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1n4O are to major tests readiness tor college: the ACT ard the SaT; AcT scores Aecoret reported om Tecenl Yealshae scale irom to 36, Tha distribution 0f ACT been ro...

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1n4O are to major tests readiness tor college: the ACT ard the SaT; AcT scores Aecoret reported om Tecenl Yealshae scale irom to 36, Tha distribution 0f ACT been rouahly Notrnal 'lth meam mu 1600 Gat #comaeh 70.9 and standard devlaticn sigma 4.8. Sat scoras Jre reported on Len Townh {clettmom Normal with mean Mu 40zt Standaid derialon sdma Sat scores that make Up the top 28 of AM Gcores range (rom 1600 (kound KDur unswer the nearest whole number; )Iceea$ mar

1n4O are to major tests readiness tor college: the ACT ard the SaT; AcT scores Aecoret reported om Tecenl Yealshae scale irom to 36, Tha distribution 0f ACT been rouahly Notrnal 'lth meam mu 1600 Gat #comaeh 70.9 and standard devlaticn sigma 4.8. Sat scoras Jre reported on Len Townh {clettmom Normal with mean Mu 40zt Standaid derialon sdma Sat scores that make Up the top 28 of AM Gcores range (rom 1600 (kound KDur unswer the nearest whole number; ) Iceea$ mar



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The ACT and SAT are two college entrance exams. The composite score on the ACT is approximately normally distributed with mean 21.1 and standard deviation $5.1 .$ The composite score on the SAT is approximately normally distributed with mean 1026 with standard deviation $210 .$ Suppose you scored 26 on the ACT and 1240 on the SAT. Which exam did you score better on? Justify your reasoning using the normal model.

Problem eating the distribution of the Cetus. The scores is approximately normal with a mean of 500 and the sender the patient of 100 for party. We want to calculate the proportion of set scores at least then 400 in the normal distribution which is centered at the mean 500. We want to calculate the area that is less than 400. You want to calculate this area, we can use the school to do so That equals X minus new, divided by six in birthday. X year is 400. Then Z equals 400 minus 500 divided by 100 which equals minus one. Then in desert skate or the score we have Year zero, we have here minus one. We want to calculate the area to the left of the school. We have the standard normal distribution tables which he gives the area to the left of the school. Let's enter these tables. Why is it equals minus one? We have here that equals minus form. We get the first value because it's minus one point. Oh, Then the area to the left of the school is about 16%. Then the perversion or the probability to have X less than 400 X, the notes said. School equals a 0.1 58 66 This is a proportion of the set scores less than 400 for birth. We want to calculate the proportion for the set. Scores are greater than 650. We do the same calculations. We have normal distribution centered at 500 went to calculate this area. You can calculate the school four x equals 650 minus immune. Divided by sigma. This gives 1.5. We enter the tables again to calculate the area to the left of the school. She means we will get this area then to calculate the proportion for X is greater than 650. It equals one minus the area that would be calculated using this score 1.5. Let's enter to get the area to the left of 1.5. The area to the left 1.5 is here. We take the first value because it's 1.5 zero one boy Playful Oh, 0.9331 line over 29 3319 It equals Oh, point Oh 6681 This is a proportion for the set scores that is greater than 650 for varsity. Now we knew that the area is 20% because we need to calculate the minimum set score to be in the highest 20% of the population. The highest 20% of the population law is here, which means this area equals 0.2 four. The area to the left of this value, which is X, is appointed. And to do so, we will make the same procedure for the for the previous points. But in a reverse order we have here the area to the left of the school. We will enter the table by this area to get the correspondence in schools that equals, Let's enter the tables. To get this value, we will search the table for a value of a point. We have 4.79 here we have 4.79955 which is very close to a point. It this means the value of the school is a 0.8 four point it four. This is the value of that, and using does that score formula we can get X is is it multiplied by sigma plus me equals Oh 0.84 deployed by 100 close £5 then equals 584. This is the minimum set. School could be in the highest 20% of the population for body. We want to calculate the minimums that score to be in the 40% to be accepted in the State college in the same as Parsi. We want to calculate X here, which is which has an area to the right of it as all 0.4 or the area to the left of it as 4.6. We enter the table by this value to get the corresponding. Is that the school? We enter the tables to get the value corresponding to a 0.6 you can see right here open 59 this 0.59871 This is a 0.6 oh three. We can take one of them the average anyway, which means it approximately equals 4.25 There's a score corresponding to an area of all of open six is all going to five z equals 4.25 Then X equals that Sigma plus me. That is, Oh 0.25 deployed by 100 plus 500. It equals 525. This means we need to get a set score of 525 to be in the 40 or the top 40% to be accepted in the State college and for to be conservative we can use that equals open to sex to get this value to be conservative. This means this is 2.6 and the school is 500 and 26. We need to get I said score of 526 to be accepted in the State College.

