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Three classes took test, with histograms of the scores shown below;OfetlcentscsnudeniorStudentyClassClassClzss(a) Which class(es) had the highest mean score? Both c...

Question

Three classes took test, with histograms of the scores shown below;OfetlcentscsnudeniorStudentyClassClassClzss(a) Which class(es) had the highest mean score? Both classes 2 and Class Both classes and 2 ClassClass(b) Which class(es) had the highest median score? Class Both classes and 2 ClassClassBoth classes and 3(c)(i) For which classes are the mean and median similar? All 3 classes have similar means and medians Classes andClasses andClasses and(ii) In the class for which the mean and median d

Three classes took test, with histograms of the scores shown below; Ofetlcents csnudeni orStudenty Class Class Clzss (a) Which class(es) had the highest mean score? Both classes 2 and Class Both classes and 2 Class Class (b) Which class(es) had the highest median score? Class Both classes and 2 Class Class Both classes and 3 (c)(i) For which classes are the mean and median similar? All 3 classes have similar means and medians Classes and Classes and Classes and (ii) In the class for which the mean and median differ the most, which is higher? The mean is higher than the median_ since the scores are left skewed The median is higher than the mean_ since the scores are left skewed: The median is higher than the mean, since the scores are right skewed The mean is higher than the median, since the scores are right skewed (d) Which class(es) had the smallest standard deviation? Class Both classes and 2 Class Class Both classes 2 and 3 (e) Which class(es) had the smallest IQR? Both classes and 3 Class Class Both classes 2 and Class



Answers

When data are summarized in a frequency distribution, the median can be found by first identifying the median class, which is the class that contains the median. We then assume that the values in that class are evenly distributed and we interpolate. Letting n denote the sum of all class frequencies, and letting m denote the sum of the class frequencies that precede the median class, the median can be estimated as shown here: ( lower limit of median class ) + ( class width ) $$(n+12)-(m+1)$$ frequency of median class ) Use this procedure to find the median of the frequency distribution given in Exercise 29. How does the result compare to the median found from the original list of data, which is 33.0 years? Which value of the median is better: the value computed using the frequency table or the value of 33.0 years?

So the question here is about um talking about data analyses more specifically. Um We're looking at statistical studies so we have statistical um studies here. So we're given the data plot representing the final examination grades for an elementary statistics course here. So for part A here it wants us to conduct a stem and leaf plot. So I'm gonna do a sample here and we'll let you do the rest. But a stem and leaf plot is essentially as follows. So this part is going to be the stem and that is gonna be the leaf. Whereas for example if we have a value of let's say 12 Essentially what you put is that for example in our value we have the smallest number here is going to be 10. So we add a zero here. The second smallest is going to be five and then seven. So we read it like such so that's gonna be 10 Um 15 and 17. So for the 20s we can do let's say a couple of numbers, we can do let's say three. Um we can say five and that's gonna be read as 23 25. And that's essentially how you do it, so you just go on with it so 45 and so on. For be here you want to construct construct a relative frequency um graham hist a gram here, so essentially you can have for example a value within the tens to twenties and then you can have like for example this valid being five. So there by doing something like such, you can say that between tens and twenties there's gonna be five different values that lie between those and then between 20 to 30 there could be four, this is just an example and you can just do this for the own data that you collect or that's actually based on this question here and that's essentially what you do for that part. Um For c here it wants us to calculate the mean, median and stem um standard deviation. So the mean is essentially just the average value here. Do you take every single value divided by the number of data sets or data? Um the number of data points rather. So in this case the mean is going to be a value of 65.67. Um the median is essentially just going to be the middle number once you've ordered the data set and this is going to be 71.5 here, as this is going to be an even number set. And then for the standard deviation we can see that this is essentially just going to be how much the data kind of ranges Um just based on a set of calculations in this particular case and what we get from our calculators, that this is going to have a value of 21.9 and that's the answer to a question here.

Problem. We are given a new formula for the median of a grouped data set on and were asked to use the data from the foreign born population for each state from number 16. So here's back data on and then this formula is ah lot. So here's what each, um, letter represents. So and we have our number and are frequent seats. So remember our frequency. All we need to do is add everything. I mean, we're talking about the number of foreign born people for each state. So when we out all of this 26 plus 11 plus four plus five plus two plus one plus one on, we should get 50. So here, we know are in is 15 then CF. Here is the cumulative frequency of the class immediately preceding the median class. So we need to figure out what the median class is. Um, and we do that by, um, using 50 and dividing it by two. So a median class will be 50 divided by two, which will give us 25. So then whichever frequency has this number included on, then that will be our class. So our median class is this 1st 1.8 to 4.4. Um and so are cumulative frequency of the class immediately preceding the median cross. Well, we don't have a class before 0.8 to 4.4, so this value is going to be zero on. Then we want the width of our median class. So all we have to do is 4.4, minus 2.8, and we will get 3.6, and then our frequency of the median class is right here. We know that is 26 and then we want the lower bound for our median class, and that is 0.8. So once we recognized each of these values, all we need to do is play it into our formula. So here we're gonna have that the median MD equals. We know that our in is 50. So offifty, divided by two minus zero, divided by R value of F, which is 26. And then we're gonna multiply this by our with of 3.6 and we're subtracting or sorry. We're adding, are lower bound of zero point eight. And when we playing all of those values in, we will get a median for a groups data of 4.26

