Let's go through each part of the question:

Given dataLet's go through each part of the question:

Given

Part (a): Finding the Mean

To find the mean of the frequency distribution, use the formula:

$\text{Mean} = \frac{\sum (\text{Value} \times \text{Frequency})}{\sum \text{Frequency}}$

1. Calculate each term $\text{Value} \times \text{Frequency}$:

   • $623 \times 13 = 8099$
   • $532 \times 6 = 3192$
   • $599 \times 11 = 6589$
   • $582 \times 14 = 8148$
   • $585 \times 7 = 4095$
   • $609 \times 3 = 1827$

2. Sum of $\text{Value} \times \text{Frequency}$: $8099 + 3192 + 6589 + 8148 + 4095 + 1827 = 31950$

3. Sum of Frequencies: $13 + 6 + 11 + 14 + 7 + 3 = 54$

4. Calculate the Mean: $\text{Mean} = \frac{31950}{54} \approx 591.7$

So, the mean is approximately 591.7.

Part (b): Finding the Median

To find the median, locate the cumulative frequency that corresponds to the middle position in the dataset. Since the total frequency is 54, the median will be around the 27th and 28th values.

1. Cumulative Frequencies:

   • For 623: $13$
   • For 532: $13 + 6 = 19$
   • For 599: $19 + 11 = 30$

   The 27th and 28th values fall within the frequency interval for 599, meaning the median value is 599.

Part (c): Finding the Mode

The mode is the value with the highest frequency.

• Checking the frequencies: 623 has 13, 532 has 6, 599 has 11, 582 has 14, 585 has 7, and 609 has 3.
• The highest frequency is 14, which corresponds to the value 582.

Therefore, the mode is 582.

Summary of Answers:

• (a) Mean: $591.7$
• (b) Median: $599$
• (c) Mode: $582$

Would you like more details on any part of the solution?

Further Practice Questions:

1. What is the range of the given values?
2. If the frequency of 609 increased by 2, how would the mean change?
3. Can there be multiple modes in a dataset? Why or why not?
4. How would removing the smallest value (532) affect the median?
5. Explain why the mean might be misleading if there is an outlier.

Tip:

When finding the median in frequency distributions, always check the cumulative frequency to identify the correct interval.