To solve this problem, we need to find the probability that a randomly selected student will score between 85 and 90 on a normally distributed exam with a mean of 74 and a standard deviation of 4.

Steps to Solve:

1. Convert the raw scores to z-scores: The z-score formula is given by:

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

   where:

   • $X$ is the raw score,
   • $\mu$ is the mean,
   • $\sigma$ is the standard deviation.

2. Calculate the z-scores for 85 and 90.

   • For $X = 85$:

     $z_{85} = \frac{85 - 74}{4} = \frac{11}{4} = 2.75$

   • For $X = 90$:

     $z_{90} = \frac{90 - 74}{4} = \frac{16}{4} = 4.00$

3. Find the probabilities corresponding to these z-scores: You can use the standard normal distribution table or a calculator.

   • The probability corresponding to $z = 2.75$ is approximately 0.9970.
   • The probability corresponding to $z = 4.00$ is approximately 0.99997.

4. Calculate the probability of scoring between 85 and 90:

   $P(85 \leq X \leq 90) = P(z_{90}) - P(z_{85}) = 0.99997 - 0.9970 = 0.00297$

Final Answer:

The probability that a randomly selected student will score between 85 and 90 is approximately 0.00297, or 0.297%.

Would you like more details or have any other questions?

Related Questions:

1. How do you interpret a z-score in a normal distribution?
2. What is the cumulative distribution function (CDF) and how is it used?
3. How would the probability change if the standard deviation were larger?
4. What is the empirical rule for normal distributions?
5. How do you find the probability of a score less than a certain value?
6. What is the significance of the mean in a normal distribution?
7. How can you use the z-score to find percentiles?
8. What if the scores were not normally distributed? How would that affect the calculations?

Tip: The z-score tells you how many standard deviations away a value is from the mean. This makes it easier to compare scores from different normal distributions.