To solve the problem and find the mean (rata-rata) from the histogram provided, follow these steps:

Step 1: Identify the Class Intervals and Frequencies

From the histogram, we can observe the class intervals and their corresponding frequencies:

• 40–43: Frequency = 5
• 43–46: Frequency = 17
• 46–49: Frequency = 14
• 49–52: Frequency = 10
• 52–55: Frequency = 4

Step 2: Find the Midpoints of Each Class Interval

To find the mean, we first need to calculate the midpoints ($x_i$) of each class interval:

• Midpoint of 40–43: $\frac{40 + 43}{2} = 41.5$
• Midpoint of 43–46: $\frac{43 + 46}{2} = 44.5$
• Midpoint of 46–49: $\frac{46 + 49}{2} = 47.5$
• Midpoint of 49–52: $\frac{49 + 52}{2} = 50.5$
• Midpoint of 52–55: $\frac{52 + 55}{2} = 53.5$

Step 3: Calculate the Mean

The mean can be calculated using the formula:

$\text{Mean} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

Where $f_i$ is the frequency and $x_i$ is the midpoint for each class interval.

• $(41.5 \times 5) = 207.5$
• $(44.5 \times 17) = 756.5$
• $(47.5 \times 14) = 665$
• $(50.5 \times 10) = 505$
• $(53.5 \times 4) = 214$

Step 4: Sum of the Products and Frequencies

Sum of products $f_i \times x_i$: $207.5 + 756.5 + 665 + 505 + 214 = 2348$

Sum of frequencies $\sum f_i$: $5 + 17 + 14 + 10 + 4 = 50$

Step 5: Final Calculation

Now, calculate the mean: $\text{Mean} = \frac{2348}{50} = 46.96$

Answer:

The mean (rata-rata) is 46.96.

Would you like more details or have any questions?

Here are five related questions to expand on this concept:

1. How would the mean change if the frequency of the class interval 43-46 were doubled?
2. What is the median class interval, and how do you find it from this histogram?
3. How does the mean compare to the mode in this distribution?
4. How would you calculate the variance and standard deviation from this data?
5. What is the impact on the mean if an outlier is added to this data set?

Tip: Remember that the mean is sensitive to extreme values (outliers), while the median is more robust in such cases.