To find the percentage of employees with IQs between 92 and 104, we need to calculate the z-scores for both IQ values and then find the area under the standard normal curve between these z-scores.

First, calculate the z-score for $IQ = 92$:

$z_1 = \frac{92 - 95}{7.5} = \frac{-3}{7.5} = -0.4$

Next, calculate the z-score for $IQ = 104$:

$z_2 = \frac{104 - 95}{7.5} = \frac{9}{7.5} = 1.2$

Now, using the standard normal distribution table, find the cumulative probabilities for these z-scores:

• For $z = -0.4$, the cumulative probability is approximately 0.3446.
• For $z = 1.2$, the cumulative probability is approximately 0.8849.

To find the percentage of employees with IQs between 92 and 104, subtract the cumulative probability for $z = -0.4$ from the cumulative probability for $z = 1.2$:

$P(92 < IQ < 104) = 0.8849 - 0.3446 = 0.5403$

Convert this to a percentage:

$0.5403 \times 100 = 54.0\%$

So, the percentage of employees who would have IQs between 92 and 104 is 54.0%.

Would you like any further details or have any other questions?

Here are some more questions you might find useful:

1. What is the probability that an employee has an IQ less than 85?
2. How do you calculate the z-score for an IQ of 110?
3. What percentage of employees have IQs above 100?
4. Can you explain how to find the cumulative probability using the standard normal distribution table?
5. How do you interpret z-scores in the context of normally distributed data?

Tip: When working with normally distributed data, always check that your mean and standard deviation are correctly applied in your z-score calculations to ensure accurate probability results.