We are given a frequency distribution table and asked to find the value of $a$ given that the arithmetic mean is 5.

Frequency Distribution Table:

Step-by-Step Solution:

The formula for the arithmetic mean of a frequency distribution is:

$\text{Arithmetic Mean} = \frac{\sum f_i x_i}{\sum f_i}$

Where:

• $f_i$ are the frequencies,
• $x_i$ are the variables.

1. Calculate the sum of $f_i x_i$ (weighted sum): $\sum f_i x_i = (a - 1) \times 2 + (a + 1) \times 4 + (a + 1) \times 6 + (2a - 5) \times 8$

   Expanding: $(a - 1) \times 2 = 2a - 2$ $(a + 1) \times 4 = 4a + 4$ $(a + 1) \times 6 = 6a + 6$ $(2a - 5) \times 8 = 16a - 40$

   Adding these up: $\sum f_i x_i = (2a - 2) + (4a + 4) + (6a + 6) + (16a - 40)$ $\sum f_i x_i = 2a + 4a + 6a + 16a - 2 + 4 + 6 - 40$ $\sum f_i x_i = 28a - 32$

2. Calculate the sum of frequencies $\sum f_i$: $\sum f_i = (a - 1) + (a + 1) + (a + 1) + (2a - 5)$

   Simplifying: $\sum f_i = a - 1 + a + 1 + a + 1 + 2a - 5$ $\sum f_i = 5a - 4$

3. Substitute into the formula for the arithmetic mean: We know the arithmetic mean is 5, so: $5 = \frac{28a - 32}{5a - 4}$

4. Solve the equation: Cross-multiply: $5(5a - 4) = 28a - 32$ $25a - 20 = 28a - 32$

   Simplifying: $-20 + 32 = 28a - 25a$ $12 = 3a$ $a = 4$

Thus, the value of $a$ is $\boxed{4}$.

Would you like more details, or do you have any questions?

Related Questions:

1. What is the arithmetic mean formula for grouped data?
2. How does changing $a$ affect the frequency distribution?
3. What is the difference between mean and median in a frequency distribution?
4. How would the solution change if the arithmetic mean were a different value?
5. Can you explain the concept of weighted averages in other contexts?

Tip:

Always ensure that you expand terms carefully when calculating weighted sums in frequency distributions to avoid errors in simplifying the final equation.