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Q2) The following table gives the Examination distribution of marks sccured by some students in & certain Marks 5-10 Number of Students 10-I5 45-20 20-25 25-30 ...

Question

Q2) The following table gives the Examination distribution of marks sccured by some students in & certain Marks 5-10 Number of Students 10-I5 45-20 20-25 25-30 30-35 35.40 40-45Find the_Third Quartile and Decile and Percentile marks of students in the examination.

Q2) The following table gives the Examination distribution of marks sccured by some students in & certain Marks 5-10 Number of Students 10-I5 45-20 20-25 25-30 30-35 35.40 40-45 Find the_Third Quartile and Decile and Percentile marks of students in the examination.



Answers

The frequency distribution of the marks obtained by 28 students in a test carrying 40 marks is given below.
$$ \begin{array}{|l|c|c|c|c|} \hline \text { Marks } & 0-10 & 10-20 & 20-30 & 30-40 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Students } \end{array} & 6 & \mathrm{x} & \mathrm{y} & 6 \\ \hline \end{array} $$
If the mean of the above data is 20 , then find the difference between $\mathrm{x}$ and $\mathrm{y}$. (1) 3 (2) 2 (3) 1 (4) 0

So a teacher has predicted, Ah, great distribution for the final exam and it was predicted that 25% will be aces, 30% will be beast, 35% will be seized and 10% will be DS. Then the exam is actually given and the great distribution is posted. And it turns out that seven people earned an A seven people earned a B five people earned sees and one person earned a d. So those would be classified as our observed frequencies. And we want to calculate the Chi Square test statistic on this so that we could run a goodness of fit test. And if we're going to run a chi square goodness of fit test and find that test statistic, we're going to need some additional information. The formula to find a test statistic is the sum of the quantity of observed minus expected squares over expected. So we now have our observed. The 775 and one are observed values. So we need to generate our expected frequencies. And if we were to add up our observed frequencies, they're going to add up to 20 students taking this test. So we expected 25% of those 20 students or we expected five A's. We would expect 30% of those 20 students to get the bees, and that's going to give us six students. We would expect 35% to get C's, and that generates seven students. And then we would expect 10% of the 20 students to get the DS, and that would be to students. Now, technically, we could not run a goodness of fit test on this because are expected, values must be greater than five. And our last category just turned out to be a two. But since we're not truly running an actual goodness of fit test on this, we're going to keep going. So we have our observed and we have are expected. So now what we need to do is we need to take each of those observed, subtract the expected, then square that value and then divided by the expected. So the first one is going to be seven, which is observed minus five. We're going to square that, and then we're going to divide that by five, and when we do that, we end up with 4/5 or 0.8, We're gonna do it for the second ones. We're gonna have seven minus six squared, divided by six. And we're going to end up with 16 and then we'll have five minus seven squared, divided by seven, which is 4/7. And I'm leaving these infraction just for the sake. They are decimals that don't clean up nice and neat. And then for the last one, we're going to have one minus two squared, divided by two, which would end up being one half. Okay, so can you see where all those values came from? And the test statistic is what we get when we add up that column. So now we wanna add up this whole column here. And when we do that, we get an answer of approximately 2.0 three 81 So if we were able to run a goodness of fit test, that is how we would find our chi squared test statistic. So this would be the answer to this problem

Hello. So in discussion we are given a graph, cumulative frequency graph and we need to find out the number of students who scored less than or equal to 50% marks. In this graph there are two Cubs, one girl represent the greater than community frequency and the other one represents the less than community frequency. So by applying less than community frequency concept we have learned. We will look at this girl lesson, Let them come into the frequency of these are the person days scored by the students on the X axis and on the Y axis are the number of students. So the number of students who scored less than or equal to 50% marks will be 50. It's clear from the ground and that's the answer. Thank you.

Right in this problem, you have a teacher making predictions on the distribution of grades, so that could be classified as you're expected, proportions or values. And the great distribution could be an A, a B, a C or a D. And the teacher is expecting 25% or 250.251 quarter of the grades to be ace, 30% to be bees, 35% to be seized and 10% to be DS, the actual distribution for a class of 20 people. So this is going to be the observed values This particular teacher saw that she had seven days. There were seven bees, there were five sees and there was only one D. And what we're trying to do in this particular scenario is we're trying to see how wealthy observed grades modeled what was expected, or how well the observed grades fit what was expected. So we would be running a goodness of fit test. And in order to run that goodness of fit test, you have to create two different hypotheses. The first one is called your null hypothesis, and we're going to use H sub zero to represent null hypothesis and you're no hypothesis is going to say that the expected grade distribution is 25% A's 30% bees, 35% sees and 10% DS. And then the other hypothesis you have to generate is called your alternative hypothesis. And we're going to use H sub A to represent the alternative hypothesis and the alternative to the distributions being 25% a 30 bees, 35% season, 10% D's would be that the distribution of grades differs from the expected distribution. And just to make it clear, all it takes is at least one has to be different. So that is going to represent our two hypotheses, the null hypothesis and the alternative hypothesis for this goodness of fit test.

