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Assume member i$ selected at random from the population member selected at random is from the shaded represenled by the graph Find the probability that the area of ...

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Assume member i$ selected at random from the population member selected at random is from the shaded represenled by the graph Find the probability that the area of the graph Assume the variable is normally distributedS4T Critcal | Reading Scores200 < x < 375ScateTha probablay that he member selecled at random is from the shaded area of the graph Is (Round to four decimal places as needed )

Assume member i$ selected at random from the population member selected at random is from the shaded represenled by the graph Find the probability that the area of the graph Assume the variable is normally distributed S4T Critcal | Reading Scores 200 < x < 375 Scate Tha probablay that he member selecled at random is from the shaded area of the graph Is (Round to four decimal places as needed )



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A member is selected at random from the population represented by the graph. Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable $x$ is normally distributed. (Check your book to see graph)

So to solve this problem, we're going to need to compute the C values for both 204 50 which are the bounds that we're looking between. So our Z score for 200 you can you see above equation. So we have acts which in this case is 200 minus the population means she's given us for 88. And that's all divided by the standard deviation which is given this 1 14. So the Z score for 200. Once you calculate that should be negative. 2.5 to 6 Repeat the same process for 4. 50 except instead of 200 we'll replace the X value Where will 74 50 instead of 200 for X. So right here saw it the same way. And we should get a Z value of negative 0.333 Okay, Now, when we look these values up on the see table, um, that tells you the probability that value is going to be to the left of that or less than that. And it's the area under the standard normal curve to the left. So our Z value for 4. 50 is going to give us all of this area that I've shaded in green. The one for 200 will just give us this part. So in order to get our blue area, we're going to take the corresponding probability or area for a Z score for 4. 50 which again is the negative 500.333 And the more subtracting off the smaller one. So subtracting off this excess the tail which is given with the probability given from the sea score 200 so we can look up both these values and rz table. You should get negative. 0.333 should give you a an area or probability of 0.37 and we're subtracting off. One for 200 r Z score is negative. 2.5 to 6 in that area should be 0.6 And you just subtract this. You should get an answer of 0.364 technical, and that's our answer

So to solve this problem, we're going to need to compute two different see values one for 6 71 for 800 which are the bounds of the region were looking for for the C score for 6 70. We're going to use the equation I included. Above we have our X, which is the 6 70 minus the population mean, which is 5 14. It's all divided by the standard deviation, which is 1 17. That should give us an answer of 1.333 We can repeat the same process for our second value, the Z score of 800. Except we're plugging in 800 instead of 6 70 for our X value. But then you compute it all the same, and you should get a Z score of 2.44 now a Z score. When you look it up on the table, that gives you the area under the curve to the left of that. So when we look up 6 70 it's our sorry when we look up 800 that's going to give us this entire green region, which includes the blue region. When we look up the 6 70. It's again going to be this green region, but this time it's only up until this left bound because it's on Lee to the left of the 6 70. So to get this blue region, we're going to take the entire area, which is the one given by our larger value 800. And then we're subtracting off the smaller one. So everything that's outside of our bound, which is given by the Z score for 6 70. So first we look up the corresponding area for 2.44 our 800 z score in the Z table, and you should get 0.993 So we have 0.993 and then we're subtracting off the area, corresponding with fizzy score for 6 70. So we look up 1.333 and you should get 0.909 Subtract thes and you get 0.84 And this is the probability that value is going to fall between these two bounds

So to solve this problem, we're going to need to compute to Z scores one for each of our balance. So to 22 55 first, the Z score for 2. 20. We can use the equation I included above. So are the score is going to be equal to X, which is the to 20 minus the population mean, which is 205 divided by the standard deviation just 37.8. Now you can solve that equation and you should get a Z score of 0.397 How we conduce this same process for our second bound to 55. Now the only thing that's going to change is we're plugging in to 55 instead of to 20 for X right here. Otherwise, it's salt the same. And then you should get a Z Valley Z score of 1.323 Now, when we look up one of these see scores on the table, it gives us on the Z table. It gives us the area under the normal curve to the left of that otherwise known, which represents the probability of number of being less than that much so for 2. 55 the Z score of 1.323 It can represent this whole green area through the blue, and we'd only want the blue area. So to get that, we're going to take this ceased, um, corresponding value for the 1.323 So this whole green area, then we're subtracting off up to the 2 20 Which is this thescore? The 2. 20 z score, which is the 200.397 So first we look up 1.323 are upper bound, the one that corresponds with 2 55. We look that up in RZ table and we get an area of 0.907 Now we want to subtract off what's not this blue area. So we look up the Z score for 2. 20 which is this 0.397 on RZ table, and you should get an area of 0.654 are you subtract thes. You should get 0.253 and this is our answer

