Q: What z value corresponds to an area of 0.9881 being to left of z?

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sigma represents standard deviation. In a normal distribution, +/- 1 sigma from the mean, for instance, corresponds to approximately 67% of the area under the normal distribution. +/- 2 sigma corrresponds to 95% of the area and +/- 3 sigma from the mean corresponds to 99% of the area under a normal distribution. The area that is covered under +/- six sigma from the mean corresponds to nearly 100% -- that is, the part of the area NOT under that +/- 6 sigma is in the 10^-15 range or 1/1,000,000,000,000,000. The six sigma name borrows from this to suggest that the method gives this degree of certainty: that in 999,999,999,999,999 in 1,000,000,000,000,000 cases the result will be predictable. It does nothing, however, to explain how an agricultural process management methodology applies to other fields.

A rectangle has no value - experimental or otherwise. Its area has a value, its perimeter, its aspect have values.

A triangle does not have any particular value. There are the magnitudes of its size, its angles, its area, its perimeter etc, but no single VALUE.

It is 100*(Measured Value/True Value - 1)

Most shapes have different perimeter than area, as far as value.

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I believe your question is to find a range going from the mean to a z-value on the standard normal distribution that corresponds to 17% of the area. A normal distribution goes from values of minus infinity to positive infinity. A standard normal distribution has a mean of 0 and an standard deviation of 1. It is usually best if you draw a diagram, in this case a bell shape curve with mean = 0. The area to the left of the mean is 50% of the total area. We find a z value that corresponds to 67% (50% + 17%) of the area to the left of this value. This can be done either with a lookup table or a spreadsheet program. I prefer excel, +norminv(0.67) = 0.44. The problem could also be worded to find the area going from a z-value to the mean. In this case, we must find a z-value that corrsponds to 33% (50-17). Using Excel, I calculate +norminv(0.33) = -0.44.

2.82

There cannot be such a value since the total area, being a probability, is 1.

There are 300+ Zip Codes in area code 202, which is Washington, D.C.

3.06

At Z = 0.25, the area (larger) to the left of the line is 0.5987. The smaller area to the right of the line is 1 - 0.5987 = 0.4013. Therefore, the smaller section corresponds to 40.13% of the area.

It is three units out of every eight. Its value will depend on what is being measured: area, length, mass, etc and the units being used.

Ancient Thrace corresponds to an area that today would cover Southern Bulgaria, North Eastern Greece and European Turkey.

Counterpart-A person or thing holding a position or performing a function that corresponds to that of another person or thing in a different area.

"Alta California". Which nowadays corresponds to the U.S. state of California.

I will try to answer your question, although I am unclear about "where most competitors operate." The area under the normal curve from beginning to end (minus infinity to positive infinity) equals 1. The area from minus infinity to the median (also the mean) is always 0.5, which corresponds to a 50% probabability. If I want to know the probability of a value of x or less occurring, this would the area from negative infinity to x.

sigma represents standard deviation. In a normal distribution, +/- 1 sigma from the mean, for instance, corresponds to approximately 67% of the area under the normal distribution. +/- 2 sigma corrresponds to 95% of the area and +/- 3 sigma from the mean corresponds to 99% of the area under a normal distribution. The area that is covered under +/- six sigma from the mean corresponds to nearly 100% -- that is, the part of the area NOT under that +/- 6 sigma is in the 10^-15 range or 1/1,000,000,000,000,000. The six sigma name borrows from this to suggest that the method gives this degree of certainty: that in 999,999,999,999,999 in 1,000,000,000,000,000 cases the result will be predictable. It does nothing, however, to explain how an agricultural process management methodology applies to other fields.