https://file.aiquickdraw.com/mathbot/file/1726443509531rgxok14q.jpg

Let's analyze the histogram to answer both questions.

### Part (a): **Finding the Median Number of Servings of Fruit per Day**
1. **Identify the total number of subjects:**  
   By summing up the heights of all the bars in the histogram:
   - 0 servings: 15 subjects
   - 1 serving: 15 subjects
   - 2 servings: 12 subjects
   - 3 servings: 10 subjects
   - 4 servings: 8 subjects
   - 5 servings: 4 subjects
   - 6 servings: 3 subjects
   - 7 servings: 2 subjects
   - 8 servings: 1 subject

   Total subjects = \( 15 + 15 + 12 + 10 + 8 + 4 + 3 + 2 + 1 = 70 \)

2. **Find the median position:**  
   The median is the value at the middle position when the data is arranged in ascending order. For 70 subjects, the median will be the average of the 35th and 36th positions.

3. **Locate the median in the distribution:**  
   - The first 15 subjects are in the "0 servings" group.
   - The next 15 subjects (positions 16 to 30) are in the "1 serving" group.
   - The next 12 subjects (positions 31 to 42) are in the "2 servings" group.

   Since both the 35th and 36th subjects fall into the "2 servings" group, **the median number of servings is 2**.

### Part (b): **Calculating the Mean of the Distribution**
The mean (or average) is calculated by summing the products of each value (number of servings) and its frequency, then dividing by the total number of subjects.

Mean = \[ \frac{\sum{(x_i \cdot f_i)}}{N} \], where \(x_i\) is the number of servings, \(f_i\) is the frequency, and \(N\) is the total number of subjects.

\[
\text{Mean} = \frac{(0 \cdot 15) + (1 \cdot 15) + (2 \cdot 12) + (3 \cdot 10) + (4 \cdot 8) + (5 \cdot 4) + (6 \cdot 3) + (7 \cdot 2) + (8 \cdot 1)}{70}
\]

\[
= \frac{0 + 15 + 24 + 30 + 32 + 20 + 18 + 14 + 8}{70} = \frac{161}{70} \approx 2.3
\]

So, the **mean number of servings of fruit per day is approximately 2.3**.

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Would you like further clarification on any part of this solution?

### Here are five related questions:
1. How would the median change if the number of subjects with "0 servings" increased by 5?
2. What does the mean represent in this histogram?
3. How can you calculate the mode from a histogram like this one?
4. If the distribution were skewed, how would you expect the mean and median to compare?
5. How would the histogram look if the variance in fruit servings increased?

### Tip:
In a histogram, the **median** is less sensitive to extreme values than the **mean**, making it a better measure of central tendency for skewed distributions.

Find the median number of servings of fruit per day from the histogram. Explain your method clearly. Calculate the mean of the distribution. Show your work.

Learn how to calculate the median and mean number of fruit servings per day using a histogram. This problem covers key statistical concepts like central tendency and distribution analysis. Suitable for students in Grades 6-8.

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