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.
X	Frequency
0	3
1	15
2	33
3	28
4	11
5	9
6	1
Find the number that is 1.5 standard deviations BELOW the mean. (Round your answer to three decimal places.)


To solve this problem, we need to find the number that is 1.5 standard deviations below the mean of the data set. Here's the process:

### Step 1: Calculate the mean (μ)
The mean is given by the formula:

\[
\mu = \frac{\sum (X \times f)}{\sum f}
\]

Where:
- \( X \) is the number of days per week (0, 1, 2, 3, 4, 5, 6),
- \( f \) is the corresponding frequency of each \( X \).

Using the provided dataTo solve this problem, we need to find the number that is 1.5 standard deviations below the mean of the data set. Here's the process:

### Step 1: Calculate the mean (μ)
The mean is given by the formula:

\[
\mu = \frac{\sum (X \times f)}{\sum f}
\]

Where:
- \( X \) is the number of days per week (0, 1, 2, 3, 4, 5, 6),
- \( f \) is the corresponding frequency of each \( X \).

Using the provided 
\[
\mu = \frac{(0 \times 3) + (1 \times 15) + (2 \times 33) + (3 \times 28) + (4 \times 11) + (5 \times 9) + (6 \times 1)}{3 + 15 + 33 + 28 + 11 + 9 + 1}
\]

I'll calculate this.

### Step 2: Calculate the standard deviation (σ)
The standard deviation is given by the formula:

\[
\sigma = \sqrt{\frac{\sum f \cdot (X - \mu)^2}{\sum f}}
\]

This measures the spread of the data around the mean.

### Step 3: Find the value 1.5 standard deviations below the mean
The value that is 1.5 standard deviations below the mean is given by:

\[
\text{Value} = \mu - 1.5 \cdot \sigma
\]

Now, let's compute all of these.The mean (\(\mu\)) of the data set is 2.6, and the standard deviation (\(\sigma\)) is approximately 1.265. 

The number that is 1.5 standard deviations below the mean is approximately 0.703.

Do you have any questions or need further clarification?

Here are five related questions:
1. How would the calculation change if the frequencies were doubled?
2. How is the standard deviation affected if the data is more spread out?
3. What does it mean when a value is a certain number of standard deviations from the mean?
4. How can the concept of standard deviation be applied in quality control?
5. How would we calculate the number that is 1.5 standard deviations above the mean?

**Tip:** The standard deviation provides insight into how much variation or dispersion exists from the average (mean) value in a data set.

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.
X	Frequency
0	3
1	15
2	33
3	28
4	11
5	9
6	1
Find the number that is 1.5 standard deviations BELOW the mean. (Round your answer to three decimal places.)

Learn how to calculate the value that is 1.5 standard deviations below the mean for a sample of 100 clients at an exercise facility, based on their frequency of visits per week. This problem covers key statistical concepts such as mean, standard deviation, and z-scores.

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