Step 1: The median of a grouped data is given by the formula: \[ \text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \] where:
- \( \ell \) is the lower boundary of the median class, - \( N \) is the total number of observations,
- \( F \) is the cumulative frequency before the median class,
- \( f \) is the frequency of the median class,
- \( h \) is the class width. From the problem, we are given:
- Median class interval: 12-18, - Median class frequency \( f = 12 \), - \( \ell = 12 \),
- Median = 14,
- Number of students with marks less than 12 is 18.
Step 2: Using the formula: \[ 14 = 12 + \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] Simplifying the equation: \[ 14 - 12 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \frac{\frac{N}{2} - 18}{2} \] \[ 4 = \frac{N}{2} - 18 \] \[ \frac{N}{2} = 22 \quad \Rightarrow \quad N = 44 \]
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
---|---|---|---|---|---|---|
Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).