Suppose the scores of students on an exam are normally distributed with a mean of 393 and a standard deviation of 57. According to the empirical rule, what percentage of students scored between 279 and 507 on the exam
Don't know what the "empirical rule" is, but in real normal distribution calculations, I got
95.45%
http://davidmlane.com/normal.html
To find the percentage of students who scored between 279 and 507 on the exam using the empirical rule, we need to calculate the z-scores for these two values and then use a z-table to find the corresponding probabilities.
Step 1: Calculate the z-score for 279:
The formula for calculating z-score is:
z = (x - μ) / σ
where x is the individual value, μ is the mean, and σ is the standard deviation.
For 279:
z = (279 - 393) / 57
z = -2
Step 2: Calculate the z-score for 507:
For 507:
z = (507 - 393) / 57
z = 2
Step 3: Use the z-table to find the probabilities:
The z-table gives you the proportion of values below a certain z-score. To find the probability between the two z-scores (-2 and 2), we need to find the area under the normal curve between these z-scores.
From the z-table, the probability for each z-score can be found:
For z = -2, the probability is 0.0228.
For z = 2, the probability is 0.9772.
To find the proportion between -2 and 2, we subtract the probability of -2 from the probability of 2:
0.9772 - 0.0228 = 0.9544
Step 4: Convert to percentage:
To convert the proportion to a percentage, multiply by 100:
0.9544 * 100 = 95.44%
Therefore, approximately 95.44% or 95.4% of the students scored between 279 and 507 on the exam according to the empirical rule.