Question

Assume that the results of your class' statistics test are normally distributed. You know that you scored 1 standard deviation above the mean of 69/100. Rounded to the nearest whole number, what percentage of test scores will be below yours?

Answer 1

84%

Explanation:

We have to remember that z-scores are values to find probabilities for any normal distribution using the standard normal distribution, a conversion of the normal distribution to find probabilities related to that distribution. One way to find the above z-scores is:

`\\ z = (x - \mu)/(\sigma)`

As a result, we can say that one standard deviation above the mean is equal to a z-score = 1, or that one standard deviation below the mean is equal to a z-score = -1, to take some examples.

The corresponding cumulative probability for a z-score = 1 (one standard deviation above the mean) can be obtained from the cumulative standard normal table, that is, the cumulative probabilities from z= -4 (four standard deviations below the mean) to the value corresponding to this z-score = 1.

Thus, for a z-score = 1, the cumulative standard normal table gives us a value of P(x<z=1) = 0.84134 or 84.134. In other words, below z = 1, there are 84.134% of cases below this value.

Applying this for the case in the question, the percentage of test scores below 69 (one standard deviation above the mean) is, thus, 84.134%, and rounding to the nearest whole number is simply 84%.