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3. IQ is 4 score that is transformed 80 that it has normal distribution with mean 100 and standard deviation 16.Let I be the IQ of @ randomly chosen person. (e) Wha...

Question

3. IQ is 4 score that is transformed 80 that it has normal distribution with mean 100 and standard deviation 16.Let I be the IQ of @ randomly chosen person. (e) What percentage of people have an IQ less than 112? (6) What percentage of people have an IQ between 96 and 1202 What percentage of people have an IQ above 1252 (d) Mensa 'Welcomes people from every walk of life whase IQ is in the top 2% of the population What would the cut off IQ be for this organization?

3. IQ is 4 score that is transformed 80 that it has normal distribution with mean 100 and standard deviation 16.Let I be the IQ of @ randomly chosen person. (e) What percentage of people have an IQ less than 112? (6) What percentage of people have an IQ between 96 and 1202 What percentage of people have an IQ above 1252 (d) Mensa 'Welcomes people from every walk of life whase IQ is in the top 2% of the population What would the cut off IQ be for this organization?



Answers

$\forall$ IQ Scores Mensa is a club for people who have high IQ scores. To qualify, your IQ must be at least $132,$ putting you in the top $2 \%$ of the general population. If a group of 10 people are chosen at random, what is the probability that at least 2 of them qualify for Mensa?

This problem is about membership in MENSA, and your membership in MENSA is based on the Wexler I Q test, and we were provided some information about that. We were told that scores are normally distributed. We're also told that the average score is 100 and the standard deviation would be 15. We're told that the top 2% get into MENSA, so because it was normally distributed, we contro that bell shaped curve, and the top 2% would be like right in here. So if 2% is on the right side, that means there's 98% of the curve on the left side. And what we're going to try to find in part a is what the cut off score is or what the minimum requirement is. What were 1st 1st going to do is we're going to find the Z score associated with that cut off score, and then we're going to transition into the X score or the raw score or the actual score needed to get into that Mensah. In order to find the Z score, we're going to use your inverse norm feature on your graphing calculator. I'm using the T I 84 or the T I 83 calculator, you can use your standard normal table as well. Um, the calculators just a little bit faster. When you use your inverse norm, you have to follow it up with some parameters. You're going to put the area of the curve. That's to the left of the cut off score you're going to then use your average and you're going to include your standard deviation. So in our problem, the area that goes into the left tail is 0.98 Because right now we're finding the Z score. The average of a Z score is zero, and the standard deviation is one. So I'm gonna bring in my calculator and to access the inverse norm, you're going to hit your second button. You're then going to hit your VA R s button and you're going to see a drop down menu. And we're going to select universe Norm and we're typing in the area to the left, the average and the standard deviation. And when you hit enter, we're finding out that we have a Z score of 2.5 So if we know my Z scores 2.5 and we know a formula for Z score to be Z equals X minus mu over sigma. We can put the value of 2.5 in for your Z. We're trying to calculate the X. We know our average to be 100 and our standard deviation was 15. So I'm going to place a one under the 2.5 so that I have a proportion and I'm going to calculate the cross products. So I'm gonna cross multiply this way and I'm going to get X minus 100. And when I cross multiply the other way, I am going to get a value of 37 5 and then going to solve the algebraic equation by adding 100 to both sides of the equal sign. And I'm going to get 130.75 But because the Wechsler IQ test scores do not come infraction form, we're going to round that up and we're going to get a 131. So therefore the minimum score for acceptance into MENSA would be 131. There is another part to this problem, So let's go into part B and in part B, you're asked if four randomly selected battles take the Wexler, find the probability that their means scores. That means X bar is, at least which would mean greater than or equal to 131. So in this particular problem, we're working with a sample with the sample size is four randomly selected adults. We are going to need to calculate the average of those sample means and the standard deviation of the sample means. In order to calculate those we can use the central limit theorem and the central Limit. Theorem says that the average of the sample means is the same as the average of the population, and in this case it was 100. And the standard deviation of the sample means is equivalent to the standard deviation of the population divided by the square root of the sample size. So in this case, it would be 15 divided by the square root of four. Now, the probability that X bar is greater than or equal to 131 we would technically need to use continuity correction in order to use the normal distribution to approximate this answer. So what we're going to do is we're going to switch it on, get a close enough answer, and we're going to just say the probability that X is greater than 131. So because we were provided information that we were dealing with normal distribution, we control our bell shaped curve. We're going to put our average in the center and our average was 100. And we're trying to figure out the probability that the four selected people average greater than a 1 31. So we will need a Z score. So the Z score formula associated with sample means would be X bar minus the average of the sample means divided by the standard deviation of the sample means. So we are going to find the Z score, associate it with 1 31. So we're going to subtract 100. And in place of the standard deviation of the sample means we're going to put the expression 15 divided by the square root of four. And when we do that, we get a Z score of 4.1. So we're trying to calculate the probability that our X bar was greater than 1 31 well being greater than 1 31 If 1 31 is comparable to a 4.1 Z score, we could say that this problem can be rewritten as the probability that Z is greater than 4.1 and our standard normal table in the back of the book and table A to always talks about the area or the probability into the left tail of the bell shaped curve. And as you can see our pictures going into the right tail, so we're going to have to rewrite our problem as one minus the probability that Z is less than 4.1. And when you use your standard normal tail or a table, the probability for anything greater than a 3.5 is going to be 0.9999 So we're going to do one minus that 10.9999 and we're going to get a probability of 0.1 So to summarize Part B, the probability of selecting four people and their average Wexler Ike test score being at least 131 is approximately 0.1 now there's one more part to this problem, so we're going to start Part C and in part C. The question is asking you if four subjects take the Wexler test and they have a mean of 132 but the individual scores are lost. Can we conclude all four of them are eligible for MENSA? And the answer to that is no. It is possible that the four subjects where people have a mean of 1 32 while some of them have scores below the 131 test score requirement to get into MENSA here would be an example. We want for test scores. We want them to average 1 32. So maybe we have 1 33 1 31 a 1 35 and a one 30. No, actually won 29. Let's go with that. So these are going to average to be a 1 32. But as you can see, this guy would not have qualified ferments

