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	12
2	30
3	28
4	11
5	7
6	9
Find the number that is 1.5 standard deviations BELOW the mean.

To find the number that is 1.5 standard deviations below the mean, we'll go through the following steps:

### 1. Find the Mean (\(\mu\))
The mean is calculated as:

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

Where \(X\) is the number of days per week, and \(f\) is the frequency.

Let's compute:

\[
\mu = \frac{(0 \cdot 3) + (1 \cdot 12) + (2 \cdot 30) + (3 \cdot 28) + (4 \cdot 11) + (5 \cdot 7) + (6 \cdot 9)}{100}
\]

\[
\mu = \frac{(0) + (12) + (60) + (84) + (44) + (35) + (54)}{100} = \frac{289}{100} = 2.89
\]

### 2. Find the Variance (\(\sigma^2\))
Variance is calculated using the formula:

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

First, calculate \(\sum{(X^2 \cdot f)}\):

\[
\sum{(X^2 \cdot f)} = (0^2 \cdot 3) + (1^2 \cdot 12) + (2^2 \cdot 30) + (3^2 \cdot 28) + (4^2 \cdot 11) + (5^2 \cdot 7) + (6^2 \cdot 9)
\]

\[
= 0 + 12 + 120 + 252 + 176 + 175 + 324 = 1059
\]

Now, calculate the variance:

\[
\sigma^2 = \frac{1059}{100} - (2.89)^2 = 10.59 - 8.3521 = 2.2379
\]

### 3. Find the Standard Deviation (\(\sigma\))
Standard deviation is the square root of the variance:

\[
\sigma = \sqrt{2.2379} \approx 1.496
\]

### 4. Find the Number 1.5 Standard Deviations Below the Mean
To find the number that is 1.5 standard deviations below the mean:

\[
\text{Value} = \mu - 1.5 \cdot \sigma = 2.89 - 1.5 \cdot 1.496 \approx 2.89 - 2.244 = 0.646
\]

Thus, the number that is 1.5 standard deviations below the mean is approximately **0.646**.

Would you like further clarification or details?

Here are 5 related questions you could explore:
1. How do you interpret the mean in this context?
2. What is the significance of standard deviation in such a distribution?
3. How would you calculate the number that is 1.5 standard deviations above the mean?
4. What does a negative standard deviation represent in real-world scenarios?
5. How does increasing the sample size affect the standard deviation?

**Tip:** When analyzing distributions, the mean and standard deviation give you a strong sense of the data's central tendency and variability. Always consider both to understand the dataset comprehensively.

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 12 2 30 3 28 4 11 5 7 6 9 Find the number that is 1.5 standard deviations BELOW the mean.

Learn how to calculate the number that is 1.5 standard deviations below the mean from a dataset of 100 clients using an exercise facility. This solution includes key statistical concepts such as mean, variance, and standard deviation. Suitable for students in Grades 10-12.

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