Area to the left of z = -1.72 = area to the right of z = 1.72
That is ALL the "working" that you will be able to show - unless you are into some serious high level mathematics. Most school teachers and many university lecturers will not be able to integrate the standard normal distribution: they will look it up in tables. (I have an MSc in Mathematical Statistics and I could do it but not without difficulty).
Pr(z < -1.72) = 0.042716
0.088508
One standard deviation
The Z value is negative, but area is always positive.
It is 0.877
The area under the normal distribution curve represents the probability of an event occurring that is normally distributed. So, the area under the entire normal distribution curve must be 1 (equal to 100%). For example, if the mean (average) male height is 5'10" then there is a 50% chance that a randomly selected male will have a height that is below or exactly 5'10". This is because the area under the normal curve from the left hand side up to the mean consists of half of the entire area of the normal curve. This leads us to the definitions of z-scores and standard deviations to represent how far along the normal curve a particular value is. We can calculate the likelihood of the value by finding the area under the normal curve to that point, usually by using a z-score cdf (cumulative density function) utility of a calculator or statistics software.
0.088508
One standard deviation
-1.43 (approx)
The Z value is negative, but area is always positive.
It is 0.839
It is 0.877
The area is 0.008894
In a normal distribution half (50%) of the distribution falls below (to the left of) the mean.
Approx 78.88 % Normal distribution tables give the area under the normal curve between the mean where z = 0 and the given number of standard deviations (z value) to its right; negative z values are to the left of the mean. Looking up z = 1.25 gives 0.3944 (using 4 figure tables). → area between -1.25 and 1.25 is 0.3944 + 0.3944 = 0.7888 → the proportion of the normal distribution between z = -1.25 and z = 1.25 is (approx) 78.88 %
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 area under the normal distribution curve represents the probability of an event occurring that is normally distributed. So, the area under the entire normal distribution curve must be 1 (equal to 100%). For example, if the mean (average) male height is 5'10" then there is a 50% chance that a randomly selected male will have a height that is below or exactly 5'10". This is because the area under the normal curve from the left hand side up to the mean consists of half of the entire area of the normal curve. This leads us to the definitions of z-scores and standard deviations to represent how far along the normal curve a particular value is. We can calculate the likelihood of the value by finding the area under the normal curve to that point, usually by using a z-score cdf (cumulative density function) utility of a calculator or statistics software.
Standard deviation helps you identify the relative level of variation from the mean or equation approximating the relationship in the data set. In a normal distribution 1 standard deviation left or right of the mean = 68.2% of the data 2 standard deviations left or right of the mean = 95.4% of the data 3 standard deviations left or right of the mean = 99.6% of the data