2
Approx 47.7%
100%. And that is true for any probability distribution.
Between z = -1.16 and z = 1.16 is approx 0.7540 (or 75.40 %). Which means ¾ (0.75 or 75%) of the normal distribution lies between approximately -1.16 and 1.16 standard deviations from the mean.
A researcher wants to go from a normal distribution to a standard normal distribution because the latter allows him/her to make the correspondence between the area and the probability. Though events in the real world rarely follow a standard normal distribution, z-scores are convenient calculations of area that can be used with any/all normal distributions. Meaning: once a researcher has translated raw data into a standard normal distribution (z-score), he/she can then find its associated probability.
The standard deviation (SD) is a measure of spread so small sd = small spread. So the above is true for any distribution, not just the Normal.
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 %
3
99.7% of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution.
100%. And that is true for any probability distribution.
z-scores are distributed according to the standard normal distribution. That is, with the parameters: mean 0 and variance 1.
They are said to be Normally distributed.
If the distribution is Gaussian (or Normal) use z-scores. If it is Student's t, then use t-scores.
The IQs of a large enough population can be modeled with a Normal Distribution
Between z = -1.16 and z = 1.16 is approx 0.7540 (or 75.40 %). Which means ¾ (0.75 or 75%) of the normal distribution lies between approximately -1.16 and 1.16 standard deviations from the mean.
Assuming that you are refering to the standard normal distribution and the z-scores, the answer is 99.73%. If the assumption is incorrect, please resubmit the questionwith more information.
-1.28
You calculate the z-scores and then use published tables.
95% of the area falls between Z = -1.96 & 1.96.