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Problsmpoints) Let X be the number of undamaged items out 0f 146 units, where the probability that a item undamaged 0.316Appronmaie the pnon probabilty using nommal...

Question

Problsmpoints) Let X be the number of undamaged items out 0f 146 units, where the probability that a item undamaged 0.316Appronmaie the pnon probabilty using nommal disinbution:Note that X hasdistlibulion By normal approrimation(ollowsP(X < 45) ~ P (Y <=P(z < 0.1132) 0.45493where X(non-slandard) normma andom varlable wilh Meanand standard deviaiionthe slandard noma (andom vanabieRct YoU can earn partial credit On Its problern,

Problsm points) Let X be the number of undamaged items out 0f 146 units, where the probability that a item undamaged 0.316 Appronmaie the pnon probabilty using nommal disinbution: Note that X has distlibulion By normal approrimation (ollows P(X < 45) ~ P (Y < =P(z < 0.1132) 0.45493 where X (non-slandard) normma andom varlable wilh Mean and standard deviaiion the slandard noma (andom vanabie Rct YoU can earn partial credit On Its problern,



Answers

$X$ is a normally distributed random variable with mean 0 and standard deviation $0.75 .$ Find the probability indicated. a. $\quad P(-4.02<X<3.82)$ b. $\quad P(X>4.11)$

Okay, Number 95.1 point three. Um, they give us a porcelain distribution of scratches, and we are supposed to look at the data. And from that data, we're supposed to figure out a couple of things. Um, so in the blue box there, we've got the number of scratches per item, so you'll notice that there's, like, 37 that had one scratch. There's 42 that had to etcetera. Um, so when you find out your end, it's actually 100 and 50 because you have to add up all those numbers to figure out how money actual observations that were right. So it's not just eight because there's, you know, eight numbers. That's not at all. It's like there's that's the money. That's the number of times that it happened. Okay, so in a post on distribution, the sample, mean is a great, unbiased estimator. Let's go ahead and use that. So we go ahead and we use this here with and is 1 50 we come up with 2.11 for our mu. All right? And they give us little hint on the next one, right? What? They're actually looking for though, is what's the standard error? Well, if we know this to be true, then we know that this is true, right? And then we also know that this is true and algebraic Lee, this will be true. So for the the estimated standard error, we can go ahead and use this. Take the square root of 2.11 divided by the square root of 150. And after a brief visit from your calculator, you should come up with 0.119 All right. Excellent. Thanks.

This time I want to find the probability that my Z is less than 1.45 So if I draw my normal distribution, if I draw my normal distribution, this is somewhere 1.45 days. This is 1.45 So I want to find the probability that my C is less than 1.45 This is zero. This means I want this shaded reason that is to the left off 1.45 Okay, so what I can do is I know that the area under the normal distribution is when under the scope is one I can do one miners the area to the right off 1.45 which is the thing with a P value of 1.45 If I use the calculator to find the value of 1.451 pale, this is 0.735 So I can do one minus 0.735 This value is turning out to be 0.9. Do 65 This is the probability that my Z is less than 1.45 This is my answer

So for this problem, the first thing that we're going to do is we need to know are actual equation that we're going to need. So we're doing a standard deviation and standard deviation can be denoted by this little funny guide and we're doing the standard deviation of X because this sub script says that we're doing it for X, so it is the square root of E. In the in parentheses you have X, and this is a capital X. That matters minus mu squared. And um so you need to know you first because for this problem we don't know what you is. And the way you find that is you do E to the X. M. X. And so what that looks like for this problem is U equals negative two times 0.3. Close three times two plus five times point for. And so what this is you're taking your exes and you're multiplying them by their function and then what you're doing is is at the end you are mhm, adding them together. And so, Let me fix this, this is .3 times 0.3. And these are all in parentheses, of course you're doing pimp dos. Um and so what your answer for this problem, and this is not the complete answer, 2.3. And so now, you know what mu is next, We're going to now put this all together. So we know that our formula for standard deviation is this. But so it's kind of write it again, the standard deviation of X is the square root of E. And the more times you write this, the more times you can kind of memorize this X minus ive X squared. And so what we're going to do is is what this turns into is is the square root E. And then in princes, you have x minus mu squared. And so then you can break this down even more. And if this looks funny, you can google on why this happens the proof for this. But you can put the summation of X times x minus mu squared times. Eh Fedex little eggs. Those are both little excess because that matters. A big X denotes a random variable. And so what that looks like, because I know that that looks like a handful. But when we put our numbers in there and remember you Is equal to 2.3. So we'll have negative too minus 2.3 squared times 0.3. Close and we can put curly brackets there, tended either. And then we're going to have more curly brackets and then we'll have The 3 -2.3 squared times 0.3 class, More curly brackets, five -2.3. That is not a 15, that's a a brace squared times 0.4 in curly brackets. And when you add these three numbers together this, you actually get 2.93, you go around To like 2.9 because you're only going to that figure in this problem. But if you want to be more precise, I would at least around like the 1000th place, especially if you're in a higher level class.


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