Final answer:
The middle 50% of exam scores for a normal distribution with a mean of 81 points and a standard deviation of 4 points ranges from 78.3 to 83.7 points. This is found by using the z-scores for the 25th and 75th percentiles and converting them back to exam scores.
Step-by-step explanation:
To find the range of scores that represents the middle 50% of the distribution (also known as the interquartile range), we need to find the 25th and 75th percentiles of a normal distribution with a mean μ of 81 points and a standard deviation σ of 4 points. This task involves using the standard normal distribution and z-scores.
First, we find the z-scores that correspond to the 25th and 75th percentiles. These are typically around -0.675 and +0.675, respectively. Next, we convert these z-scores back to the original scores on the exam by using the formula:
X = μ + (z * σ)
For the 25th percentile (z = -0.675):
Lower Bound
X = 81 + (-0.675 * 4) = 81 - 2.7 = 78.3
For the 75th percentile (z = +0.675):
Upper Bound
X = 81 + (+0.675 * 4) = 81 + 2.7 = 83.7
Therefore, the range of scores that represents the middle 50% of students is from 78.3 to 83.7 points.