The answer is about 16%
Using the z-score formula(z = (x-u)/sd) the z score is 1.
This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
You can't do this without knowing the distribution of scores.
-1.28
When putting the scores in, you use the normal distribution graph, which is the best start.
You calculate the z-scores and then use published tables.
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].
2
The mean of a distribution of scores is the average.
100%. And that is true for any probability distribution.
99.7% of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution.
3
Credit scores are used to determine loan percentages when a person applies for a loan. If a person has a low credit score, the percentages of interest are higher, whereas higher credit scores result in lower loan percentage rates.
You can't do this without knowing the distribution of scores.
Tests are curved by adjusting scores to a predetermined distribution, such as a bell curve, to determine final grades. This helps account for variations in difficulty and ensures fairness in grading.
They are said to be Normally distributed.
The distribution is skewed to the right.
If the distribution is Gaussian (or Normal) use z-scores. If it is Student's t, then use t-scores.
This simply means that if you plot a histogram of the scores it will be asymmetric.