The mean mark of the class is 48.2, as provided in the image.

For the standard deviation of the 25 students:

Formula:

$\sigma_{\text{combined}} = \sqrt{\frac{(n_1 - 1)\sigma_1^2 + (n_2 - 1)\sigma_2^2}{n_1 + n_2 - 1} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\mu_1 - \mu_2)^2}$

Variables:

• $n_1 = 15$, $\mu_1 = 47.6$, $\sigma_1 = 11.2$
• $n_2 = 10$, $\mu_2 = 49.1$, $\sigma_2 = 15.4$

Steps:

1. Calculate $(n_1 - 1)\sigma_1^2$ and $(n_2 - 1)\sigma_2^2$: $(15 - 1)(11.2)^2 = 14 \cdot 125.44 = 1756.16$ $(10 - 1)(15.4)^2 = 9 \cdot 237.16 = 2134.44$

2. Combine variances: $\frac{1756.16 + 2134.44}{15 + 10 - 2} = \frac{3890.6}{23} = 169.16$

3. Adjust for the difference in means: $\frac{15 \cdot 10}{(15 + 10)^2} \cdot (47.6 - 49.1)^2 = \frac{150}{625} \cdot (-1.5)^2 = 0.24 \cdot 2.25 = 0.54$

4. Add the components and take the square root: $\sigma_{\text{combined}} = \sqrt{169.16 + 0.54} = \sqrt{169.7} = 13.0 \, (\text{3 s.f.})$

Final Answer:

• Mean = 48.2
• Standard Deviation = 13.0

Let me know if you'd like an explanation or detailed steps! Here are 5 related questions and a tip:

1. What happens to the standard deviation when the group sizes are unequal?
2. How do outliers affect the mean and standard deviation?
3. Why is it necessary to account for the difference in means when combining data?
4. How would the standard deviation change if all scores increased by 5?
5. Can you combine variances if the distributions are dependent?

Tip: Always check units and ensure consistent rounding in final answers!