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The problem you shared describes a scenario where the distribution of weights is approximately normal with:

- Mean \(\mu = 160\)
- Variance \(\sigma^2 = 400\), which implies that the standard deviation \(\sigma = \sqrt{400} = 20\)

We are asked to find the value for which about 99% of the people can weigh themselves using the bathroom scales. In other words, we need to determine the value that covers the upper bound of the central 99% of this normal distribution.

### Steps to solve:

1. **99% of people** corresponds to 99% of the area under the normal distribution curve. To find this, we can look for the z-score that corresponds to a cumulative probability of 99%. This leaves 0.5% on either tail of the distribution.
   
2. Using a standard normal distribution table, the z-score corresponding to 99% is approximately \(z = 2.33\).

3. The value we are looking for can be found using the formula:
   \[
   x = \mu + z \cdot \sigma
   \]
   Substituting the values:
   \[
   x = 160 + 2.33 \cdot 20
   \]
   \[
   x = 160 + 46.6 = 206.6
   \]

Therefore, the bathroom scale should be calibrated to approximately **206.6 units** (most likely pounds or kilograms, depending on the context) so that 99% of the people can weigh themselves.

Would you like further details, or do you have any questions on this?

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Here are 5 related questions that expand on this concept:

1. How would the calibration change if we wanted only 95% of people to weigh themselves?
2. Can you explain how the z-score is derived from a standard normal distribution?
3. What does the variance of a distribution tell us about the data?
4. How does changing the mean of the distribution affect the calibration of the scale?
5. How would the result change if the variance were doubled?

**Tip:** In normal distribution problems, knowing how to find and interpret z-scores is key to understanding probabilities related to data points!

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?

Discover how to calibrate a bathroom scale using normal distribution concepts, with a mean of 160 and variance of 400. Learn how to find the z-score for 99% probability and apply the formula to ensure accurate scale calibration for 99% of users.

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