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Worked example finding area under density curves

Worked example finding area under density curves.

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  • piceratops seed style avatar for user Abdelrhman Adel
    tell now i dont really understand why the area always 1 , if we consider our set of data to be (1,3,4) and another set to be (321,4,21,4,51,3,321) how the two will confirm the same area of 1 . i understand the concept of the 100% that the curve represented the entire data of my set but why the area will be always 1 ?!
    (3 votes)
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  • blobby green style avatar for user Ahmed Nasret
    could you help me get out of that paradox please? i know that 1 represents 100% in general but if we have intervals of 10 (mean 60-70 and 70-80 and 80-90) for example then the area under the relative frequency graph will be 100% of that 10 means 10 not 1.
    say:
    10(20%)+10(15)+10(65%) this will sum up an area of 10 not 1
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      If we were to make a density curve from this bar graph we would assume that the 20% that are in the 60-70 bucket are spread out evenly, so there are 2% between 60-61, 2% between 61-62, and so on...
      The same process would be done to the 70-80 and 80-90 buckets, and we would end up with the sum
      10(2%) + 10(1.5%) + 10(6.5%) = 1
      (4 votes)
  • spunky sam blue style avatar for user ved  prakash
    How do you deduce that height of A is 0.25 and of B is 0.50. It is simply not known.
    Better way would have been to divide area into rectangle & triangle and calculate it.
    (2 votes)
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    • leaf green style avatar for user James Buchanan
      The y-axis shows that three grid lines up corresponds to 0.75, which means one grid line up corresponds to 0.25. Looking at the diagonal line, we see that the leftmost point is one grid line up, and the midpoint is two grid lines up, meaning those points are at a height of 0.25 and 0.50, respectively.
      (2 votes)
  • blobby green style avatar for user Vanessa
    how do you know it was 0.25 i dont understand that
    (2 votes)
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  • blobby green style avatar for user Lol Hij
    How do you find Q1 and Q3 in this case?
    (2 votes)
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    • blobby green style avatar for user daniella
      To find Q1 and Q3 (the first and third quartiles), you would need to integrate the density curve to determine the cumulative distribution function (CDF) and then locate the points where the CDF equals 0.25 and 0.75, respectively. This involves finding the values of x corresponding to these cumulative probabilities.
      (1 vote)
  • leaf green style avatar for user Junsang
    is there a mathematical reason why the vertical lines are dotted instead of drawn in a line? (perhaps because a density 'curve' must be a function?)
    (1 vote)
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    • leaf green style avatar for user cossine
      It indicates the boundary. The dotted lines imply '1' and '3' are not part of the probability density function.

      As such it doesn`t matter in terms of finding the area as including '1' and '3' will make an infinitely small difference to the area.

      In other word including 1 and 3 has no noticeable impact on the percentage for the probability density function.
      (2 votes)
  • purple pi purple style avatar for user reeeimastegosaurus
    Right so in the first problem, when you find the area of the trapezoid, won't you be finding the area of the curve when x is more than or equal to two instead of when x is more than two?
    (1 vote)
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    • blobby green style avatar for user daniella
      The area of the trapezoid represents the total area under the density curve from x=1 to x=3, inclusive. Since the problem asks for the percent of the area where x is more than two, the entire area of the trapezoid is considered, including the portion where x equals two.
      (1 vote)
  • aqualine ultimate style avatar for user Rohan Suri
    Suppose we have the data : 10,10.5,11,17,19 and we map it into two intervals 10-15 and 15-20. So, the interval 10-15 will contain 3/5 data points and the interval 15-20 will contain 2/5 data points.
    Now if we calculate the area (as sum of areas of two rectangles), it comes out to be:
    area=3/5*5 + 2/5*5 =5 but not 1.
    Can anyone explain this to me?
    Thanks.
    (1 vote)
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    • blobby green style avatar for user daniella
      In your example, if you have mapped the data into two intervals and calculated the area as the sum of the areas of two rectangles, it's important to note that the total area under a density curve should indeed be 1. However, your calculation seems to be based on rectangular approximations rather than directly integrating the density curve. The discrepancy in the area calculation might arise from this approximation method. To accurately represent the density curve, you would need to integrate the density function within each interval and then sum up the areas.
      (1 vote)
  • blobby green style avatar for user Austin In
    Does anyone know how you could find the area of different percentiles if the hight was not given?

