https://file.aiquickdraw.com/mathbot/file/172628534203679l8meow.jpg

https://file.aiquickdraw.com/mathbot/file/1726285342037h93yslqo.jpg

Let's break down the two images you uploaded and provide guidance:

### Image 1 (Sentence with Drop-downs):
It seems to be asking about two statistical measures: one that decreases more in the presence of outliers and another that is resistant to outliers. Here's the solution:

- The **mean** decreases more than the **median** because the **mean** is not resistant to outliers, while the **median** is resistant to outliers.

This is because outliers (extremely high or low values) disproportionately affect the mean, pulling it toward the outlier values. The median, being the middle value, remains unchanged as long as the outliers do not change the order of the data.

### Image 2 (Quiz Grades Problem):
This image presents a list of Joey's quiz grades and asks for the following calculations:

1. **Mean quiz grade of the first 13 quizzes:**
   To find the mean, sum the 13 quiz grades and divide by 13:
   - Grades: 82, 93, 77, 79, 90, 82, 85, 85, 95, 73, 79, 83, 89
   - Sum: \(82 + 93 + 77 + 79 + 90 + 82 + 85 + 85 + 95 + 73 + 79 + 83 + 89 = 1092\)
   - Mean: \( \frac{1092}{13} \approx 84\)

2. **Median after an unexcused absence (14th quiz is 0):**
   Adding a zero to the data set affects the median. Here's the updated list:
   - Grades: 0, 73, 77, 79, 79, 82, 82, 83, 85, 85, 89, 90, 93, 95
   The median is the average of the 7th and 8th values: \( \frac{82 + 83}{2} = 82.5 \).

3. **Mean after the unexcused absence (14th quiz is 0):**
   - New sum: \(1092 + 0 = 1092\)
   - New mean: \( \frac{1092}{14} \approx 78\)

Would you like a more detailed explanation of any of these steps? 

### Related Questions:
1. Why does the mean change more than the median when there are outliers?
2. How do you find the mean for a larger data set?
3. What is the difference between resistant and non-resistant statistical measures?
4. How does adding a score of 0 (like Joey's unexcused absence) affect central tendency?
5. Can you identify when it’s more appropriate to use the median instead of the mean?

### Tip:
The median is especially useful when data contains outliers because it provides a better sense of the "central" value without being skewed by extremes.

Why does the mean change more than the median when there are outliers? Also, calculate the mean and median for a set of quiz grades including an outlier score.

Understand why the mean is more affected by outliers than the median through an analysis of quiz grades. Calculate the mean and median before and after including an outlier score, with a step-by-step breakdown. Suitable for students in Grades 8-10.

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