To find the proportion of students who receive a score less than 366, we need to use the properties of the normal distribution.

Step 1: Standardize the score using the Z-score formula

The Z-score tells us how many standard deviations a score is away from the mean. The formula for the Z-score is:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 366$ (the score we are interested in)
• $\mu = 308$ (the mean score)
• $\sigma = 50$ (the standard deviation)

Substituting the values:

$Z = \frac{366 - 308}{50} = \frac{58}{50} = 1.16$

Step 2: Find the cumulative probability for Z = 1.16

Now, we need to look up the cumulative probability for $Z = 1.16$ in the standard normal distribution table or use a calculator that provides the cumulative probability.

For $Z = 1.16$, the cumulative probability is approximately 0.8770.

Step 3: Conclusion

The proportion of students who score less than 366 is approximately 0.8770.

So, the answer is: $\boxed{0.8770}$

Let me know if you need further explanation or if you'd like to explore more! Here are some related questions:

1. How would you find the proportion of students scoring above 366?
2. What happens if the distribution is not normal?
3. How can you use Z-scores to compare scores from different exams?
4. What is the proportion of students scoring between 308 and 366?
5. How would you find the score corresponding to a certain percentile?

Tip: If you ever need to find a percentile for a Z-score, simply look up the Z-score in the standard normal table or use a cumulative probability calculator.