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Solving a Variance Problem
In today’s session, we will be solving a variance problem for a discrete frequency distribution and finding the value of ‘a’ when Var(X) is given as 16. Let’s dive into the solution!
We are tasked with finding the value of ‘a’. We are provided with the table values for \( x_i \) and \( f_i \).
1. Calculating \( f_i \) \( x_i \)
– Multiply as follows: \( 2a, 3a, 4a, 5a, 6a \).
2. Calculating \( f_i x_i^2 \)
– The results will be: \( 2a^2, 4a^2, 9a^2, 16a^2, 25a^2, 36a^2 \).
3. Calculating \( N \)
– The sum of all frequencies gives us \( N = 7 \).
4. Calculating \( \sum f_i x_i \)
– This results in \( \sum f_i x_i = 20 \) and \( \sum f_i x_i^2 = 92a^2 \).
Variance Formula
You might already know the variance formula:
\[
\text{Variance} = \frac{1}{N} \sum f_i x_i^2 – \left(\frac{1}{N} \sum f_i x_i\right)^2
\]
Given that the variance is 160, we plug in the values:
\[
160 = \frac{92a^2}{7} – \left(\frac{20}{7}\right)^2
\]
Now, expand and combine the terms. You will find:
\[
160 = \frac{168a^2}{49}
\]
From this, we can find the value of ‘a’:
\[
a = \sqrt{49} = 7
\]
Presentation
var ex 5Video
Conclusion
And that’s how we solve for the value of ‘a’! Make sure to practice this step-by-step approach to master variance calculations in frequency distributions.
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