We are informed that I. Q. S have a bell shaped distribution with a mean of 100 the standard deviation of 15. The first thing you want to answer is what percent of people have an I. Q. Between 70 and 1 30. 1st off, we need to note that 70 and 1 30 are both two standard deviations away from the mean, 70 is two standard deviations below and 1 30. 2. Standard deviations above. Since we're speaking in terms of standard deviations, and we're trying to figure out what percent of people fall within these guidelines. We know that we need to make use of Chevy Chefs inequality, which says that one minus one over a case where times 100% of people fall within casing deviation of the mean. For this particular problem, we're using K equals two. So we can compute the percent of people having the psych us as one minus 1/4 times 100% or 75% of people have these like us. Next question we want to answer is what percent of people have like us under 70 or above 1 30? We see that since in the last problem we identify people with like us between these numbers that everyone else in the population has to have like us under 70 or above 1 30. So we can simply use our results from a and say that if 100% of people have accused period and 75% of people have accused between 70 and 1 30 then it must be that 100 minus 75 equals 25% of people have accused under 70 or above 1 30. Finally the last question we want to answer is what percent of people have accused above 1 30? Again, we can rely on our answer to the previous part of this question. Remember specifically that bell curves are symmetric about their mean, Since we uh are trying to identify where people, or rather what proportion of people have an I. Q. Above 1 30. And the number of people should be equal to the number of people who have accused under 70. Given the symmetric nature, the bell care. That means the number of people with I. Q. Above 1 30 is simply half a result from B or 12.5%.

For this question, we're gonna be using our normal distribution R. Z score and then your standard normal probability table from the back of your book. So I've already drawn our bell curve. The scores on the test are normally distributed with a mean of 1:10 and a standard deviation of 25. The first question asked this, If we score 150 at what percentile is this? Well, in other words, what percent of the people do we beat? If we score 1 50 so we're gonna get our Z score, which will be our observed value minus the mean divided by the standard deviation, And that'll be 40, divided by 25, which is equal to 1.6, And then we'll take that 1.6 and look that up on our standard normal probability table in the back of the book, a Z score of 1.60 equates to a probability of .9452. So 94.52, fight Is the percentile. You would be 94.52% of the people. If you score 150. Our next question asked about the percentage of people that score between 1 25 and 1 50. So here is about 1 25. I'm gonna mark that in green. Here's our 1 25. So we're going to do the same thing we need are Z score at 1 25 25 -1 10, divided by 25, gives us a Z score of 606 And when we look up that Z score in the back of the book, 0.60. Equates to a probability of 7.7257 7257 to the left. So we want to figure out this space in here, well, we know to the left of 1 50 is 94.52 to the left of 1 25 at 72.57 So if we simply subtract those two percentages, that will leave us the percent in between the 1 25 and 1 50 marker, Which is 21.95 Between 1 25 and 150. In our last question, if we want to score in the top two Way down here on our Bell Curve, the top 2%, what score is this On our number line? Well, keep in mind if we want to score in the top 2%, we have beat 98% of the other scores. So we need to look .98 up in our standard normal probability table And get our Z score that corresponds to that. So inside of the table looks like the closest I can get his .9798, which corresponds to 2.05 as our Z score. Yeah. Now we need to figure out what score on the exam this equates to. So we're going to use our Z. Score formula Z is equal to our observed value, which we don't know minus the mean, divided by the standard deviation. So to solve this equation will do 2.05 times 25 And then add 110 and that gives us A value of 161.25. So 161.25 will put you in the top 2% of I. q scores.

The problem, we're going to be discussing Z scores and how to divide the bounce of um Z scores and the date set with the data set. So we're given that um for I. Q. Scores that the mean is 100 With a standard deviation of 15. So I've driven I drew a sort of normal distribution which is associated mean and standard deviation. It's asking if unusual I. Q. Scores are plus or to our two standard deviations away from the mean, what is the score? So for part a you know that you know if the if the if if um if it's too strange deviations away from the mean disease square must also be too right. So it's negative two and positive too because you can have um a sort of minimum I. Q. That's unusual and a maximum like you that's unusual and to find the associated um the final associated like you for each of the z scores you can just sort of visualize it, right? So the for the for the minimum bounds um you are two standard deviations below the mean, so you do 100 which is the mean of the data set -2 times the standard deviation Right? And this gives us a lower limit of 70. You repeat the same algorithm for the positive to to determine what the upper limit of an unusual of the usual I. Q. Score is. So you do 100 100 Plus two times 15 Which gives you 1:30. So this tells us if your I. Q. Is below 70 or above 1 30 then you have an unusual like you.


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