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For a given Z-score the formula calculates the value of Q, the area under the upper tail of the Normal Distribution.
Best to use values of Z > 0.

For a given probability Q the corresponding Z-score is evaluated.
Best to use Q < 0.5.

Do NOT solve for X.

All answers are approximate.

UTPN
Q = ABS( ( ( ( ( ( 1.330274429 * ( 1 ÷ ( 1 + 0.23164419 * ABS( Z ) ) ► X ) ) ) – 1.821255978 ) * X + 1.781477937 ) * X – 0.35653782 ) * X + 0.31938153 ) * X ÷ ( SQRT( 2 * π ) * EXP( SQ( Z ) ÷ 2 ) )
(11-21-2014 05:47 PM)Gerald H Wrote: [ -> ]For a given Z-score the formula calculates the value of Q, the area under the upper tail of the Normal Distribution.

This implements the well known 1955 approximation from C. Hastings' "Approximations for digital computers", resp. Abramovitz & Stegun 26.2.18. It was also used in various HP software pacs, e.g. for the 41 series.

Please let me add three remarks:
  1. Two coefficients are wrong — the correct ones are 0,2316419 and 0,356563782.
  2. This approximation has an absolute (!) error within 7,5 ·10–8.
  3. It works fine if z is not too large. At z=3 the result still has about 4 valid digits, but for z=6 only two digits remain.

So how about your own improved custom-designed approximation ?-)

Dieter
(11-21-2014 08:50 PM)Dieter Wrote: [ -> ]
(11-21-2014 05:47 PM)Gerald H Wrote: [ -> ]For a given Z-score the formula calculates the value of Q, the area under the upper tail of the Normal Distribution.

This implements the well known 1955 approximation from C. Hastings' "Approximations for digital computers", resp. Abramovitz & Stegun 26.2.18. It was also used in various HP software pacs, e.g. for the 41 series.

Please let me add three remarks:
  1. Two coefficients are wrong — the correct ones are 0,2316419 and 0,356563782.
  2. This approximation has an absolute (!) error within 7,5 ·10–8.
  3. It works fine if z is not too large. At z=3 the result still has about 4 valid digits, but for z=6 only two digits remain.

So how about your own improved custom-designed approximation ?-)

Dieter

Many thanks for the correction of the erroneous coeffficients.

The accuracy of the approximation is really surprisingly good.

Sadly my own maths abilities are inferior to the task of producing a formula attaining the same level of simplicity & accuracy.
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