Consider the following frequency distribution:
Value | 4 | 5 | 8 | 9 | 6 | 12 | 11 |
---|---|---|---|---|---|---|---|
Frequency | 5 | $ f_1 $ | $ f_2 $ | 2 | 1 | 1 | 3 |
Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.
For the given frequency distribution, let:
Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-I
List-II
(P) → (3), (Q) → (2), (R) → (5), (S) → (4)
Step 1: Use total frequency = 19
\[ 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \Rightarrow f_1 + f_2 = 7 \quad \cdots (1) \]
Step 2: Median = 6 ⇒ Cumulative frequency up to 6 is ≥ 9.5
Frequency table (partial):
Value | 4 | 5 | 6 | 8 | 9 | 11 | 12 |
---|---|---|---|---|---|---|---|
Freq | 5 | \( f_1 \) | 1 | \( f_2 \) | 2 | 3 | 1 |
Cumulative frequency up to 6: \[ 5 + f_1 + 1 = 6 + f_1 \geq 9.5 \Rightarrow f_1 \geq 4 \] From (1): \( f_2 = 7 - f_1 \)
Try \( f_1 = 4 \Rightarrow f_2 = 3 \)
Step 3: Calculate (P)
\[ 7f_1 + 9f_2 = 7 \cdot 4 + 9 \cdot 3 = 28 + 27 = 55 \Rightarrow (P) \to (5) \]
Step 4: Construct full distribution
Value (x) | 4 | 5 | 6 | 8 | 9 | 11 | 12 | Total |
---|---|---|---|---|---|---|---|---|
Freq (f) | 5 | 4 | 1 | 3 | 2 | 3 | 1 | 19 |
\( xf \) | 20 | 20 | 6 | 24 | 18 | 33 | 12 | 133 |
Mean: \[ \bar{x} = \frac{133}{19} = 7 \]
(Q): Mean deviation about mean (\( \alpha \))
\[ \alpha = \frac{1}{19} \sum f_i |x_i - 7| = \frac{48}{19} \Rightarrow 19\alpha = 48 \Rightarrow (Q) \to (3) \]
(R): Mean deviation about median (\( \beta \))
\[ \beta = \frac{1}{19} \sum f_i |x_i - 6| = \frac{47}{19} \Rightarrow 19\beta = 47 \Rightarrow (R) \to (2) \]
(S): Variance \( \sigma^2 \)
Given: \[ \sum f(x - 7)^2 = 146 \Rightarrow \sigma^2 = \frac{146}{19} \Rightarrow 19\sigma^2 = 146 \Rightarrow (S) \to (1) \]
Final Matching:
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:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is