The mean (μ) of the scale is 55 and the standard deviation (σ) is 7. Assuming that the scores are normally distributed, what PERCENTAGE of the population scores are above 59?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Multiply by 100.
To find the percentage of the population scores that are above 59, we can use the properties of the normal distribution.
First, we need to standardize the value 59 to a z-score. The formula for calculating a z-score is:
z = (x - μ) / σ,
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
Plugging in the values:
z = (59 - 55) / 7
= 4 / 7 ≈ 0.5714.
Next, we need to find the area under the standard normal curve to the right of this z-score. We can use a standard normal distribution table or a statistical calculator.
Using a standard normal distribution table, we find the area to the right of 0.57 is approximately 0.2852.
Since we want to find the percentage above 59, we can multiply this area by 100:
Percentage = 0.2852 * 100 ≈ 28.52%.
Therefore, approximately 28.52% of the population scores are above 59.