Example: Cubic Splines with Known Data Points
The area below a given ordinate of the normal curve does not vary linearly. Because a table describing the normal curve states the area at fixed increments, you can use the Cubic Splines program to interpolate the area accurately.
Example
A table of areas for the normal curve is listed blow. Find the area for z = 0.8.
| z | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 | 1.1 |
|---|---|---|---|---|---|---|
| P(z) | 0.725747 | 0.0.758036 | 0.815940 | 0.841345 | 0.864334 | |
| P″(z) | -0.1999347 | -0.2396373 |
Procedure |
Press |
Display |
|---|---|---|
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| Select the program | [ RUN ] { MTH } { INT } { CUB } { NEW } |
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| Enter the number of points | 5 { m } { EOD } |
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| Enter the first point | .6 [ x~t ] .725747 { ENT } |
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| Enter the second point | .7 [ x~t ] .758036 { ENT } |
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| Enter the third point | .9 [ x~t ] .815940 { ENT } |
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| Enter the fourth point | 1 [ x~t ] .841345 { ENT } |
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| Enter the fifth point | 1.1 [ x~t ] .864334 { ENT } |
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| Proceed with program | { EOD } |  |
| Emter second derivative at the endpoints | .1999347 [ +/- ] { g1 } .2396373 [ +/- ] { gm } { EOD } |
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| View the derivatives | { YES } |  |
| { NXT } |  |
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| { NXT } |  |
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| { NXT } |  |
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| { NXT } |  |
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| { NXT } |  |
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| Proceed with program | { NO } |  |
| Proceed with program | { YES }.8 { x } |
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The mathematics of the normal curve calculate the actual value as 0.78814467. Compared to this, the result of 0.7881405682 has an error of 0.00052%.
The most commonly used interpolation is linear. By linear interpolation ((.8 -.7)/(.9-.7)x(.815940-.758036)+.758036), you get a less accurate answer (.786988). Compare to the actual value, this result has an error of 0.147%.
The error is calculated from the equation %error = (( observed value / actual value ) - 1) x 100%
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