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The normal distribution problem

I don't understand why I can't simply add 1.5 standard deviations to get the answer.

If 1 standard deviation is 10kg and the mean is 400kg, then 415kg is 1.5 standard deviations.

So I calculated it like this: .3413 + ((.4772-.3413)/2) = 0.40925

This equation takes one half of the difference between two standard deviations and one standard deviation, then adds it to the first standard deviation.

Why does this not work? Why do I have to use the table provided?

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    $\begingroup$ Where did you get the numbers 0.3413, 0.4772 and 0.3413? In any case, the reason you can't just add 1.5 SD is because they asked you for the area under the curve, for which you have to use the normal table that gives you such areas. $\endgroup$ – mb7744 Dec 7 '15 at 14:03
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    $\begingroup$ @Ben The problem states that the standard deviation is 10. Unless you mean that you computed some value which corresponds to 1 standard deviation, and computed another value that corresponds to 2 standard deviations. But if so, that's what mb7744 was asking. $\endgroup$ – Jelsema Dec 7 '15 at 14:13
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    $\begingroup$ @jelsema yes the right side of 1 standard deviation comprises of 34.13% of results. The second standard deviation comprises of 47.72% of results. Because its only 1.5 standard deviations we take the difference between the first standard deviation and the second standard deviation and get half of it since its only 1.5 standard deviations. $\endgroup$ – Ben Dec 7 '15 at 14:29
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    $\begingroup$ @Ben, you are linearly interpolating. That is invalid, by which I mean it is not exactly correct, because the area under the curve (say from the mean) is not a linear function of the number of standard deviations; if it were, then the 2 standard deviation figure of 0.4772 would be double the 1 standard deviation figure of 0.3413, which it is not. $\endgroup$ – Mark L. Stone Dec 7 '15 at 14:42
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    $\begingroup$ I want to know 5 squared. So why can't I just take 10 squared (which is 100) and halve it? $\endgroup$ – user253751 Dec 8 '15 at 1:38
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The reason that we cannot (linearly) interpolate between 0.3413 and 0.4772 is because the pdf of the Normal distribution is not uniform (flat at a single value).

Consider this more simple example, where we can use geometry to find the areas.

enter image description here

The total area of the plot is 1 (it's a square cut diagonally, with the two pieces rearranged to be a triangle). Using Base*Height/2 we can find that the area of region A is 0.5, and the total area of regions B and C is also 0.5.

But the areas of B and C are not equal. The area of region C is 0.5*0.5/2 = 0.125, and therefore the area of region B is 0.375. So even though regions B and C are equally wide along the x-axis, since the height is not constant, they have different areas.

The Normal distribution that you are dealing with in your exercise is similar, but with a more complicated function for the height instead of a simple triangle. Because of this, the area between two values can't be solved as simply - hence the use of Z-scores and a table to find probabilities.

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    $\begingroup$ Can I use linear interpolation to guess the multiple choice answer? Since they are so far apart? Is that a bad idea? $\endgroup$ – Ben Dec 7 '15 at 23:07
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    $\begingroup$ @Ben That's generally a great idea. Multiple choice tests are built for this kind of manipulation. Being able to estimate the size of an answer is a very useful ability throughout life. Still it's also good that you ask about and understand this problem because as you go on there will be more and more questions like this that have no formula for an answer. $\endgroup$ – rhaskett Dec 7 '15 at 23:19
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    $\begingroup$ @rhaskett though the presence of the "none of the above" answer kind of spoils that strategy. $\endgroup$ – David Z Dec 8 '15 at 6:57
  • $\begingroup$ @Ben Once it is test time, of course, use whatever method you can to get the right answer. However, one problem with multiple choice tests is they allow guess and check strategies to circumvent the necessity to actually solve for a value. In the long run, unless this is your terminal math class, the reliance on that strategy will hurt you more than it helps. $\endgroup$ – Dean MacGregor Dec 8 '15 at 15:53
  • $\begingroup$ @DavidZ True. I missed that option. $\endgroup$ – rhaskett Dec 8 '15 at 18:21
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Just to provide a different illustration on the same topic...

In your initial calculation you would be treating the normal curve as a uniform distribution, in which case your initial approach would be the correct mathematical calculation for the double hatched rectangle in the plot below (with different actual values), simply because you'd be able to express the area as a simple linear dependency of the $x$ axis distance:

$A_{1.5\,SD} =\large\frac{A_{2\,SD} - A_{1\,SD}}{2} = \small height * \large\frac{X_{2\,SD} - X_{1\,SD}}{2} $

But you want to calculate the diagonally hatched area under the curve of the Gaussian distribution, which as stated before wouldn't keep a linear relationship with the distance along the $x$ axis even if the distribution was triangular:

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The formula for the Gaussian distribution is:

gauss equation

Where sigma = std deviation and mu = mean

(stolen from wikipedia)

When you are asking for the area, you are integrating this function over the range specified. This integral does not have a "closed form" solution: there is no way to come up with an expression using "normal" math functions like factorial, multiplication, exponentiation, roots, etc. that equals that integral.

It's just like logarithms or trigonometric functions: you can't produce a closed form equation for them using other algebraic functions (you can use infinite series, but that's not "closed"). So you use a table (if you are feeling retro, or a calculator, which simply uses a table for you behind the scenes embedded in its processor as a starting point) when you need to actually calculate it.

In fact, the parallel with logarithms is quite apt: one can also define a logarithm by an integral, namely ln(x) = integral of (1/x) from 0 to x.

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Geometrically, .4772 - .3413, represents the area under the graph between 1 standard deviation and 2 standard deviations. If you split this region half way across horizontally, the part to the left of the split will be the area between 1 and 1.5 standard deviations, as you want. Fine so far.

However when you take (.4772 - .3413) / 2 you're getting half the area, but not necessarily what you were looking for, which is however much area was half way across horizontally. With this graph, that left part of the split isn't half of the area - the line is sloping downward (going from the top left to the bottom right) so there's more space in the left part than the right part. If this graph was a straight horizontal line, then the area you were splitting would be a rectangle, and half the area really would be half way across.

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