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The problem asks to find the 47th percentile, \( P_{47} \), in a normally distributed dataset where the mean \( \mu = 0^\circ C \) and the standard deviation \( \sigma = 1^\circ C \). 

To solve this:

1. **Z-Score for 47th Percentile**: In a normal distribution, the percentile corresponds to a specific z-score. You can use a standard normal distribution table or a calculator to find the z-score associated with the 47th percentile.

   The 47th percentile corresponds to a z-score of approximately \( z = -0.075 \) (since the z-score at 47% is slightly negative as it is below the mean).

2. **Convert Z-Score to Temperature**: Use the z-score formula:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   where:
   - \( X \) is the temperature corresponding to the 47th percentile,
   - \( \mu = 0^\circ C \) is the mean, and
   - \( \sigma = 1^\circ C \) is the standard deviation.

   Rearranging for \( X \):
   \[
   X = z \cdot \sigma + \mu
   \]

   Substituting \( z = -0.075 \), \( \mu = 0 \), and \( \sigma = 1 \):
   \[
   X = (-0.075) \cdot 1 + 0 = -0.075^\circ C
   \]

Thus, the correct temperature reading at the 47th percentile is approximately \( -0.075^\circ C \), not \( -0.412^\circ C \), as shown in the original answer. The answer of \( -0.412^\circ C \) seems to be incorrect based on the calculation.

Would you like more details or clarifications?

Here are 5 related questions to explore further:
1. How do you find z-scores for different percentiles?
2. What is the significance of a z-score in a normal distribution?
3. How would the percentile change if the standard deviation increased?
4. How would the calculation change if the mean were different from 0°C?
5. How can we compute percentiles using a statistical calculator?

**Tip**: For quick z-score lookups, use statistical software or online z-score tables to find exact values for different percentiles.

Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P47, the 47th percentile. This is the temperature reading separating the bottom 47% from the top 53%.

Learn how to calculate the 47th percentile in a normal distribution with a mean of 0°C and a standard deviation of 1.00°C. This problem demonstrates the use of z-scores and percentiles, perfect for students in Grades 11-12.

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