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We are given a frequency table representing the number of days per week that a sample of 100 clients use an exercise facility. We are asked to find the value that is 1.5 standard deviations below the mean.

### Steps:

1. **Calculate the mean**: 
   The mean (\(\mu\)) can be calculated using the formula:
   \[
   \mu = \frac{\sum (X \times f)}{\sum f}
   \]
   where \(X\) is the number of days, and \(f\) is the corresponding frequency.

2. **Calculate the variance and standard deviation**:
   The variance (\(\sigma^2\)) is calculated as:
   \[
   \sigma^2 = \frac{\sum f(X - \mu)^2}{\sum f}
   \]
   and the standard deviation (\(\sigma\)) is the square root of the variance.

3. **Determine the value 1.5 standard deviations below the mean**:
   This is given by:
   \[
   \mu - 1.5\sigma
   \]

Let's calculate it. I'll start by computing the mean and standard deviation from the provided data.Here are the results:

- The mean (\(\mu\)) is approximately \(2.82\).
- The standard deviation (\(\sigma\)) is approximately \(1.410\).
- The value that is 1.5 standard deviations below the mean is approximately \(0.705\).

Would you like more details on any part of this process? 

Here are some related questions you might explore:
1. How would the result change if the frequencies were different?
2. What is the significance of using standard deviations in data analysis?
3. How would we calculate values above the mean using standard deviations?
4. How does variance differ from standard deviation, conceptually?
5. What does it mean for data to be normally distributed around the mean?

**Tip**: Standard deviation is a measure of how spread out numbers are in your dataset. The larger it is, the more spread out the numbers are from the mean.

A sample of 100 clients of an exercise facility was selected. Let X = the number of days per week that a randomly selected client uses the exercise facility. Find the number that is 1.5 standard deviations below the mean.

Learn how to calculate the number of days per week a randomly selected client uses an exercise facility, 1.5 standard deviations below the mean. This solution includes steps to find the mean, variance, and standard deviation from a given frequency distribution.

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