According to her doctor, Mrs. Cooper's cholesterol level is higher than only 15% of the females aged 50 and over. The cholesterol levels among females aged 50 mg mg and over are approximately normally distributed with a mean of 240 and a standard deviation of 25 . What is Mrs. Cooper's cholesterol level? Carry dL dL your intermediate computations to at least four decimal places. Round your answer to one decimal place. 0 ? mg dL

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### Educational Content Transcription

---

**Problem Description:**

According to her doctor, Mrs. Cooper's cholesterol level is higher than only 15% of the females aged 50 and over. The cholesterol levels among females aged 50 and over are approximately normally distributed with a mean of 240 mg/dL and a standard deviation of 25 mg/dL. 

**Question:**
What is Mrs. Cooper’s cholesterol level? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.

---

**Explanation Area:**

- **Given Data:**
  - Mean (μ): 240 mg/dL
  - Standard Deviation (σ): 25 mg/dL
  - Percentile: 85% (since cholesterol level higher than only 15% means Mrs. Cooper belongs to the 85th percentile)

- **Relevant Formula:**
  To find Mrs. Cooper's cholesterol level, we need to use the z-score formula:
  \[
  Z = \frac{X - \mu}{\sigma}
  \]
  Where \( Z \) is the z-score, \( X \) is the cholesterol level, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

- **Steps to Find Cholesterol Level:**
  1. Find the z-score corresponding to the 85th percentile.
  2. Use the z-score formula to calculate the cholesterol level \( X \).

**Computation Area:**

1. **Finding z-score:**
   - For the 85th percentile, refer to the z-table or use a standard normal distribution calculator.
   - \( Z \approx 1.0364 \) for the 85th percentile.

2. **Calculating Cholesterol Level:**
   - Rearrange the z-score formula to solve for \( X \): 
     \[
     X = Z \cdot \sigma + \mu
     \]
   - Substitute the values:
     \[
     X = (1.0364 \cdot 25) + 240
     \]
   - Perform the calculation:
     \[
     X \approx 25.91 + 240 \approx 265.9
     \]

**Final Answer:**
Mrs. Cooper’s cholesterol level is approximately \( 265.9 \) mg/dL.

---

**Interaction Options:**
- **Explanation**: Offers
Transcribed Image Text:### Educational Content Transcription --- **Problem Description:** According to her doctor, Mrs. Cooper's cholesterol level is higher than only 15% of the females aged 50 and over. The cholesterol levels among females aged 50 and over are approximately normally distributed with a mean of 240 mg/dL and a standard deviation of 25 mg/dL. **Question:** What is Mrs. Cooper’s cholesterol level? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place. --- **Explanation Area:** - **Given Data:** - Mean (μ): 240 mg/dL - Standard Deviation (σ): 25 mg/dL - Percentile: 85% (since cholesterol level higher than only 15% means Mrs. Cooper belongs to the 85th percentile) - **Relevant Formula:** To find Mrs. Cooper's cholesterol level, we need to use the z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Where \( Z \) is the z-score, \( X \) is the cholesterol level, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. - **Steps to Find Cholesterol Level:** 1. Find the z-score corresponding to the 85th percentile. 2. Use the z-score formula to calculate the cholesterol level \( X \). **Computation Area:** 1. **Finding z-score:** - For the 85th percentile, refer to the z-table or use a standard normal distribution calculator. - \( Z \approx 1.0364 \) for the 85th percentile. 2. **Calculating Cholesterol Level:** - Rearrange the z-score formula to solve for \( X \): \[ X = Z \cdot \sigma + \mu \] - Substitute the values: \[ X = (1.0364 \cdot 25) + 240 \] - Perform the calculation: \[ X \approx 25.91 + 240 \approx 265.9 \] **Final Answer:** Mrs. Cooper’s cholesterol level is approximately \( 265.9 \) mg/dL. --- **Interaction Options:** - **Explanation**: Offers
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