The value is 0.3055
z = ±0.44
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.
THe 75th percentile
Interpolation
It is the data value that is observed most often.It is the data value that is observed most often.It is the data value that is observed most often.It is the data value that is observed most often.
z value=0.44
z = ±0.44
Percent deviation is a measure of how much a value deviates, or differs, from a standard or expected value. It is calculated by taking the absolute difference between the measured value and the standard value, dividing by the standard value, and then multiplying by 100 to express it as a percentage.
A z score of .5 corresponds to 19% of the data between the mean and z. P( 0 < z < .5) = .19
Percent deviation is a measure of the difference between an observed or measured value and a true or accepted value, expressed as a percentage of the true value. It is calculated by dividing the absolute difference between the two values by the true value, then multiplying by 100. Percent deviation helps to quantify the accuracy or precision of measurements.
The answer is 47
It is 84.3%
It i the smallest value in the data set and corresponds to the value of the left-most end of the whisker. Unless there were outliers, in which case it will be an "X" to the left of the left-whisker.
RANGE IS the diffrence between the greatest value data and the least value of data
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.
THe 75th percentile
the difference between the highest value and the lowest value in your data