So we're looking at the data and we can see that the data goes from 27 all the way up to 34. And if we look at the history, Graham Instagram, uh, looks to be, you know, pretty bell shape. So we would think that this is approximately a normal distribution. Then part B were to look at the normal Quanta will plot and remember, for a normal quantum plot, you're probably not ever asked to make one long hand. I never had my students do that. But you are looking at those values and you're hoping that you're seeing some linearity and so approximately linear. Therefore, the tendency is for the distribution to be approximately normal, not perfect, but not too bad. Then we want to look at on part, see what the inter quartile ranges and, uh, we need We have 50 numbers, so if you have them all listed down, um, if you put them all on your calculator, your calculator will give it all to you with 11 should put one variable stat. But if you have 50 numbers and then you need to count to find the 25th number 26 number, and that number will be your median and that number ends up coming out to be 30. And then we have 25 numbers who are below and so 25 numbers below. If I take 25 divide it by two, I get 12.5. So there are 12 numbers here, 12 numbers there. We need the 13th number in the list counting this way, and we also will need the 13th number counting this way, this one will be our Q one and Q one comes out to be 29 and therefore Q three counting comes out to be 31. And so our inner core tell range is the difference between Q three and Q one is two. And so, in order to find where those whether we have outliers or not, we want to take the Q one, and we want to subtract away 1.5 boxes. So 29 minus three anything below 26 is going to end up being an outlier, and there are none. No low outliers, and then we need to take Q three and add on 1.5 boxes, box wits or recurs. So 31 plus three is 34 anything higher than 34 is going to be an outlier. And our highest number is 34. So there are none. There are no outliers. Yeah, we don't count those that are right at that limit. And then we need to find the Pearson uh, index. And there are two methods for finding that one is the ski Eunice, where we take the mean and subtract away the mode and then divided by the standard deviation. And I have that the, uh I mean, when I calculated was this my mode most frequently occurring number was 30 and then dividing it by the standard deviation which I had had that day to put in my calculator. And I get this as the scariness. So it is positive it's pretty close to zero. So if it's skewed, it's just a little bit skewed to the right. And if we use the second index formula, that's three times we take the mean minus the median divided by the standard deviation. And in that case, it ends up that the median and the mode are the same. So I just finished writing here. So basically three times that quantity that we had up here and this Kunis number would be 30.265 So again, if there is Kunitz, it's skewed just a little bit to the right. And so it appears that this distribution is pretty close to being a normal distribution. Not too bad. You can tell that really from the 22 earlier plots that there's not too much skin this.

Okay, So in this problem, we want to try a different approach to computing the media. So it won't be exactly the median, but we want to try if we can get good approximation of the median using a frequency table. So here we have the frequency table off the time, the waiting time in seconds at a fast food restaurant. And we need to identify what's the median class first? So what is the median class? Well, it's the class that contains the media, so we have. In total, we have 50 50 different waiting times. So 11 plus 24 is 35 plus 10 is 45 3 48 plus 2 50. So we have 40 50 different observations and where is the medium? Well, the median would be in the class with the 25th observation. So we have 11 observation in the first class and 24 in the second class. That mean in total we have 35 of observation in that in the first two class class is therefore the the media is in the second class. So the media, the medium classes that class 125 2nd 274 seconds. Therefore, the lower limit. The lower limit of medium class is 125. So we have the lower land. Was the class with? Well, it's the number of of information in the class, the number of the range covered by the class. So in that case, we cover 50 seconds. So the class with is 50. We have fish and what is N and is the total frequency. So ah, and is 50 because we have 50 different observation. Mm. Is the number of frequencies, uh, in the class, the some of the frequencies before off the classes that precede the median class. So the medium, the medium class is 125 174. There, for the only class that precedes it is 75 to 124 which has 11 entries, 11 values. So and the sum of 11 added to nothing is 11. So we have em and, um, the frequency of the medium class. So frequency of eso we have basically and and m and the frequency of medium classes than them is what what you would find in the table. So it's, um, 24. So we have 24 values in the 125 and 174 class, so we can put everything in this formula we have here. So 124 101 125 plus 50 times 50 plus one, divided by two minus 11 plus one, divided by 24. And let me zoom out a little bit to give myself room to breathe. Yes, that seems a bit better, Which gives us a median of 153 0.125 eso That's the media using this formula based on the frequency table. And if, um if you use the data set in Excel and you want to find the true median, uh, so you simply use this command. So the data, the full data set is available on the book website, so it's a two to a 51 and that is 150 0.2 for all Together. Those two there there's, like 2.5 seconds 2 to 3 seconds difference in the two median, which means they're basically so that the formula we used with the frequency table is accurate is accurate enough. Although you would never use this method if you had access toe all the data. Like if you have data set 25 available to you you for any purpose, you might need the median. You will always compute the rial media Theis formula that we use here is used for only if you only have the frequency table. It's the only data available to you then that that formulas Good enough. But of course, if you have the raw data, the full the full data set, you will compute the rial media. Yeah.


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