We are given that the final exam scores follow a normal distribution with:

• Mean ($\mu$) = 155
• Standard deviation ($\sigma$) = 6.3

The problem asks for the percentage of students scoring below 136.1 or above 142.4.

To solve this, we need to convert these raw scores into z-scores and use the properties of the standard normal distribution to find the corresponding probabilities.

Step 1: Calculate the z-scores

The z-score is calculated using the formula:

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

For 136.1:

$z = \frac{136.1 - 155}{6.3} = \frac{-18.9}{6.3} \approx -3$

For 142.4:

$z = \frac{142.4 - 155}{6.3} = \frac{-12.6}{6.3} \approx -2$

Step 2: Find the probabilities

We use standard normal distribution tables or a calculator to find the probabilities corresponding to these z-scores.

• The z-score of $-3$ corresponds to approximately 0.00135, which means about 0.135% of students score below 136.1.
• The z-score of $-2$ corresponds to approximately 0.0228, which means about 2.28% of students score below 142.4.

Step 3: Calculate the percentage of students

Since we are looking for the percentage of students who score below 136.1 or above 142.4, we add the two probabilities together:

$\text{Total percentage} = 0.135\% + 2.28\% = 2.415\%$

Thus, approximately 2.42% of students score either below 136.1 or above 142.4.

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Would you like further details on how to interpret these probabilities or additional clarification?

Here are some related questions:

1. How do we interpret z-scores in the context of normal distributions?
2. What is the empirical rule, and how does it relate to this problem?
3. Can we use the empirical rule for this problem instead of z-scores?
4. What percentage of students score between 136.1 and 142.4?
5. How would the calculation change if the distribution was not normal?

Tip: The z-score represents how many standard deviations a value is from the mean. Negative z-scores indicate values below the mean.