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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.
You calculate the z-scores and then use published tables.
-1.28
When putting the scores in, you use the normal distribution graph, which is the best start.
Assuming a normal distribution, the proportion falling between the mean (of 8) and 7 with standard deviation 2 is: z = (7 - 8) / 2 = -0.5 → 0.1915 (from normal distribution tables) → less than 7 is 0.5 - 0.1915 = 0.3085 = 0.3085 x 100 % = 30.85 % (Note: the 0.5 in the second sum is because half (0.5) of a normal distribution is less than the mean, not because 7 is half a standard deviation away from the mean, and the tables give the proportion of the normal distribution between the mean and the number of standard deviations from the mean.)
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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.
You calculate the z-scores and then use published tables.
-1.28