On a chemistry exam, Student A scored a 81 and Student B scored a 70. The mean for the class was 75 and the standard deviation was 3.8.
What percentage of students scored in somewhere in between Student A and Student B?
about 15 percent if you believe it was a normal distribution. Seldom it is.
http://davidmlane.com/hyperstat/z_table.html
Here is another method.
If it approximates a normal distribution, you can use
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
To find the percentage of students who scored somewhere between Student A and Student B, we first need to convert their scores into z-scores.
The z-score measures the number of standard deviations a particular data point is from the mean. It allows us to standardize data in order to find the proportion or percentage of values in a given range.
To find the z-score for each student, we can use the formula:
z = (x - μ) / σ
Where:
- x is the individual score
- μ is the mean of the population
- σ is the standard deviation of the population
Let's calculate the z-scores for Student A and Student B:
For Student A:
z_A = (81 - 75) / 3.8 = 1.58
For Student B:
z_B = (70 - 75) / 3.8 = -1.32
Now that we have the z-scores, we can refer to the standard normal distribution table (also known as the Z-table) to find the percentage of students between Student A and Student B.
However, since Student A scored higher than Student B, we need to find the proportion of students that scored less than Student A and subtract the proportion that scored less than Student B.
Using the Z-table, we can find these proportions as follows:
For Student A (z_A = 1.58):
- Look up the z-score (1.58) in the Z-table.
- The table gives us the proportion corresponding to the z-score, in this case, 0.9441. This means that 94.41% of students scored below Student A.
For Student B (z_B = -1.32):
- Look up the z-score (-1.32) in the Z-table.
- The table gives us the proportion corresponding to the z-score, which is 0.0934. This means that 9.34% of students scored below Student B.
To find the percentage of students who scored between Student A and Student B, we subtract the proportion of students who scored less than Student B from the proportion of students who scored less than Student A:
Percentage = (94.41% - 9.34%) = 85.07%
Therefore, approximately 85.07% of students scored in between Student A and Student B.