Ihe popualon mean umouni waler Included I-qallon botileSps (Round =

boltled water disinbutor wants ostimate Ihe amount waler contained 1-gallon bctlles purchased Irom nationally known waler boitling company The wate botling corpany'$ specifications state that the= standard devialion the amoun ol watcr equal to 0. 01 gallon_ random sampie 50 bottles selected and (nc sample mean amount waler pur Aqllon bottle 998 qallon. Comptole pana through (d) Consinuct 95% confidence interva estimale Ior Ihe popualon mean umouni waler Included I-qallon botile Sps (Round = Ivu ducandI pucos ncuded |



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Water Dispensing Machine A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume the population of volumes is normally distributed. (a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a $95 \%$ confidence interval for the population mean. Assume the population standard deviation is 3 milliliters. (b) Repeat part (a) using an error tolerance of 2 milliliters. (c) Which error tolerance requires a larger sample size? Explain.

A bottle of water contains 12.5 fluid ounces with a standard deviation of 0.1 ounces. Defined the random variable X and words because we know we have 12.5 fluid ounces with a standard deviation of 0.1 ounces. The interest here is the number of ounces of water in a bottle, so we're going to say that X is equal to the number of ounces of water in a bottle.

This question We're told a machine that fills two little bottles. The standard deviation is .002 and the mean is set to whatever amount the machine is set to deliver. So in part they are mean is set to two leaders. And we asked what proportion will contain at least two later. So probability that excess more than r equals to two. That's one minus the probability X is less than this is less than two are mean, here is too, so this is just one minus the probability Z is less than zero, this value is 00.5. So our probability here is just 0.5. Now in beaver asked for the minimum setting, So the mean amount, The mean amount so that at least 99% of the bottles who hear the areas .99 will contain two L. So we want the probability that X is less than or equal to two. B 1 -19. Let's .1 and converting that to our Z value so that's all it is. The lesson are opposed to minus mu over the standard deviation Is Point for one. This is star value from the table is actually 2 -2.33, so we have 2 -9 over. The standard deviation is 2.33, and Our value of the mean is 2.005. Yeah.

Problem. We're going to be considering a packing machine that delivers a precise amount of liquid and that the amount is Pennst Always has a standard deviation of 0.07 ounce. And so we need to calibrate the machine By operating it 50 times And we find that the main amount is 6.02oz. So that's the sample mean, And the sample standard deviation is 0.04 ounce. No. Using that information, we're going to construct a 99.5% confidence interval for the mean amount delivered at that fixed setting. Now, not all the information here will be needed. So for us to get a confidence interval we have to use the formula ex ba plus or minus the margin of error. Now, imagine of error Can be obtained in two ways. One if the sample population standard deviation is known, we shall use it. But if we don't have the sample population standard deviation, we use this sample mean, uh population uh the sample standard division. So in this case we only need to use uh the population uh sample some population standard deviation and not example, uh standard deviation. So in other words, we shall not use the 0.04 oz. Instead we lose the remaining information no for us to work it out and get the margin of error, we have to get the margin of error. And for this case the margin of error um Relies on 99.5% confidence interval, Which means that the level of significance, alpha is 1 -0.995, which equals 0.005. The critical value of that that corresponds to the given level of significance is 2.807. Now we're ready to substitute the values into the given formula. So we're going to have 6.0 to plus or minus 2.807, Multiplied by 0.07 Divided by the Square Root of 50. And when he saw that you get 6.0 to plus or minus 0.03 when you run off to to their small places. And so this is the 99.5% confidence interval for the population mean for the given data.

For this exercise, we are given a continuous random variable that is uniformly distributed On the interval. 374- 380. And for part A were asked for the mean and standard deviation of the variable for a uniformly random variable, uniformly distributed random variable, the mean is equal to A plus B or two, where is the bottom of the interval and be at the top of the interval, And this is 377. And for the variance given by b minus A squared over 12, This is going to be six squared over 12 or 36/12, which is three or B. We are asked for the probability That X is less than 375. So let's first look at our density function for this random variable, it's equal to one divided by B -A. In other words, one of our six Forex between 374 and 380. So the probability is the area underneath the density curve, you're, the density curve is a rectangle because it's uniformly distributed. So the height is once 1/6, and the width is going to be 375 -374, that's the width of the area that we're trying to calculate. And so this comes out to 1/6, and next, for part C were asked for the value of X, for which 95% of the distribution exceeds it. So basically the probability that X is greater than what Is equal to 0.9 And actually that's not .9 but .95. So for what value of X does, 95% of the distribution lie above that value, so proceeding in a similar manner to part the the relevant area of the distribution that lies above X Is going to have a width of 380 minus x mm. And solving for X here Gives 374.3. So 95% of the distribution lies above the value of 374.3.


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