To solve this problem, we need to use the concept of the median in a grouped frequency distribution. Given that:
The formula for finding the median in a grouped data set is:
\(Median = L + \left( \frac{N/2 - F}{f} \right) \times h\)
Substitute the known values into the formula:
\(14 = 12 + \left( \frac{N/2 - 18}{12} \right) \times 6\)
Solving for \(N\):
\(14 - 12 = \left( \frac{N/2 - 18}{12} \right) \times 6\)
\(2 = \left( \frac{N/2 - 18}{12} \right) \times 6\)
\(2/6 = \frac{N/2 - 18}{12}\)
\(1/3 = \frac{N/2 - 18}{12}\)
Convert this to fraction and solve:
\(12 \times 1/3 = N/2 - 18\)
\(4 = N/2 - 18\)
\(N/2 = 4 + 18\)
\(N/2 = 22\)
\(N = 44\)
Therefore, the total number of students is 44. This matches with the given correct answer option.
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: \[ 14 - 12 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \frac{\frac{N}{2} - 18}{2} \] \[ 4 = \frac{N}{2} - 18 \] \[ \frac{N}{2} = 22 \quad \Rightarrow \quad N = 44 \]
\(x_i\) | \(f_i\) |
|---|---|
| 0 - 4 | 2 |
| 4 - 8 | 4 |
| 8 - 12 | 7 |
| 12 - 16 | 8 |
| 16 - 20 | 6 |
Find the value of 20M (where M is median of the data)
\(x_i\) | \(f_i\) |
|---|---|
| 0 - 4 | 2 |
| 4 - 8 | 4 |
| 8 - 12 | 7 |
| 12 - 16 | 8 |
| 16 - 20 | 6 |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,