To find the percentage of adult women expected to have cholesterol levels between 164 mg/dL and 212 mg/dL, we can use the properties of the normal distribution. Assuming cholesterol levels follow a normal distribution, we must know the mean (μ) and standard deviation (σ) of cholesterol levels in adult women. Let's proceed with the steps:

### 1. Break Down the Problem
1. Identify the mean and standard deviation of cholesterol levels in adult women.
2. Standardize the values using the z-score formula.
3. Use the z-scores to find the corresponding probabilities from the standard normal distribution table.
4. Calculate the area (percentage) between the two z-scores.

### 2. Relevant Concepts
- **Z-Score Formula**:
$$
z = \frac{(X - \mu)}{\sigma}
$$
where $ X $ is the cholesterol level, $ \mu $ is the mean, and $ \sigma $ is the standard deviation.
- **Normal Distribution**: The area under the curve represents the probability of a random variable falling within a certain range.

### 3. Analysis and Detail
Assuming the mean cholesterol level (μ) for adult women is 200 mg/dL and the standard deviation (σ) is 30 mg/dL:

1. Calculate the z-scores:
- For $ X_1 = 164 $:
$$
z_1 = \frac{(164 - 200)}{30} = \frac{-36}{30} = -1.2
$$
- For $ X_2 = 212 $:
$$
z_2 = \frac{(212 - 200)}{30} = \frac{12}{30} = 0.4
$$

2. Look up the z-scores in the standard normal distribution table:
- $ P(Z

Question

To find the percentage of adult women expected to have cholesterol levels between 164 mg/dL and 212 mg/dL, we can use the properties of the normal distribution. Assuming cholesterol levels follow a normal distribution, we must know the mean (μ) and standard deviation (σ) of cholesterol levels in adult women. Let's proceed with the steps:

### 1. Break Down the Problem
1. Identify the mean and standard deviation of cholesterol levels in adult women.
2. Standardize the values using the z-score formula.
3. Use the z-scores to find the corresponding probabilities from the standard normal distribution table.
4. Calculate the area (percentage) between the two z-scores.

### 2. Relevant Concepts
- **Z-Score Formula**: 
  $$
  z = \frac{(X - \mu)}{\sigma}
  $$
  where $ X $ is the cholesterol level, $ \mu $ is the mean, and $ \sigma $ is the standard deviation.
- **Normal Distribution**: The area under the curve represents the probability of a random variable falling within a certain range.

### 3. Analysis and Detail
Assuming the mean cholesterol level (μ) for adult women is 200 mg/dL and the standard deviation (σ) is 30 mg/dL:

1. Calculate the z-scores:
   - For $ X_1 = 164 $:
   $$
   z_1 = \frac{(164 - 200)}{30} = \frac{-36}{30} = -1.2
   $$
   - For $ X_2 = 212 $:
   $$
   z_2 = \frac{(212 - 200)}{30} = \frac{12}{30} = 0.4
   $$

2. Look up the z-scores in the standard normal distribution table:
   - $ P(Z < -1.2) \approx 0.1151 $
   - $ P(Z < 0.4) \approx 0.6554 $

3. Calculate the area between the two z-scores:
$$
P(164 < X < 212) = P(Z < 0.4) - P(Z < -1.2) 
$$
$$
= 0.6554 - 0.1151 = 0.5403
$$
This means approximately 54.03% of adult women are expected to have cholesterol levels between 164 mg/dL and 212 mg/dL.

### 4. Verify and Summarize
- Checked calculations: 
  - Z-scores are correct.
  - Proportions from the normal distribution table are verified.
- The final calculation confirms that about 54.03% of adult women are expected to have cholesterol levels in this range.

### Final Answer
Approximately **54.03%** of adult women are expected to have cholesterol levels between 164 mg/dL and 212 mg/dL.

Knowee AI · Accepted Answer