Given a mean of 60 and a standard divation of 7, what percentate of scores ae 73 or less? What is the formula for determining this value?
To determine the percentage of scores that are 73 or less, we can use the concept of z-scores and the standard normal distribution.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the individual value (73 in this case)
- μ is the mean (60 in this case)
- σ is the standard deviation (7 in this case)
Substituting the given values into the formula, we get:
z = (73 - 60) / 7
z = 13 / 7
z ≈ 1.857
Once we have the z-score, we can refer to a standard normal distribution table or use statistical software to find the corresponding area/probability.
In this case, we want to find the area to the left of the z-score. By looking up the z-score of 1.857 in the standard normal distribution table or using software, we find that the corresponding area/probability is approximately 0.9664.
To convert this probability to a percentage, we multiply it by 100:
0.9664 × 100 ≈ 96.64%
Therefore, approximately 96.64% of scores are 73 or less when the mean is 60 and the standard deviation is 7.