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The problem involves a normal distribution of weights with a given mean and variance. Here's how we can approach it:

### Problem Breakdown:
- The weights are **normally distributed** with:
  - Mean (\(\mu\)) = 160
  - Variance = 400, which gives a standard deviation \(\sigma = \sqrt{400} = 20\)
  
We are asked to determine how high a set of bathroom scales should be calibrated so that **99%** of individuals can weigh themselves. This means finding the weight corresponding to the 99th percentile of this normal distribution.

### Step-by-Step Solution:
1. **Standardization**: First, we will convert the 99th percentile to a corresponding z-score using standard normal distribution tables or a calculator.
   - For 99%, the z-score corresponding to the upper tail is approximately **2.33**.
   
2. **Unstandardizing**: Now, we convert the z-score back into the weight using the formula:
   \[
   X = \mu + z \times \sigma
   \]
   where:
   - \(\mu = 160\)
   - \(\sigma = 20\)
   - \(z = 2.33\)

Let's calculate the weight.

### Calculation:

\[
X = 160 + 2.33 \times 20
\]The set of bathroom scales should be calibrated to approximately **206.6** pounds in order for 99% of the individuals to be able to weigh themselves.

Do you want more details on the solution or have any further questions?

Here are 5 related questions to explore further:
1. How would the answer change if the variance was different, for example, 625 instead of 400?
2. What is the z-score for the 95th percentile, and how would it affect the calibration?
3. How can you calculate the weight corresponding to the 1st percentile in this normal distribution?
4. If the mean weight changed to 170 pounds, how would the calibration change for 99% coverage?
5. How can we use the cumulative distribution function (CDF) to find percentiles for different probabilities?

**Tip:** For normal distributions, you can easily calculate percentiles by using z-scores and the relationship between mean, standard deviation, and probability.

If the distribution of weights of a certain class of individuals is approximately normal, with mean 160 and variance 400, how high should a set of bathroom scales be calibrated so that about 99% of the people will be able to weigh themselves?

Learn how to calibrate bathroom scales for a population with weights normally distributed with a mean of 160 and a variance of 400. Using z-scores and standard deviation, calculate the weight that covers 99% of the population. Suitable for high school students in Grades 10-12.

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