Answer
See explanation
Work Step by Step
#### For Women:
* $N = 60$
* $\sum X = \sum (\text{Height} \times \text{Frequency}) = 3852$
* $\sum X^2 = \sum (\text{Height}^2 \times \text{Frequency}) = 248898$
* $\bar{X} = \frac{3852}{60} = 64.2$
* $s^2 = \frac{248898 - \frac{3852^2}{60}}{59} = \frac{248898 - 247292.4}{59} \approx 7.06$
#### For Men:
* $N = 60$
* $\sum X = 4200$
* $\sum X^2 = 295140$
* $\bar{X} = \frac{4200}{60} = 70.0$
* $s^2 = \frac{295140 - \frac{4200^2}{60}}{59} = \frac{295140 - 294000}{59} \approx 9.59$
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### **Descriptive Statistics Summary**
| Statistic | Women | Men |
| --------------------------- | ----------- | ----------- |
| Sample Size ($N$) | 60 | 60 |
| Sum of Scores ($\sum X$) | 3,852 | 4,200 |
| Sum of Squares ($\sum X^2$) | 248,898 | 295,140 |
| Mean Height ($\bar{X}$) | **64.2 in** | **70.0 in** |
| Sample Variance ($s^2$) | **7.06** | **9.59** |
| Sample Std. Deviation ($s$) | **2.66 in** | **3.10 in** |
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### **Interpretation**
* **Men** in their 20s are, on average, taller than women by **5.8 inches**.
* **Men's height distribution is more variable**, with a higher standard deviation (3.10 in vs. 2.66 in).
* The sample variance confirms this broader spread in men's heights compared to women.
* These findings could reflect natural biological differences and greater heterogeneity in male height.
