To find the probability that a randomly selected student scored below 73 on the quiz, you can use the properties of the normal distribution, as the distribution of scores is typically assumed to be normal.
Given:
- Mean (\(\mu\)) = 80
- Standard deviation (\(\sigma\)) = 8
First, you need to calculate the z-score for the score of 73 using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
Where:
- \(X\) is the score you are comparing (73 in this case),
- \(\mu\) is the mean (80),
- \(\sigma\) is the standard deviation (8).
Substituting the values:
\[ z = \frac{(73 - 80)}{8} = \frac{-7}{8} = -0.875 \]
Next, you can look up the z-score of -0.875 in a standard normal distribution table, or use a calculator or statistical software to find the corresponding probability.
The probability associated with a z-score of -0.875 is approximately 0.1915.
Thus, the probability that a randomly selected student scored below 73 is about 0.1915, or 19.15%.
Therefore, the probability that a randomly selected student scored below 73 is approximately 19.15%.