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%.

Q.27) The science test grades are posted. The class did very well. All students taking the test scored over 75. Unfortunately, 4 students were absent for the test and the computer listed their scores as 0 until the test is taken. Assuming that no score repeated more times than the 0's, what measure of central tendency would most likely give the the best representation of this data?