Consider the following frequency distribution- Class- 0-6 6-12 12-18 18-24 24-30- Frequency - a b 12 9 5- -If mean -30922 and median - 14- then the value a-b2 is equal to

x̅=∑ x_if_i∑ f_i=3a+9b+504a+b+26=30922 ⇒ 81a+37b=1018 ⋯ (1) nbsp; nbsp; Median =12+13+a+b2 (a+b)12× 6=14 nbsp; ⇒ a+b=18 ⋯ (2) nbsp; From (1) and ...

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Standard XII

Mathematics

Frequency Distributions

Consider the ...

Question

Consider the following frequency distribution:
$\begin{matrix} Class: & 0 - 6 & 6 - 12 & 12 - 18 & 18 - 24 & 24 - 30 Frequency : & a & b & 12 & 9 & 5 \end{matrix}$
If mean $= \frac{309}{22}$ and median $= 14,$ then the value $(a - b)^{2}$ is equal to

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Solution

$¯ x = \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{3 a + 9 b + 504}{a + b + 26} = \frac{309}{22} \Rightarrow 81 a + 37 b = 1018 \dots (1)$
$Median = 12 + \frac{13 + \frac{a + b}{2} - (a + b)}{12} \times 6 = 14 \Rightarrow a + b = 18 \dots (2)$

From $(1)$ and $(2), a = 8$ and $b = 10$

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Consider the following frequency distribution: Class:0−66−1212−1818−2424−30Frequency :ab1295 If mean =30922 and median =14, then the value (a−b)2 is equal to

¯x=∑xifi∑fi=3a+9b+504a+b+26=30922⇒81a+37b=1018 ⋯(1) Median =12+13+a+b2−(a+b)12×6=14⇒a+b=18 ⋯(2) From (1) and (2), a=8 and b=10

Consider the following frequency distribution:
$\begin{matrix} Class: & 0 - 6 & 6 - 12 & 12 - 18 & 18 - 24 & 24 - 30 Frequency : & a & b & 12 & 9 & 5 \end{matrix}$
If mean $= \frac{309}{22}$ and median $= 14,$ then the value $(a - b)^{2}$ is equal to

$¯ x = \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{3 a + 9 b + 504}{a + b + 26} = \frac{309}{22} \Rightarrow 81 a + 37 b = 1018 \dots (1)$
$Median = 12 + \frac{13 + \frac{a + b}{2} - (a + b)}{12} \times 6 = 14 \Rightarrow a + b = 18 \dots (2)$

From $(1)$ and $(2), a = 8$ and $b = 10$