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:
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
Find the variance of the following frequency distribution:
| Class Interval | ||||
| 0--4 | 4--8 | 8--12 | 12--16 | |
| Frequency | 1 | 2 | 2 | 1 |
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
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Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
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