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Homework: Chapter 5 sampling distributions Score: 0.88 of 401 12 (12 complete) 5.2.10SaveHw Score: 99% _88 of 12 ptsQuestion HelpConsidor the lollowing probability ...

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Homework: Chapter 5 sampling distributions Score: 0.88 of 401 12 (12 complete) 5.2.10SaveHw Score: 99% _88 of 12 ptsQuestion HelpConsidor the lollowing probability distnbulon Complctu Dars Ihrough f bolowap(x)FindIRoundto decima placesIeeded |For a random gample oln= observations Irom Ihis dlsiribulion , Iind the gamplng distribullon Ithe 9aMcle mean ype Ihe answersascen dinq order for %(Type Integers Aintounod Iga ctions;,Find the sampln g & stribution Ithe madian M; ofa sampla of nosarvat

Homework: Chapter 5 sampling distributions Score: 0.88 of 401 12 (12 complete) 5.2.10 Save Hw Score: 99% _ 88 of 12 pts Question Help Considor the lollowing probability distnbulon Complctu Dars Ihrough f bolowa p(x) Find IRound to decima places Ieeded | For a random gample oln= observations Irom Ihis dlsiribulion , Iind the gamplng distribullon Ithe 9aMcle mean ype Ihe answers ascen dinq order for % (Type Integers Aintounod Iga ctions;, Find the sampln g & stribution Ithe madian M; ofa sampla of n osarvatione Jrom Inis populalion . Type Ihe answers ascanding ordar iorM enter Vou{anrwver Ihe edit field; and then click Check Answer nan remaining Ciear Alll Ccheck Anawer



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The College Board reported the following mean scores for the three parts of the Scholas-
tic Aptitude Test (SAT) (The World Almanac, 2009):
$\begin{array}{ll}{\text { Critical Reading }} & {502} \\ {\text { Mathematics }} & {515} \\ {\text { Writing }} & {494}\end{array}$
Assume that the population standard deviation on each part of the test is $\sigma=100$ .
a. What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test?
b. What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a).
c. What is the probability that a random sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (a) and (b).

Even standard normal distribution. We are new zero. We know and sigma is one posted steady or standard normal distribution. We have to find the value of probability between their disposed has 2.1, 1 and 4.0. So this is 2.11, another one is 4.00. And we have to find the inside area. This is the inside area. We can write for lTO ZG better than 2.1 and less than 4.00. There will be The same as for until that less than 4.00-. Girl feels that less than 2.11. This book value from the table comes out to be the last zed less than This value. 0.999 98. And another Valerie two point when one comes out to be zero point 98 645 25 cents. So this is 25 cents. Final answer will be just subtract the value. Yeah. 0.17. This is a valid European. Yes

Referring to an example in the section. Example 442 Yeah, for the by variant normal model for the random variable X, which was the S a T critical reading score and the random variable wide It was the s a T mathematics score. Define the random variable bull W to be the S a T writing score. The third component of the S A. T, we're told us is a mean of 488 and standard deviation of 114 were asked to assume that X and W So the critical reading and the writing portions heavy by various normal distribution where the correlation coefficient of the X and W is 0.5 in part a were asked to find the distribution of X plus w Well, we know that X and W are by vary. It's normal. This is what we're given as an assumption. It follows that you some X plus w has a UNA vary it normal distribution, in particular with a mean of the expected value of X plus w, which recalled the expected value of a sum is the same as thes some of expected values. So this is this expected value of X plus The expected value of w were given that the expected value of X previously Excuse me. Yeah, was 496 were given that the expected value of W is 488. So we have 496 plus 488 and this sums up 2 984. This is the mean and it has a standard deviation well determined by the variance. So we'll actually find the variance first, The variance of a some of variables. This is going to be the some of the variances so variance of X plus the variance of W plus to times the coefficients Next w, which is one and one times the co variance of X and w. This is the same as the variance of X plus the variance of W plus two times the correlation coefficient rho of X and W times the standard deviation of x times the standard deviation of w By the definition of correlation, coefficient and substituting, we have that the variance of X with standard deviation of x 114 This is would be 114 squared staring deviation was of W. Was also 114. And so this is 114 squared, plus two times the Correlation coefficient Rho, which we're told was 0.5 times the standard deviation of X, which again is 14 or 1 14. I mean times the standard deviation of W, which is also 1 14 and this adds up to 38,988. Therefore, it follows that the standard deviation of X plus w is the square root of this number. Yeah, this gives us 197.45 approximately so we have that. X Plus W is normally distributed with a mean, which we calculated to be 984 and a standard deviation of approximately 197.45 next in part B. Whereas to calculate the probability that thes some of the reading and writing scores is greater than 1200. Well, because the sum of X and W, we determined is normally distributed probability that X plus w is greater than 1200. This is the same as one, minus the probability that it's less than or equal to 1200 which is one minus five of 1200 minus the mean, which is 984 over the standard deviation, which is approximately 197.45 This reduces to approximately one minus Phi of 19 and using a table or a computer to calculate this value, we find that this is approximately 0.1379 Next, in Part C were asked to find the combined reading and writing score that separates the top 10% of the students from the rest. In other words, we're looking for the 90th percentile of this distribution. So the 90th percentile of the normal distribution with mean of 984 standard deviation of 197.45 So to find this, we'll set 0.9 for 90 percentile, equal to five of and then are variable X minus 984 over staring deviation 197.45 This is approximately five of Well, we could calculate it that way, or we could again using computer. You want to find the value of X minus 984 for 1 97.45 such that why of this number will be approximately equal 2.9. Seizing a computer, a table we have that X minus 984 over 197.45 is approximately equal to 1.28 and therefore, using some algebra, we find that X is equal to 984 plus 1.28 times 197.45 And this is approximately 1237. So this is what the 90th percentile looks like. Score of 12 37 in the combined reading and writing.

Okay, So the question here is so given a normal distribution of test scores, um, the mean is 50. And the standard deviation is it, um, so are mean is 50 and our standard deviation is equal to eight. Um, given, like, what is ours? E score. We want to find the Z score. Ah, of 38 in this normal distribution. Um, okay, so the formula for disease score is just take our value. 38 subtract the mean, which in this case is 50 and divided by the standard deviation. So this I'm actually tells us how many standard deviations above or below the mean are set of our, um, unit of data is so in this case, 38 minus 50 is negative. 12 divided by a. And we get negative three over two, which is equal to negative one and 1/2. Um, so that is our Z score. Um, and referring back to the problem that is answer choice number two. So that is your answer. Z score is negative. 1.5. Meaning it's 11.5 standard deviations below the mean, and that is choice number two

So for the given case, we have to find the area between the to the discourse that are yes, 0.68. And the 2nd 1 is 1.78. That equals to 0.68 And another one is equal to 1.7 17. We have to find this area. Well, this would be that Less than 0.60. And here 1.78. This value will be for l p o. That less than 1.78 minus quality of that, less than 0.68. This boat value from the table first value is yeah, zero point 9616 minus another value is 0.7485. And you can see abstract now local news one three 1 2. So answered. A 0.2131.


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