So we start with a teachers prediction on grade distribution for a final exam, and the teacher is going to expect 25% of the final exam grades to be AIDS and 30% to be bees, 35% to be seized and 10% of the grades to be DS. The teacher then gives the test and discovers that seven or eight days seven or bees five our seas and one is a day, and we want to know how well the expected compares to what she actually observed. So we're going to end up running a chi square goodness of fit test to see how well the actual frequencies fit to the expected frequencies. Now the value 775 and one would be referred to as Theobald served frequencies. So in order to run the goodness of fit test, we're going to have to find a Chi Square test statistic which calls for a formula to add up or the sum of observed minus expected quantity squared, divided by expected. So because of that formula, we're now going to need an additional column on our chart up here, and we're going to call that the expected frequencies. And if this teacher expected to 25% of the students and there were 20 students, then he or she would expect to have five A's. And this teacher expected 30% off the 20 students to be bees. So we would expect six bees, 35% of the 20 students. We would expect seven seas and 10% of the 20 students we would expect to DS Now, Technically, we can't run a goodness of fit test on this data. And the reason being is, when you run a goodness of fit test, all the expected values should be greater than or equal to five. And since we came up with an expected value of two, technically, our data, we don't have a large enough sample size to run an effective goodness of fit test. But we're going to do so anyway, just for the practice of running a goodness of fit test. So we're now going to create a column called Oh minus e Quantity squared, divided by E. So that means we're going to take the observed AIDS, which was seven. We're going to subtract the expected A's, which was five. We're going to square that quantity and divide by five, which is that expected value. And in doing so, we get a fraction of 4/5. Then we're gonna do that for the B. We observed seven, We expected six. We're gonna square that quantity and divide by the expected and you'll get 1/6. And then we observed five sees we expected seven seas. So we get 4/7. And then finally we observed one d we expected to, and we'll get one. And then when we add up that column, we get what is called our Chi Square test statistic on our chi square. Test statistic for this data is 2.3 81 We're then going to calculate a P value, and the P value is going to be equal to the probability that our Chi square is greater than that test statistic we just found. And a picture oftentimes helps with this. So the chi Square family of grass, for the most part, is a skewed right graph. And for this particular case, our degrees of freedom of this graph was three and we find degrees of freedom when it comes time to working with Chi squares by taking the number of categories and subtracting one. And since this teacher broke her grades up into A, B, C and D, we had four categories. So that means the degrees of freedom for this chi square test is going to be three. And we know that the average of a chi square distribution is equivalent to the degrees of freedom. So the average of this is going to be three, and you always find the average slightly to the right of the peak. So we're gonna have a three right here and what we're doing when we're running this test is we're calculating the P value and the P value would be what is the probability that the chi square value is greater than two point 0381 So we're going to have to use a feature in the calculator in order to generate that area to the right of 2.381 for a chi square distribution with a degree of freedom of three. So what we're going to use is we're gonna use archives square C D f or cumulative density function in your graphing calculator to calculate this and when you do so it asks you for the lower boundary, the upper boundary and then the degrees of freedom. So in our example, the lower boundary is going to be 2.381 Thea upper boundary is going to be just a super super large number. We're going to just say 10 to the 99th Power and our degrees of freedom here was three. So we're going to bring in my graphing calculator to show you or to calculate that p value. So we're going to hit the second button. We're going to hit the Bears button or the distributions and we're going to select in my calculator. It's number eight. We're gonna type in that lower boundary 2.381 We're gonna follow it up with a comma. We're then going to type in a super large number 10 to the 99th Power, and then we're gonna follow that up with the degrees of freedom, and we're going to get a P value of 0.5645 So what does that refer to in our chart that refers to the area where the proportion of the curve that is shaded that is greater than a 2.381 Now, we could have gotten that another way using your graphing calculator, which also gives us the actual graph or images. Well, so I'm gonna bring my calculator in one more time. Let me be clear what we had there. And we're going to hit that stat button, and we're gonna select tests, and we're going to do a chi Square goodness of fit test, which in my calculator is Letter D. And in doing so, we're going to have to tell it the degrees of freedom, which in this case, was three. And you could see we have our, um, total in there. The 2.381 and try again tests goodness of fit test. And now you can see the P value is 0.5645 and you can see what the picture looks like. Okay. And the picture on the calculator kind of mirrors the picture that I have drawn on the white board is Well, okay, so from there, I'm gonna get rid of the graphing calculator at this point, and we're ready to draw a conclusion. So we would have to write a hypothesis and are no hypothesis is going to be stating that the expected distribution of grades on this final exam are as what is stated, which was 25% A's 30% bees, 35% sees and 10% DS three. Alternative hypothesis is going to be that the distribution of grades on the final exam differs from what is stated, and all it has to do is differ by one. So it might be that the A's are 25% but the bees might be 25% as well, so as long as one of the categories differs than would fall into that category. So when you're running a chi square test to make your decision, you are going to do a comparison, and you are going to compare the level of significance to the P value. And if your level of significance is greater than your P value than the decision has to be to reject the null hypothesis. So in our case, we're going to be running our test at a 5% significance level. So we're saying 0.5 is greater than the P value and R P value is 0.5645 and that is false. That's not a true statement. So because that's not true, we're not going to reject the null hypothesis. So our decision is going to be to fail to reject the null hypothesis. And if we fail to reject the null hypothesis, then we have to draw this conclusion. And the conclusion is going to be that there is not enough evidence at the 5% significance level to state that the distribution is different from the stated proportions. 25% A's 30% bees, 35% sees and 10% titties. So there's not enough evidence to say it's not that now. We're not saying it is that we're saying there's just not enough evidence to dispute that it isn't that so that's how we run a chi square goodness of fit test


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