To find the mean, median, and mode from the given frequency distribution histogram, we first extract the necessary data from the histogram.

Frequency Table (based on the histogram):

• $x = 1$, frequency $f = 1$
• $x = 2$, frequency $f = 3$
• $x = 3$, frequency $f = 4$
• $x = 4$, frequency $f = 2$
• $x = 5$, frequency $f = 1$

1. Mean:

The mean is calculated as:

$\text{Mean} = \frac{\sum (x \times f)}{\sum f}$

Steps:

• Sum of $x \times f$: $(1 \times 1) + (2 \times 3) + (3 \times 4) + (4 \times 2) + (5 \times 1) = 1 + 6 + 12 + 8 + 5 = 32$
• Sum of frequencies ($\sum f$): $1 + 3 + 4 + 2 + 1 = 11$
• Now, the mean is: $\text{Mean} = \frac{32}{11} \approx 2.91$

2. Median:

To find the median, we first need the cumulative frequencies:

• Cumulative frequency at $x = 1$: 1
• Cumulative frequency at $x = 2$: 1 + 3 = 4
• Cumulative frequency at $x = 3$: 4 + 4 = 8
• Cumulative frequency at $x = 4$: 8 + 2 = 10
• Cumulative frequency at $x = 5$: 10 + 1 = 11

Since the total frequency is 11 (odd), the median corresponds to the $\frac{11 + 1}{2} = 6$-th value.

Looking at the cumulative frequencies, the 6th value is within the category where $x = 3$. Therefore, the median is 3.

3. Mode:

The mode is the value of $x$ that has the highest frequency.

From the table, $x = 3$ has the highest frequency of 4. Hence, the mode is 3.

Summary:

• Mean: $\approx 2.91$
• Median: $3$
• Mode: $3$

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

Here are 5 related questions to deepen your understanding:

1. How do changes in the frequency distribution affect the mean?
2. What happens to the median if the frequencies of two middle values are equal?
3. How is the mean influenced by extremely high or low values?
4. What are the conditions for a dataset to have no mode?
5. How do you determine the quartiles from a frequency distribution?

Tip: The median is often more reliable than the mean when dealing with skewed distributions since it is less affected by extreme values.