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? Please check work
70-65/8 = .625 = .2357 = 1.-.2357 = .7643
ANSWER 76% ????
No, since the value is above the mean, it would be less than 50%.
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 in a class test, you can use z-scores and the standard normal distribution.
First, calculate the z-score for a score of 70 using the formula:
z = (x - μ) / σ
where x is the score (70), μ is the mean (65), and σ is the standard deviation (8).
z = (70 - 65) / 8
z = 0.625
The z-score of 0.625 tells us the number of standard deviations a score of 70 is away from the mean.
Next, refer to the standard normal distribution table, which provides the probabilities associated with different z-scores. Look up the z-score of 0.625 in the table.
The z-score of 0.625 corresponds to a cumulative probability of approximately 0.7357. This means that approximately 73.57% of the students scored below 70.
To find the percentage of students who scored more than 70, subtract this value from 1 (since the sum of probabilities is 1).
1 - 0.7357 = 0.2643
So, approximately 26.43% of the students scored more than 70 in the class test, not 76% as you calculated.
Therefore, the correct answer is around 26.43%, not 76%.