Let’s tackle this step by step!

We are given:

• Mean $\mu = 100$
• Standard deviation $\sigma = 20$
• We want to find the probability that a randomly selected score $X$ falls between 80 and 100:

$P(80 \leq X \leq 100)$

Step 1: Convert raw scores to z-scores

The z-score formula is:

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

• For $X = 80$:

$z = \frac{80 - 100}{20} = \frac{-20}{20} = -1$

• For $X = 100$:

$z = \frac{100 - 100}{20} = \frac{0}{20} = 0$

Step 2: Find the probabilities using the z-table

• The probability corresponding to $z = -1$ is approximately 0.1587.
• The probability corresponding to $z = 0$ is 0.5000.

Step 3: Calculate the area between the two z-scores

The probability that $X$ is between 80 and 100 is the difference between these two probabilities:

$P(80 \leq X \leq 100) = P(Z \leq 0) - P(Z \leq -1)$

$P(80 \leq X \leq 100) = 0.5000 - 0.1587 = 0.3413$

Final Answer:

The probability that a randomly selected score will be between 80 and 100 is:

$P(80 \leq X \leq 100) \approx 0.3413 \text{ or } 34.13\%$

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Would you like me to dive deeper into any part of the explanation, like how to use the z-table or why we use z-scores?

5 Related Questions:

1. How do you find the probability of a score falling above a certain value in a normal distribution?
2. What is the probability that a randomly selected score will be less than 80 in this distribution?
3. How would the probabilities change if the standard deviation were smaller or larger?
4. Can we calculate the probability of a score falling exactly at the mean in a continuous distribution?
5. How do we use the empirical rule (68-95-99.7) to approximate probabilities without using z-tables?

🔍 Tip:

Z-scores allow you to standardize any normal distribution, making it easy to compare different distributions or use a single z-table for all normal distributions!