The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.67 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.54 inches. a) If a man is 6 feet 5 inches tall, what is his z-score? z = 2.88 b) What percentage of men are SHORTER than 6 feet 5 inches? 0.9980 c) If a woman is 5 feet 8 inches tall, what is her z-score? z = 1.26

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**Title: Understanding Z-Scores and Percentiles in Height Distributions**

The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.67 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.54 inches.

**a) Calculating the Z-Score for Men:**

- **Problem:** If a man is 6 feet 5 inches tall, what is his z-score?
- **Solution:** 
  - Height in inches: 6 feet 5 inches = 77 inches
  - Z-score: \( z = \frac{(77 - 69.3)}{2.67} = 2.88 \)

**b) Men's Height Percentile:**

- **Problem:** What percentage of men are shorter than 6 feet 5 inches?
- **Solution:** 
  - From the z-table, a z-score of 2.88 corresponds to 0.9980.
  - Thus, 99.80% of men are shorter than 6 feet 5 inches.

**c) Calculating the Z-Score for Women:**

- **Problem:** If a woman is 5 feet 8 inches tall, what is her z-score?
- **Solution:** 
  - Height in inches: 5 feet 8 inches = 68 inches
  - Z-score: \( z = \frac{(68 - 64.8)}{2.54} = 1.26 \)

**d) Women's Height Percentile:**

- **Problem:** What percentage of women are taller than 5 feet 8 inches?
- **Solution:** 
  - From the z-table, a z-score of 1.26 corresponds to approximately 0.8962.
  - Percentage taller = 1 - 0.8962 = 0.1038 or 10.38% of women are taller than 5 feet 8 inches.

These calculations help illustrate how to use z-scores to determine relative standings within a normal distribution.
Transcribed Image Text:**Title: Understanding Z-Scores and Percentiles in Height Distributions** The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.67 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.54 inches. **a) Calculating the Z-Score for Men:** - **Problem:** If a man is 6 feet 5 inches tall, what is his z-score? - **Solution:** - Height in inches: 6 feet 5 inches = 77 inches - Z-score: \( z = \frac{(77 - 69.3)}{2.67} = 2.88 \) **b) Men's Height Percentile:** - **Problem:** What percentage of men are shorter than 6 feet 5 inches? - **Solution:** - From the z-table, a z-score of 2.88 corresponds to 0.9980. - Thus, 99.80% of men are shorter than 6 feet 5 inches. **c) Calculating the Z-Score for Women:** - **Problem:** If a woman is 5 feet 8 inches tall, what is her z-score? - **Solution:** - Height in inches: 5 feet 8 inches = 68 inches - Z-score: \( z = \frac{(68 - 64.8)}{2.54} = 1.26 \) **d) Women's Height Percentile:** - **Problem:** What percentage of women are taller than 5 feet 8 inches? - **Solution:** - From the z-table, a z-score of 1.26 corresponds to approximately 0.8962. - Percentage taller = 1 - 0.8962 = 0.1038 or 10.38% of women are taller than 5 feet 8 inches. These calculations help illustrate how to use z-scores to determine relative standings within a normal distribution.
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