To okay for this problem we're giving, given some s east as a a C T scores and going to convert them to s A T scores. And there's a formula for that that were provided to the most important information from a reading. Our problem is that the S A T scores on their scale equals to the A C T scores times 40 plus 151. Break this into two steps. So I'm going to think it had a little bit here and figure out what I want to get the lowest score and the third quarter I'll score and the mean and the median. They all get basically put into that formula. Um, and we'll figure that out using our calculator. So what I want to find is all these scores. And for a second step, I'm gonna do the same thing. But for the standard deviation and for the inter portal range, the i Q. R. A little bit different for a linear transformation. So we call this a linear transformations a little bit different for that. So let's just do first things first. Let's use our calculator over here, and we're gonna take each of the scores that we have, for example, are problems. Statement told us that I think 40 times that 19. That was my lowest score for the easy T. If I take that times 40 and then I add 1 50 970. I like to go in order from lowest to highest. So the mean score is second. So 40 I mean, score. Everything is adjusted by a factor of 40 and everything gets 150 added to it. Same thing for the median. These measures of center and finally for Q three, otherwise known as their quartile brief scored a 30 in it E c T. That translates into this number for the S A. T s levy a scaled score of 13 50. We take a second rate goes down so we know the lowest score for the S a T B clear. Here's the S A T scores What we wanted to find. The Louis translates to a scaled score of 9 10 The mean is a 12 30. The median of the 12 70. Yeah, and the third quartile, which we abbreviate is the Q three often and statistics is 13 50 they all get multiplied and shifted. I have thought about that on the graph. All right. Now, for the next two, I'm gonna go over here and I'm gonna take these two different numbers with standard deviation. Now, for these measures have spread. The standard deviation does not get added by 1 50 Because the standard deviation is just a, um do you think about its a difference between a low on a high between the lower and higher scores? We take the difference so we could add the 1 50 But we subtracted away for both of them. So really, what you need to know is that you just take for standard deviation. You just multiply by 30 and don't have the 1 50 for standard deviation. That's gonna be 400. And for the in a court. Delloreen, same thing. I'm gonna just multiply, Ray, Wait. Let me know will stick going back up. It's most by the 40. The standard deviation were given us three and the Inter quartile Ranger given six again. But both measures have spread, so you don't have the constant number. You just multiply by that multiplying factor. Important effect, you know So for the standard deviation, my end result for the S A. T. The standard deviation would be 120 and minor quartile range is going to be 240. And that's everything we need there. So nice and clearly Books my answers make a little observation in my final as a T scores, measures have spread and all these other scores five number summaries and measures of center.

In this exercise were given a sample of 20 act scores of college freshmen who are taking calculus. We're part they were asked to use an appropriate graph to determine if it's plausible that the scores were taken from a normal population or a normal distribution. So what we want to make is a normal probability plot for our sample data. So I have done this in R, and I'll show you how to do that. So the first step is to enter the data of the sample, which I'm doing with this command here. And then we can make a Q a normal probability plot by using the QQ norm. Command comes out like that, and we can also add a straight line to the plot with this command to help us determine if it's if the outcome is a linear relationship to determine the linearity of the scatter, that is Yeah, And as we can see from our plot, this data, the scatter here follows a pretty good line. So it is pretty plausible that the data comes from a normal distribution. So we could say that based on the mm mm, yeah, and then for B, we want to calculate a 95% confidence interval for the population mean now in 95%. Confidence interval is given as follows, so it will be the sample mean plus or minus the critical value times the standard error, which is the sample standard deviation divided by the square root of em. So we should first calculate the sample average. Now there are 20 data so back to our so we can simply enter the mean of score. That's 22.5 and now for the sample standard deviation that command is S D. It's about 2.69 We know that Anna's 20 and the critical value can be looked up in the table or calculated using our and it's equal to 2.93 So if we plug these values into our formula for the confidence interval and this comes out to confidence interval ranging from 23.79 to 26.31 so we can say that we are 95% confident that the true mean lies between these two values. Another way to say that is 95% of confidence interval is constructed in this manner would contain the true mean. So remember our sample was a C T scores for freshmen who were taking calculus in Part C. We're told that the A C T average score for freshmen in general was 21. So are the students taking calculus better than average, as measured by the And the answer is yes, because with 95% confidence, we predict that the true meaning is between these numbers, which are quite a bit higher than 21. This suggests that the calculus students have higher I a CD scores. MM mhm.


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