Complete the sentences with the correct preposition or phrase. Also, answer the math question about modal height.

$$
4.5, 3.5, 4.8, 5.1, 4.5, 5.0, 4.5
$$

The height 4.5 appears 3 times, which is more than any other height.
*   **A. 4.5**

$$
45, 50, 60, 65, 70, 75, 82
$$

There are 7 marks, so $n=7$.

For an odd number of data points, the median is the $\left(\frac{n+1}{2}\right)$-th value.
$$
\text{Median position} = \frac{7+1}{2} = \frac{8}{2} = 4
$$

The 4-th value in the ordered list is 65.
*   **D. 65**

$$
\text{Total marks for 3 subjects} = \text{Average} \times \text{Number of subjects}
$$
$$
\text{Total marks for 3 subjects} = 48 \times 3 = 144
$$

Assuming the average pass mark of 50% refers to the average across all four subjects.
$$
\text{Required total marks for 4 subjects} = \text{Required average} \times \text{Number of subjects}
$$
$$
\text{Required total marks for 4 subjects} = 50 \times 4 = 200
$$

$$
\text{Marks in 4th subject} = \text{Required total marks} - \text{Total marks from 3 subjects}
$$
$$
\text{Marks in 4th subject} = 200 - 144 = 56
$$
*   **C. 56**

$$
\text{Total marks for males} = \text{Number of males} \times \text{Average mark of males}
$$
$$
\text{Total marks for males} = 40 \times 75 = 3000
$$

$$
\text{Total marks for females} = \text{Number of females} \times \text{Average mark of females}
$$
$$
\text{Total marks for females} = 60 \times 82 = 4920
$$

$$
\text{Total marks for all students} = \text{Total marks for males} + \text{Total marks for females}
$$
$$
\text{Total marks for all students} = 3000 + 4920 = 7920
$$

$$
\text{Total number of students} = \text{Number of males} + \text{Number of females}
$$
$$
\text{Total number of students} = 40 + 60 = 100
$$

$$
\text{Average mark of all students} = \frac{\text{Total marks for all students}}{\text{Total number of students}}
$$
$$
\text{Average mark of all students} = \frac{7920}{100} = 79.2
$$
*   **C. 79.2**