Question:

Observe the following table and find the Mean: \[ \text{Assumed Mean (A)} = 300 \] \[ \begin{array}{|c|c|c|c|} \hline \text{Class Interval} & \text{Class Mark } (x_i) & \text{Frequency } (f_i) & f_i d_i \\ \hline 200 - 240 & 220 & 5 & -400 \\ 240 - 280 & 260 & 10 & -400 \\ 280 - 320 & 300 & 15 & 0 \\ 320 - 360 & 340 & 12 & 480 \\ 360 - 400 & 380 & 8 & 640 \\ \hline \text{Total} & & \Sigma f_i = 50 & \Sigma f_i d_i = 320 \\ \hline \end{array} \]

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When using the Assumed Mean Method, choose a class mark near the center of the data as the assumed mean to simplify calculations. \[ \bar{X} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} \]
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Solution and Explanation

Step 1: Recall the formula for Mean (Assumed Mean method).
\[ \bar{X} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} \] Step 2: Substitute the given values.
\[ A = 300, \quad \Sigma f_i d_i = 320, \quad \Sigma f_i = 50 \] Step 3: Substitute in the formula.
\[ \bar{X} = 300 + \frac{320}{50} \] Step 4: Simplify.
\[ \bar{X} = 300 + 6.4 = 306.4 \] Step 5: Conclusion.
Therefore, the mean of the given data is \(306.4.\)
Final Answer: \[ \boxed{\bar{X} = 306.4} \]
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