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H 2 Cas5 The lengths possible: Attempis ~Xk3la) 8 peofected; fhd de huas probability that couOU class unifotn length distribution betwccn 2 Sand 514 l 52.0 fonc qu...

Question

H 2 Cas5 The lengths possible: Attempis ~Xk3la) 8 peofected; fhd de huas probability that couOU class unifotn length distribution betwccn 2 Sand 514 l 52.0 fonc quch

H 2 Cas5 The lengths possible: Attempis ~Xk3la) 8 peofected; fhd de huas probability that couOU class unifotn length distribution betwccn 2 Sand 514 l 52.0 fonc quch



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A machine produces large fasteners whose length must be within 0.5 inch of 22 inches. The lengths are normally distributed with mean 22.0 inches and standard deviation 0.17 inch. a. Find the probability that a randomly selected fastener produced by the machine will have an acceptable length. b. The machine produces 20 fasteners per hour. The length of each one is inspected. Assuming lengths of fasteners are independent, find the probability that all 20 will have acceptable length.

Were given a uniform probability density function F of x equals 1.25. And it's defined for the range 74.6- 75.4. For part any were asked for the probability that X Is less than 74.8. So that's the integral of From 74.6 to 74.8 of our density function, Just equal to 1.25 times x, evaluated from 74.6 To 74.8. And this comes out to one quarter for part B. We want the probability That X is less than 74.8 Or X is greater than 75.2. Now these ranges are mutually exclusive. So we can say that this probability is equal to the probability Of X being less than 74.8 plus. The probability that X Is greater than 75.2. Also note that this is a uniform distribution, it's always a density of 1.25 Between 74.6 and 75.4. So it's centered on 75. So it looks something like this. So it's flat across that domain And its height is 1.25. Now, 74.8 and 75.2 are symmetrical. Mhm. About the middle of this distribution. They both point to away from 75. So, we're essentially looking for this area and we've already found the probability that it's between 74.8. Sorry, we're actually looking for the area outside of this domain. So this area. What's this area. We've already found this area in part a By symmetry. This area is also going to be .25 and the result in probability is one half. And for part C, we are essentially asked what is the probability that X is between 74.7 And 75.3? And that gives us a probability of .75. The probability is .75, that X is between 74.7 and 75.3.

And this problem FX is too so it's a constant and we have the domains 2.3-2.8. Denver's would calculate a communicative distribution function Is from X- 2.3. FXDX. So the answer is being two X from X to 2.3 and it's two x minus 4.6. So if you write it down it should be zero eggs. It's more equal than 2.3. two eggs from 2.3 to 2 point mhm. Two point eight and one if actually screwed. Or it could and 2.8. And then we can calculate the Probability that X is greater than 2.7. So recorded if it's square than 2.7, it should be smaller than 2.8 and it's going to be 2.7. It's more than X, more than 2.8. So the answer is the So F 2.8 -F 2 7. And the answer would be to times .1 and the other is .2. So the problem is .2

Okay, so for this question, we're told it's got a uniform random variable between 32 44. We want to know what the mean or the expectation is s. That's gonna be halfway between these two, which is 38. And then we want to know the probability that X is between 38 and 40. And if we try this out so we can avoid doing any actual calculations, we see that this is 12 across. And if we did 38 40 it would be do across. So this can just be to over 12 which is 1/6. So these are our final answers.

So we know that lumber these cuts are normally distributed And we know that the mean is length is 96", But the standard deviation of .5 for this lumber. And we want to know if you take an individual board, What is the likelihood that is longer than 96.25"? So we have in our little normal distribution picture here for individual boards, it's centered at 96, here is about 96.5, And we're dealing on 96.25 and we're finding that. So we want to convert that to a Z value, which I can already see what that Z value is, 96.25 -96, divided by the standard deviation, that's going to be half. And I can either look that up on a table and it's actually easier for me to look it up on the table and .5 less than negative .5. That's gonna give me .3085. So it's about 31 chance of that happening Now. We change the question, we take 40 boards and we find their means. What's the likelihood that the mean will be greater than 96.25? I noticed that distribution changes. The sampling distribution would still be centered at 96. However, it's standard deviation would end up being the .5 divided by the square root of 40. So let's just see what that is. .5 divided by the square root of 40 Comes out to be about point away, it comes out to be a .079. So if you go up one standard deviation, that's up to about 96.08 And we're going all the way up to 96.25 that and higher. So when we convert this to a Z value, We have 96.25 -196. But now divided by that standard deviation or the square root of end of that, this value of zero, about .08, because remember that's a mean Being of 40 boards being 90 Being Smelt finding the average of and having a B 96.25. And so if we take that .25 divided by that value, We get this z value is all the way up to 3.162, so it's over three standard deviations above the mean, and I'm going to use my normal CDF to find that the answer and when I do, I find out that that value comes out to be about this, so it's very unlikely. So in part C we can see that an individual board, it's very likely to get an individual board being that length. But to get a mean of 40 boards being larger than that is very, very unlikely.


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