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What are the z-score boundaries for the middle of 50 percent of the distribution?
50 * * * * * z = -0.67449 to z = +0.67449
What z value corresponds to 17 percent of the data between the mean and the z value?
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.
What is the z value such that 50 percent of the total area lies to the right of the curve in a normal distribution?
The Z value is 0.
What is the only value z for which 6z - 24 equals 2z plus 50?
z = 18.5 6z - 24 = 2z + 50 Take the -24 to the other side of the equals and change its sign; take the 2z to the other side of the equals and change its sign: => 6z - 2z = 50 + 24 => 4z = 74 Divide both sides by 4: => z = 18.5
What plus what times what equals to 50?
2 + 6 x 8
