Math Problem Statement

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.

Solution

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=109282 + 93 + 77 + 79 + 90 + 82 + 85 + 85 + 95 + 73 + 79 + 83 + 89 = 1092
    • Mean: 10921384\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: 82+832=82.5\frac{82 + 83}{2} = 82.5.
  3. Mean after the unexcused absence (14th quiz is 0):

    • New sum: 1092+0=10921092 + 0 = 1092
    • New mean: 10921478\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.

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Math Problem Analysis

Mathematical Concepts

Mean
Median
Outliers
Central Tendency

Formulas

Mean = Sum of all data points / Number of data points
Median = Middle value in a sorted data set

Theorems

The mean is sensitive to outliers, while the median is resistant to outliers.

Suitable Grade Level

Grades 8-10