https://file.aiquickdraw.com/mathbot/file/1729874293097y5jf5grh.jpg

Let's analyze the histogram to calculate the mean and median number of days missed by Mr. Clark's students.

### Step 1: Frequency Distribution
From the histogram, we can extract the following dataLet's analyze the histogram to calculate the mean and median number of days missed by Mr. Clark's students.

### Step 1: Frequency Distribution
From the histogram, we can extract the following 

- 3 students missed **0 days**
- 2 students missed **1 day**
- 4 students missed **2 days**
- 1 student missed **3 days**
- 0 students missed **4 days**
- 5 students missed **5 days**

Thus, the data set of missed days is:
\[ \{ 0, 0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 5, 5, 5, 5 \} \]

### Step 2: Calculate the Mean
The mean is calculated as the sum of all values divided by the total number of students.

- Total number of students = 15
- Sum of missed days = \( (0 \times 3) + (1 \times 2) + (2 \times 4) + (3 \times 1) + (5 \times 5) = 0 + 2 + 8 + 3 + 25 = 38 \)

So, the mean is:
\[
\text{Mean} = \frac{38}{15}
\]

### Step 3: Calculate the Median
To find the median, we need to order the data and find the middle value. Since there are 15 students, the 8th value in the ordered data is the median.

The ordered data set is:
\[ \{ 0, 0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 5, 5, 5, 5 \} \]

The 8th value is **2**. Thus, the median is 2.

### Step 4: Difference Between Mean and Median
We now compute the difference between the mean and the median:
\[
\text{Difference} = \frac{38}{15} - 2 = \frac{38}{15} - \frac{30}{15} = \frac{8}{15}
\]

Thus, the mean number of days missed per student is **\(\frac{8}{15}\)** greater than the median number of days missed.

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Would you like more details or have any questions? Here are some related questions:

1. How would the median change if one student missed one additional day?
2. What effect does a large number of students missing 5 days have on the mean?
3. Can you describe how to find the mode from this histogram?
4. What happens to the mean if the number of students missing 5 days increases?
5. Why is the median less sensitive to extreme values compared to the mean?

**Tip:** The mean can be greatly affected by extreme values (like the five students missing 5 days), while the median remains more stable when dealing with such outliers.

For the data whose frequency histogram is shown, by how many days is the mean number of days missed per student greater than the median number of days missed per student for the 15 students?

Analyze how to calculate the difference between the mean and median number of days missed by Mr. Clark's students using a frequency histogram. This problem covers essential statistical concepts such as mean, median, and frequency distribution, suitable for students in Grades 7-9.

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