Answer:
84%
Step-by-step explanation:
We have to remember that z-scores are values to find probabilities for any <em>normal distribution</em> using the <em>standard normal distribution</em>, a conversion of the normal distribution to find probabilities related to that distribution. One way to find the above z-scores is:
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 (<em>one standard deviation above the mean</em>) can be obtained from the <em>cumulative standard normal table</em>, 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 <em>cumulative standard normal table</em> 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%.