Once again welcome to a new problem. Uh, always assume that if you have a random variable, there's always gonna be specific values connected to that random variable. For example, you could have a variable like safe height in terms of X, or you could have a random variable like a weight in terms off X, for example, and the reason why the random is because connected with probabilities and these probabilities also random. So, for example, what's the probability of getting a height in a class off kids in the fifth grade? Between say, you know, we can come up with value 30 inches on 60 inches, for example, You know we can come up with values like that, and then we have probabilities being assigned to this distributions. So a random variable the reason why it's random is the fact that the probabilities come in, uh, in terms off chance processes. So in this problem, or given a distribution and a lot of distributions in nature happened to be normal, for example, you could have a height distribution, which is normal and normal means that 50 50% on one side of the distribution you have ah specific number of heights. And then on the other side of the distribution, you have the same number of heights in terms off percentages and using something called empirical rule. The way the percentages air spread out is that 34% off. The distribution on each side is within one standard deviation off the mean. So on this side, you have one standard deviation on. Then on the left side, you have negative one standard deviation. And the way you measure standard deviations is through Cisco's where you take a particular school. Uh, this middle part is the main, and then you get the distance between that's called X on the mean on. Did you determine how many standard deviations go in there? So, on average, a standard deviation by definition, this is a population standard deviation. This is population standard deviation, and we can define it. Uh, you know, we do also have a sample standard deviation as so as is a sample standard deviation, and, uh, we can define it. We can define the Z score as the number of standard deviations are particular school eyes from the main. This is the population, I mean, and this is the population standard deviation. We call it Sigma. We call population standard deviation Sigma. And so what happens is were given a problem, which is the normal distribution are meaning one standard deviation 68% off. The distribution is within one standard deviation of the mean on this is called empirical role on. Then we also have have go on both sides. We have 95% off. The distribution is within two standard deviations and then we have 99.7% off. The distribution is within three standard deviations. So that's called empirical role. And in this problem, we have a distribution like that. Well, given the mean, the population mean off that distribution on the normal distribution always has a bump in the middle and it tips off towards the end. So the mean is 50 in terms off the random variable, uh, in terms off the random variable, the main is 50. The standard deviation, which is the average deviation from the mean happens to be seven and then we're given to scores. One of them is above the mean 65 s o x one is 65 that's above the mean on, then X two is below the means. So this would be maybe somewhere right there, this would be 40. So x two is 40. So in this particular problem way, want to find out? We want to find out the, um the probability, this probability or that area or that percentage. Uh, you know, what's the probability that our school is gonna lie between 40 and 65? You know, that's that's our probability that our score is gonna lie between 40 and 65 on it's a process. So we have to use Cisco and Z Table. Remember, we define the Z scores number of deviations from the main. So our target equation is Z equals two X minus mu off. Um, Sigma. So maybe we want to change this, too, because it's the left and the right side. We want to change this to x one and x two, and so x one, uh, will give us Z one. The Z score off 40 eyes, the same as it's a negative Cisco. So that means it's below the mean eso. This is approximately 1.43 That's the Cisco. And then we also have to find the Cisco of the second school, which is 65. And that's going to give us approximately. It's a positives isco because it's above the means. So it's approximately 2.14 thes airs isco's and so another way off seeing this problem is, how are we gonna find the probability for Z Scope's ISCO's between 1.43? Um and so we're looking for the area between these to Z scores were looking for the area between these two Z scores and we're gonna use the Z table. So in terms of the distribution, the first thing want to do is, uh, if you go to the table, the Z tables, the distribution table. Uh, this is still our Z hours isco off zero, and this is a was ISCO off 2.14 Uh, the distribution, this whole distribution is the same. And so the probability that, um, this whole area that Z is less than 2.1 full, that distribution from the Z table is the same as a 0.98 389838 Another way of saying it is that the percentage represented by that shaded area is 98.38%. And then also, we're gonna go ahead and and, uh, find the distribution. Find the distribution for the ZESCO. That's less than 40. So if this Z is, uh, negative 1.43 our interest is this particular area on that particular area is the same as so this this distribution there's an area off 0.764 Okay, so if I wanted to find the middle part, which is the shaded part in red, it means that I'm going to take, um I'm gonna take the probability that Z is less than 2.1 full and subtract the probability that Z is less than what are negative. 1.43 And so this is 0.9838 minus 0.764 That's going to give us zero point 90 74 So our solution off probability between negative 1.43 uh, in terms off Ciscos and 2.14 in terms of Z scores is the Samos probabilities off 40 on X. This is the raw score in 65 on, then this gives us a probability off 0.9074 which is approximately 90.74% if you wanna think of which, in terms of percentages. So once again we had a problem. And in this particular problem, we were given, um, two scores 40 and 65. We wanted to find the percentage between them or probability between them. And so we picked up on There's Isco's. This is remember, this is negative 1.43 Just to clarify that we picked up on that Cisco and then another Cisco of 2.14 We went to the Z tables and found that the probability that it's less than 2.14 is 98.38 and then they're the probability is 0.764 That's for the So this is isco to on This is the score one and so probability that ZESCO one or rather Z score two is less than 2.14 and Cisco one is less than negative. 1.43 We got those numbers and our final final solution is 0.9074 So I hope you enjoy the problem. Feel free to send any questions or comments, um, and have a wonderful day. So this is our percentages


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