A large company employs workers whose IQs are distributed normally with mean 110 and standard deviation 12.5. Management uses this information to assign employees to projects that will be​ challenging, but not too challenging. What percent of employees would have IQs less than 100​?
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Part 1
The percentage of employees expected to have IQs less than 100 is 
  
enter your response here​%.
​(Round to the nearest tenth as​ needed.)

To determine the percentage of employees with IQs less than 100, we first need to standardize the IQ score to find its corresponding z-score. The formula for the z-score is:

\[ z = \frac{X - \mu}{\sigma} \]

where:
- \( X \) is the value of interest (100 in this case),
- \( \mu \) is the mean (110),
- \( \sigma \) is the standard deviation (12.5).

Substituting in the given values:

\[ z = \frac{100 - 110}{12.5} = \frac{-10}{12.5} = -0.8 \]

Next, we use the z-score to find the corresponding percentile in the standard normal distribution. Using the standard normal distribution table, a z-score of -0.8 corresponds to a cumulative probability of approximately 0.2119.

Therefore, the percentage of employees with IQs less than 100 is:

\[ 21.2\% \]

So, the percentage of employees expected to have IQs less than 100 is **21.2%**.

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

Here are some more questions you might find useful:
1. What is the z-score for an IQ of 120?
2. How many employees have IQs between 105 and 115?
3. What is the probability that an employee has an IQ greater than 130?
4. How do you calculate the probability for a given range of IQ scores?
5. Can you explain how to use the standard normal distribution table?

**Tip:** When dealing with normally distributed data, converting raw scores to z-scores allows you to use standard normal distribution tables to find probabilities easily.

Learn how to calculate the percentage of employees with IQs less than 100 using the normal distribution and z-score. Understand the application of statistics in employee IQ assessment with a step-by-step solution and use of the standard normal distribution table.

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