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	2
1	12
2	33
3	29
4	11
5	9
6	4
Find the number that is 1.5 standard deviations BELOW the mean. (Round your answer to three decimal places.)

To solve this, we will first calculate the mean and standard deviation of the data, and then find the number that is 1.5 standard deviations below the mean.

### Step 1: Calculate the Mean (μ)

The formula for the mean is:

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

where:
- \( X \) is the number of days,
- \( f \) is the frequency of clients for each corresponding \( X \),
- \( N \) is the total number of clients, which is 100.

Let's calculate it:

\[
\mu = \frac{(0 \times 2) + (1 \times 12) + (2 \times 33) + (3 \times 29) + (4 \times 11) + (5 \times 9) + (6 \times 4)}{100}
\]

### Step 2: Calculate the Variance and Standard Deviation (σ)

The formula for the variance is:

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

The standard deviation \( \sigma \) is the square root of the variance.

### Step 3: Find the Value 1.5 Standard Deviations Below the Mean

Once we have the mean and standard deviation, the value 1.5 standard deviations below the mean is given by:

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

Let me compute this for you.The mean number of days per week that clients use the facility is \( \mu = 2.780 \), and the standard deviation is \( \sigma = 1.331 \).

The number that is 1.5 standard deviations below the mean is approximately:

\[
\mu - 1.5 \cdot \sigma = 0.783
\]

Thus, the value is \( 0.783 \) days.

Would you like any further explanation or details?

Here are five related questions to explore further:
1. What is the number that is 1.5 standard deviations **above** the mean?
2. How would the result change if we included more data points?
3. How is the standard deviation interpreted in this context?
4. What percentage of data typically lies within 1.5 standard deviations of the mean?
5. How can we interpret the result in terms of client behavior at the facility?

**Tip**: The standard deviation gives insight into how spread out the data points are from the mean—more spread means a larger deviation!

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  2 1  12 2  33 3  29 4  11 5  9 6  4. Find the number that is 1.5 standard deviations BELOW the mean. (Round your answer to three decimal places.)

Calculate the value 1.5 standard deviations below the mean for a dataset representing 100 clients at an exercise facility. This solution includes key statistical concepts like mean, variance, and standard deviation. Suitable for students in Grades 9-12.

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