z = ±0.44
1.555 With 88% confidence, there is 6% (0.06) in either tail of the standard Normal distribution. Table C will not help here. Using Table A the correct z* is about halfway between 1.55 and 1.56. According to technology, z*=1.555
The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%
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.
z = 1.28, approx.
The value is 0.3055
1.555 With 88% confidence, there is 6% (0.06) in either tail of the standard Normal distribution. Table C will not help here. Using Table A the correct z* is about halfway between 1.55 and 1.56. According to technology, z*=1.555
The critical value of z for a 96 percent confidence interval is approximately 2.05. This value corresponds to the point where the area in each tail of the standard normal distribution is 2 percent, leaving 96 percent in the center. It is typically found using z-tables or statistical software.
A 90 percent on a test typically corresponds to an "A-" grade, assuming standard grading scales.
The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%
To find the z-score where 8% of the distribution's area lies between -z and z, we first recognize that this means 4% (or 0.04) lies in each tail of the normal distribution. Therefore, we need to find the z-score that corresponds to the cumulative area of 0.04 in the left tail. Using standard normal distribution tables or a calculator, we find that the z-score for 0.04 is approximately -1.75. Thus, the positive z-score is approximately 1.75, meaning z ≈ 1.75.
It is 68.3%
95% is within 2 standard deviations of the mean.
z = 0.8416
0.13
It depends on the shape of the distribution. For standard normal distribution, a two tailed range would be from -1.15 sd to + 1.15 sd.
95 percent of measurements are less than 2 standard deviations away from the mean, assuming a normal distribution.
z = 1.28, approx.