To solve for the measures of central tendency for the given data set, let's first summarize the data:

Frequency Distribution Table:

• Data | Frequency (freq)
  • 38 | 1
  • 39 | 4
  • 40 | 3
  • 41 | 4
  • 42 | 8

1. Mean Calculation:

The mean is given by the formula: $\text{Mean} = \frac{\sum (\text{data} \times \text{frequency})}{\sum \text{frequency}}$

• $\sum (\text{data} \times \text{frequency}) = (38 \times 1) + (39 \times 4) + (40 \times 3) + (41 \times 4) + (42 \times 8) = 38 + 156 + 120 + 164 + 336 = 814$
• $\sum \text{frequency} = 1 + 4 + 3 + 4 + 8 = 20$

So, the mean is: $\text{Mean} = \frac{814}{20} = 40.7$

2. Median Calculation:

To find the median, we need to arrange the data in order (which is already done) and find the middle value(s). With a total frequency of 20, the median is the average of the 10th and 11th values in the sorted data set.

• Cumulative frequencies:
  • 38: 1
  • 39: 5 (1 + 4)
  • 40: 8 (5 + 3)
  • 41: 12 (8 + 4)
  • 42: 20 (12 + 8)

The 10th and 11th values lie in the data category 41.

Thus, the median is 41.

3. Mode Calculation:

The mode is the data value with the highest frequency.

• Data 42 has the highest frequency of 8.

So, the mode is 42.

Summary of Results:

• Mean = 40.7
• Median = 41
• Mode = 42

Would you like any more explanations or have any questions about these calculations?

Here are some related questions you might be interested in:

1. How does the weighted mean differ from the mean of ungrouped data?
2. Can there be more than one mode for a data set? If so, what does it imply?
3. What happens to the mean if the frequencies are doubled?
4. How does the median change if a new data point is added with a low frequency?
5. How do outliers affect the mean compared to the median?

Tip:

The median is particularly useful when dealing with skewed data distributions, as it is less affected by extreme values compared to the mean.