There are exactly 5 people in a bookstore at 12:00 p.m. Each person earns an annual income that is between $ 30,000 and $35,000. No one enters or leaves the bookstroes until 12:15 p.m., when a professional athlete with an annual income of more that $ 1,000,000 enters the bookstore and joins the other 5 people. The mean, median , range and standard deviation of the annual incomes of the 5 people in the bookstore at 12:00 p.m., are calculated and compared to the same 4 statisties of the annual incomes of the 6 people in the bookstore at 12:15 p.m. If it can be determind, which of the 4 statistics changed the least?

Options

A. Range
B. Mean
C. Median
D. Standard deviation

Answer

To solve the problem, we will analyze the statistics of the incomes of the people in the bookstore before and after the athlete enters.

### Step-by-step Solution:

1. **Identify the Income Range of the First 5 People:**
Let the incomes of the 5 people (A, B, C, D, E) be represented as:
- A, B, C, D, E where \( 30,000 < A, B, C, D, E < 35,000 \).

2. **Calculate the Mean, Median, Range, and Standard Deviation for the First 5 People:**
- **Mean:** The mean is calculated as:
\[ \text{Mean} = \frac{A + B + C + D + E}{5} \]
- **Median:** Since there are 5 values, the median is the middle value:
\[ \text{Median} = C \]
- **Range:** The range is calculated as:
\[ \text{Range} = E - A \]
- **Standard Deviation:** The standard deviation is calculated using the formula:
\[ \text{Standard Deviation} = \sqrt{\frac{(A - \text{Mean})^2 + (B - \text{Mean})^2 + (C - \text{Mean})^2 + (D - \text{Mean})^2 + (E - \text{Mean})^2}{5}} \]

3. **Add the Athlete's Income:**
At 12:15 p.m., a professional athlete with an income \( X > 1,000,000 \) enters the bookstore. Now we have 6 people: A, B, C, D, E, and X.

4. **Calculate the New Statistics for the 6 People:**
- **Mean:** The new mean is:
\[ \text{New Mean} = \frac{A + B + C + D + E + X}{6} \]
- **Median:** The new median will be the average of the 3rd and 4th values in the sorted list of incomes. Since \( X \) is much larger than A, B, C, D, and E, the new median will be:
\[ \text{New Median} = \frac{C + D}{2} \]
- **Range:** The new range is:
\[ \text{New Range} = X - A \]
- **Standard Deviation:** The new standard deviation is calculated similarly:
\[ \text{New Standard Deviation} = \sqrt{\frac{(A - \text{New Mean})^2 + (B - \text{New Mean})^2 + (C - \text{New Mean})^2 + (D - \text{New Mean})^2 + (E - \text{New Mean})^2 + (X - \text{New Mean})^2}{6}} \]

5. **Compare the Changes in Statistics:**
- The **mean** will increase significantly due to the addition of \( X \).
- The **range** will also increase significantly because \( X \) is much larger than \( E \).
- The **standard deviation** will likely increase because \( X \) is an outlier.
- The **median** will not change much because \( C \) and \( D \) are still close to each other.

### Conclusion:
After analyzing the changes in the statistics, we can conclude that the statistic that changed the least is the **median**.