The test scores of students in a class test have a mean of 65 and with a standard deviation of 8. What is the probable percentage of students scored more than 70?
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.
To find the probable percentage of students who scored more than 70, we can use the concept of z-scores and the standard normal distribution.
Step 1: Calculate the z-score for a test score of 70.
The z-score formula is given by:
z = (x - μ) / σ
where
z is the z-score,
x is the test score,
μ is the mean of the distribution (65 in this case),
and σ is the standard deviation (8 in this case).
Plugging in the values:
z = (70 - 65) / 8 = 0.625
Step 2: Find the cumulative probability (area) to the right of the z-score using a standard normal distribution table or a statistical software.
From the table or software, we can find that the area to the right of z = 0.625 is approximately 0.267.
Step 3: Convert the cumulative probability to a percentage.
To convert the decimal probability to a percentage, we multiply by 100.
0.267 * 100 = 26.7
Therefore, the probable percentage of students who scored more than 70 is approximately 26.7%.