https://file.aiquickdraw.com/mathbot/file/17286929755710v88fmx1.jpg

To solve this, we need to use the properties of the normal distribution.

### Given:
- Mean (μ) = 11.7 pounds
- Standard deviation (σ) = 2.9 pounds
- The weight is normally distributed.

#### (a) Find the 81st percentile of the baby weights.
The formula to find the percentile of a normal distribution is:

\[
X = \mu + Z \cdot \sigma
\]

Where:
- \( X \) is the baby weight at the 81st percentile,
- \( \mu \) is the mean (11.7 pounds),
- \( \sigma \) is the standard deviation (2.9 pounds),
- \( Z \) is the z-score corresponding to the 81st percentile.

Using a z-table or a calculator, the z-score for the 81st percentile is approximately **0.877**.

Now, substitute the values into the formula:

\[
X = 11.7 + (0.877) \cdot 2.9
\]

\[
X = 11.7 + 2.5433 \approx 14.24 \text{ pounds}
\]

Thus, the 81st percentile of the baby weights is approximately **14.24 pounds**, which is different from the answer entered as **12.6 pounds** in the image.

Do you want more details on the remaining parts (b) and (c) or clarification on how the z-score was obtained?

### Related questions:
1. What is the formula to compute percentiles in normal distribution problems?
2. How is the z-score for a given percentile found using z-tables?
3. Why does the normal distribution have symmetrical properties around the mean?
4. How can Excel functions be used to calculate z-scores or percentiles?
5. What are the common applications of normal distribution in real-world problems?

#### Tip:
When solving normal distribution problems, make sure to use a precise z-score table or a calculator with statistical functions to find accurate percentiles.

Find the 81st percentile of baby weights, given a mean of 11.7 pounds and a standard deviation of 2.9 pounds.

Learn how to calculate the 81st percentile of baby weights in a normally distributed dataset with a mean of 11.7 pounds and a standard deviation of 2.9 pounds. This example demonstrates the use of Z-scores and normal distribution properties, suitable for high school statistics.

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