Find Equation Having Frequency Table Parameter k as Root
If the mean of the data
is , then is one of the roots of the equation :
Options
A
B
C
Correct
D
Topics & Concepts
Step-by-Step Solution
To find the value of , we first calculate the class mid-points () and the corresponding frequencies () for each class interval:
\hline \text{Class Interval} & \text{Mid-point } (x_i) & \text{Frequency } (f_i) & f_i x_i \\ \hline 5-10 & 7.5 & 2 & 2 \times 7.5 = 15 \\ 10-15 & 12.5 & k & 12.5k \\ 15-20 & 17.5 & 28 & 28 \times 17.5 = 490 \\ 20-25 & 22.5 & 54 & 54 \times 22.5 = 1215 \\ 25-30 & 27.5 & k+1 & 27.5(k+1) = 27.5k + 27.5 \\ 30-35 & 32.5 & 5 & 5 \times 32.5 = 162.5 \\ \hline \end{array}$$ Now, we compute the sum of frequencies ($\sum f_i$) and the sum of products ($\sum f_i x_i$): $$\sum f_i = 2 + k + 28 + 54 + (k+1) + 5 = 2k + 90$$ $$\sum f_i x_i = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 40k + 1910$$ The formula for the mean ($\bar{x}$) of grouped data is given by: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Given that the mean $\bar{x} = 21$: $$\frac{40k + 1910}{2k + 90} = 21$$ Cross-multiplying to solve for $k$: $$40k + 1910 = 21(2k + 90)$$ $$40k + 1910 = 42k + 1890$$ $$42k - 40k = 1910 - 1890$$ $$2k = 20 \implies k = 10$$ Now, we test $x = 10$ in the given quadratic equations: For option C: $$2x^2 - 19x - 10 = 0$$ Substituting $x = 10$: $$2(10)^2 - 19(10) - 10 = 2(100) - 190 - 10 = 200 - 200 = 0$$ Thus, $k = 10$ is a root of the equation $2x^2 - 19x - 10 = 0$.