$$ X $$ is a normally distributed random variable with mean 74 and standard deviation 22 . What is the probability that $$ X $$ is between 8 and 96 ? Use the $$ 0.68-0.95-0.997 $$ rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

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To determine the probability that a normally distributed random variable $$ X $$ with mean $$ \mu = 74 $$ and standard deviation $$ \sigma = 22 $$ falls between 8 and 96, we'll use the **68-95-99.7 (Empirical) Rule**.

### Step 1: Standardize the Interval
First, convert the values 8 and 96 to their corresponding **z-scores**:
$$
z = \frac{X - \mu}{\sigma}
$$
- For $$ X = 8 $$:
$$
z = \frac{8 - 74}{22} = \frac{-66}{22} = -3
$$
- For $$ X = 96 $$:
$$
z = \frac{96 - 74}{22} = \frac{22}{22} = 1
$$

So, we're looking for $$ P(-3 \leq Z \leq 1) $$.

### Step 2: Apply the Empirical Rule
The Empirical Rule states:
- **68%** of data lies within **±1σ**,
- **95%** within **±2σ**,
- **99.7%** within **±3σ**.

Using this:
- $$ P(-3 \leq Z \leq 3) = 99.7\% $$
- $$ P(-1 \leq Z \leq 1) = 68\% $$
- $$ P(Z \leq -3) = 0.15\% $$ (since half of the remaining 0.3% lies below -3σ)

Now, to find $$ P(-3 \leq Z \leq 1) $$:
$$
P(-3 \leq Z \leq 1) = P(Z \leq 1) - P(Z < -3) 
$$
$$
P(Z \leq 1) = 84\% \quad \text{(50\% below mean + 34\% within +1σ)}
$$
$$
P(Z < -3) = 0.15\%
$$
$$
P(-3 \leq Z \leq 1) = 84\% - 0.15\% = 83.85\%
$$

### Step 3: Convert to Decimal
$$
83.85\% = 0.8385
$$
Rounded to the nearest thousandth:
$$
0.839
$$

**Final Answer:**  
The probability is approximately **0.839**.
The probability that $$ X $$ is between 8 and 96 is approximately **0.839**.

\( X \) is a normally distributed random variable with mean 74 and standard deviation 22 . What is the probability that \( X \) is between 8 and 96 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

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