    I'm working on a problem where the density curve is a perfect triangle. The base is given (1.6) and I understand that the total area is equal to 1.0. Using A=1/2(b)(h) you can find H.

    But now I'm stuck in finding the percentages of selected areas within the triangle such as: What is the percentage of values that are below 0.4?

    I see that a new triangle is formed and if you can find that area you can subtract from one and find your answer, but in this case, we have two variables and I'm stuck.

    Please help if anyone is out there!
    (1 vote)
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  • starky ultimate style avatar for user Nad Alaba
    In both examples we could work out the answer without knowing the numbers on the y-axis... It's obvious in the second example, and in the first one we could say the height at the point "3" is x so the height at the point "1" is x/3 and the area under the entire density curve would be 2.(x+x/3)/2 which equals "1" and so we solve the equation for x and it would be 3/4 or 0.75 >>> and then we calculate the area under the smaller trapezoid
    (1 vote)
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Video transcript

- [Instructor] Consider the density curve below and this density curve doesn't look like the ones we typically see that are a little bit curvier, but this is a little easier for us to work with and figure out areas. So they ask us to find the percent of the area under the density curve where x is more than two. So what area represents when x is more than two? This is when x is equal to two. So they're talking about this area right over here. And so we need to figure out the percent of the total area under the curve that this blue area actually represents. So first let's find the total area under the density curve. The density only has area, the density curve only has area from x equals one to x equals three. So it does amount to finding the area of this larger trapezoid. Let me highlight this trapezoid in red. So we want to find the area of this trapezoid right over here. And then that should be equal to one because all density curves have an area of one under the total curve. So let's first verify that. There's a couple of ways to think about it. We could split it up into two shapes or you could just use the formula for an area of a trapezoid. In fact let's use the formula for an area of a trapezoid. The formula for an area of a trapezoid is you would take the average of, you would take the average of this length. Let me do that in another color. This length and this length and then multiply that times the base. So the average of this length and this length, let's see this is 0.25. 0.25 plus this height, 0.75, divided by two. So that's the average of those two sides times the base, times this right over here which is two. And so this is going to give us as it should have, 0.25 plus 0.75 which is equal to one. So the area under the entire density curve is one which we need to be true for this to be a density curve. Now let's think about what percentage of that area is represented in blue right over here. Well we could do the same thing. We could say all right, this is a trapezoid. We want to take the average of this side and this side and multiply it times the base. So this side is 0.5 high. 0.5 plus 0.75, 0.75 high, and we're going to take the average of that divided by two times the base. Well the base going from two to three is only equal to, is equal to one. So times one. And so this is going to give us 1.25, 1.25 over two. And what is that going to be equal to? Well that would be the same thing as zero point what? Let's see 0.625. Did I do that right, Yep. If I multiply two times this, I would get 1.25. So the percent of the area under the density curve where x is more than two, this is the decimal expression of it. If we wanted to write it as a percent, it would Be 62.5% Let's do another example. Consider the density curve below. All right we have another one of these somewhat angular density curves. Find the percent of the area under the density curve where x is more than three. So we're talking about, let's see, this is where x is equal to three, x is more than three, we're talking about this entire area right over here. Well this is actually somewhat straightforward because if we're saying the area where x is more than three, that's the entire area under the density curve. And the entire area under any density curve needs to be equal to one. Or you could say find the percent of the area under the density curve. Well the whole density curve is where x is more than three. So 100%, we don't even have to go through the trouble of trying to directly calculate the area.