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You are offered two machines: Machine A: Produces normally distributed output; Machine B: Produces uniformly distributed output Your target is 3.00 inches and LSL ...

Question

You are offered two machines: Machine A: Produces normally distributed output; Machine B: Produces uniformly distributed output Your target is 3.00 inches and LSL 2.55 and USL 3.65. Normal output (Machine A) can be set for mean 3 and &d,is given as 0.2. Uniform output (Machine B) Range [2.5,3.5]. Your customer uses Taguchi method for assessing quality costs You would buy Machine b. You would buy Machine B Either machine will work for your application ad satisfy Your customer equally None of

You are offered two machines: Machine A: Produces normally distributed output; Machine B: Produces uniformly distributed output Your target is 3.00 inches and LSL 2.55 and USL 3.65. Normal output (Machine A) can be set for mean 3 and &d,is given as 0.2. Uniform output (Machine B) Range [2.5,3.5]. Your customer uses Taguchi method for assessing quality costs You would buy Machine b. You would buy Machine B Either machine will work for your application ad satisfy Your customer equally None of the above



Answers

In Exercises $32-34,$ assume a normal distribution.
Machine Accuracy A machine produces screws with a mean length of 2.5

$\mathrm{cm}$ and a standard deviation of 0.2 $\mathrm{cm} .$ Find the probabilities that a screw produced by this machine has lengths as follows.
a. Greater than 2.7 $\mathrm{cm}$
b. Within 1.2 standard deviations of the mean

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.

Yeah, were given the mean and the standard standard deviations and some bearings. Now, I'm actually going to start with the assumptions here we are assuming that this machine is making bearings that are Fitting to a normal curve. So I have drawn that assumption here. Normal curb with the center at three and the standard deviation of 3.01. Now, to answer the question about the percentages of the fractions we're going to be using the empirical rule, it says any Bearings that are less than 2.98 Or more than 302 are considered defective. Because of the empirical rule, we know within two standard deviations of the mean, We should encompass 95% of the data. So that means that the two tails left of 2.98 and right of 3.02, that leaves us 5% left over, that is going to 5% is the percentage of bearings that are defective.

For a number 30 treat. We will, and we were told to assume normal distribution. So belt here, Um, and what they tell us is that this machine produces cruise with, on average, 2.5 centimeters length. And with this tender distribution of syrup 0.2 centimeters, Um, now the problem is, we need to know what's the probability that the screws higher than 2.77 years and was the probability that it's within 1.2 standard deviations of mean That's gonna be hard to do. Um, if we keep those numbers. So what we will have to do is convert them to Zedd's course, and then once they're converted into Zedd's course, we can use the table at the end of the book. So is that score for you Dream here. I'm gonna try it on and in the said score representation. We know the main or be zero, and we know that Signet will be one, because those air dick characteristics of a zed score curve to convert, um, 2.7 centimetres. So, hey, we have X equals 2.7. Um, and we want to know what's said. That equals 2.7 minus 2.5 divided by zero point. I'm gonna write the formal. Actually, the format for said scores X minus mu divided by signal. So that's what I've been doing here. 2.7 minus 2.5, divided by sigma, um, to a point. And that's one. So what we want to know is the probability that this crew as wing created and one in said score representation. So what's the problem? That is a hard one? Well, we can get a table. Um, and so since the table only tells you the area from mine isn't feeling well, student look at one and we're going to see was the area between minus infinity and one. It is, if your table again. It's zero point 841 tree, but that's Jerry in blue and for the area and ready. We need to do to do one minus syrup with it for one tree, and I would be equal to zero 15 87. So that's the answer for problem, eh? For problem be, um, we want to know the probability that it is between minus 1.2 standard deviation and 1.2 standard deviation So this is already in the center of representation. Because if you're doing X equals one point to signal anyone converted into it said you're gonna find, um, 1.2 are Let me rephrase that. Yeah, I'm actually gonna draw this curve in black as well, because that's how it was in the previous one. So this isn't the centimeters representation. We have our meeting here, which is 2.5. Um, and then we want to know. Okay, 1.2 sigma. What is one point you sigma in the said representation in the set score representation. So 1.2 sigma right of you, that is X plus m x equal mu plus one point to signal the zed. For that will be X minus mu core sigma. So new plus 1.2 sigma minus mia over signal. As you can see, the muse cancel out. And then after that, this significant self and you get 1.2. So all of this to say that if they tell us the problem, that something is 1.2 standard deviations from me you can directly use there. That's core representation with demeanor zero and say Okay, I'm just at 1.27 All right. You know, keep Mookie to keep moving with the problem we want to know is the probably that it's within 1.2 standard deviation from the mean. So not only this area, but also on the left side. Minus one point. Was it? Probably It is contained within this red area here, um, different ways to do this. But what I suggest is you look at the table, you get the value for mine for plus 1.2, and then you subject for minus one point. You all right? And so, for one point to the value of found debut is 0.88 49 And for minus one point, do I find 0.1151 which gives a total of Syria point 76 98. So there's 76.8% that we confined. Um, this crew within 1.2 standard deviations off the meaner. I mean, being here on this being 1.2 standard deviation

Mhm. So in this question we told, the machine produces fasteners whose length must be written in .5" of 22. So 22 plus minus 220.5 is where their length should lie, Length are normally distributed with a mean of 22" and a standard deviation of .17". And were asked for the probability that randomly selected fastener will have an acceptable ones. Because the probability that it lies Between 21.5 and 22 0.5. So basically that will be our standard Normal Variable -2.94 Hands 2.94. So we take property The less than 2.9 for minus properties the less than -2.94. So that's .9984 -1016. Which gives us .996 ft answer to party Would be first told the machine produces 20 fasteners per hour. The length of each one is independent. So we know here there's a binomial random variable, there's 10 Equals to 20 fasteners, assuming that they're all independent by the probability that they will have acceptable length, so the probability that each one As an acceptable language, .996 ft. So using our binomial distribution, we basically are looking for probability that all 20 will be acceptable, so that's Equal to 20 factorial, zero factorial and the pictorial P to the power of X. Q. to the power of N -6, which gives us .9379.


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