To find the probability that a normally distributed random variable $X$ is between 3 and 61, given that the mean is 80 and the standard deviation is 11, we need to calculate the standard normal $Z$-scores for both 3 and 61.

The formula to calculate the $Z$-score is:

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

where:

• $X$ is the value of the random variable,
• $\mu$ is the mean, and
• $\sigma$ is the standard deviation.

Step 1: Calculate the $Z$-scores

For $X = 3$: $Z_1 = \frac{3 - 80}{11} = \frac{-77}{11} = -7$

For $X = 61$: $Z_2 = \frac{61 - 80}{11} = \frac{-19}{11} \approx -1.727$

Step 2: Find the probabilities from the standard normal distribution

Now, using the standard normal distribution table or a calculator, we find the probabilities associated with these $Z$-scores.

For $Z_1 = -7$, the probability is nearly 0, since the value is extremely far in the negative tail of the distribution.

For $Z_2 = -1.727$, the cumulative probability is approximately 0.042.

Step 3: Calculate the final probability

The probability that $X$ is between 3 and 61 is:

$P(3 < X < 61) = P(Z_2) - P(Z_1) = 0.042 - 0 \approx 0.042$

Thus, the probability that $X$ is between 3 and 61 is approximately 0.042, rounded to the nearest thousandth.

Would you like more details on this, or do you have any questions?

Here are 5 related questions to consider:

1. What is the probability that $X$ is less than 50 in this distribution?
2. How does changing the standard deviation affect the probability calculation?
3. What is the probability that $X$ is greater than 100?
4. How would the probability change if the mean were 90 instead of 80?
5. How do we interpret the standard normal distribution in real-world applications?

Tip: When working with normal distributions, always calculate the $Z$-score to transform data into the standard normal distribution before finding probabilities.