Seven baseball players are trying out for a team and the coach wants to see who can throw a baseball the farthest. The distance each player threw the baseball is recorded in the list below.

74 yards,  88 yards,  94 yards,  80 yards,  95 yards,  102 yards,  91 yards
•  Determine the median and mean distances for the 7 throws, to the nearest yard. Show how you got your answers.

The coach allowed two additional players to try out for the team, and each player threw the ball a distance of 91 yards.

•  What effect will these two additional distances have on the median and mean distances you calculated? Show your work or explain how you got your answers.

Let's solve this problem step-by-step.

### Step 1: Calculate the Median and Mean of the Initial 7 Distances

Given distances are:
\[
74, \ 88, \ 94, \ 80, \ 95, \ 102, \ 91 \ \text{yards}
\]

#### 1.1 Calculating the Median

To find the median, we first need to sort the distances in ascending order:

\[
74, \ 80, \ 88, \ 91, \ 94, \ 95, \ 102
\]

Since there are 7 numbers, the median is the middle number, which is the 4th value in the sorted list.

\[
\text{Median} = 91 \ \text{yards}
\]

#### 1.2 Calculating the Mean

The mean (average) is calculated by summing all the values and dividing by the number of values.

\[
\text{Mean} = \frac{74 + 80 + 88 + 91 + 94 + 95 + 102}{7}
\]

Calculating the sum:

\[
74 + 80 + 88 + 91 + 94 + 95 + 102 = 624
\]

Now, divide by 7:

\[
\text{Mean} = \frac{624}{7} \approx 89 \ \text{yards}
\]

So, the median is **91 yards**, and the mean is **89 yards**.

### Step 2: Effect of Adding Two More Distances of 91 Yards

The new distances are:

\[
74, \ 80, \ 88, \ 91, \ 94, \ 95, \ 102, \ 91, \ 91
\]

#### 2.1 Recalculating the Median

First, sort the new list:

\[
74, \ 80, \ 88, \ 91, \ 91, \ 91, \ 94, \ 95, \ 102
\]

With 9 values, the median is the 5th value:

\[
\text{Median} = 91 \ \text{yards}
\]

The median remains **91 yards**.

#### 2.2 Recalculating the Mean

Now, we recalculate the mean with the two additional throws:

\[
\text{New Mean} = \frac{74 + 80 + 88 + 91 + 91 + 91 + 94 + 95 + 102}{9}
\]

Calculating the new sum:

\[
74 + 80 + 88 + 91 + 91 + 91 + 94 + 95 + 102 = 806
\]

Now, divide by 9:

\[
\text{New Mean} = \frac{806}{9} \approx 90 \ \text{yards}
\]

The mean increases to **90 yards**.

### Conclusion

- **Median**: Remains the same at **91 yards**.
- **Mean**: Increases from **89 yards** to **90 yards**.

### Summary
Adding two distances that are equal to the original median did not change the median, but it slightly increased the mean because the added values were closer to the upper half of the original data set.

Would you like more details on any part of this solution, or do you have any questions?

Here are 5 related questions for further exploration:
1. What is the mode of the updated list of distances?
2. How would the mean change if the additional throws were 85 yards instead?
3. What is the range of the original list and the updated list?
4. How does adding more values close to the mean affect the standard deviation?
5. If another throw was added at 100 yards, how would the mean and median change?

**Tip:** When analyzing data, it's essential to understand how additional data points can affect measures like the mean and median to interpret results correctly.

Seven baseball players are trying out for a team and the coach wants to see who can throw a baseball the farthest. The distance each player threw the baseball is recorded in the list below. 74 yards, 88 yards, 94 yards, 80 yards, 95 yards, 102 yards, 91 yards • Determine the median and mean distances for the 7 throws, to the nearest yard. Show how you got your answers. The coach allowed two additional players to try out for the team, and each player threw the ball a distance of 91 yards. • What effect will these two additional distances have on the median and mean distances you calculated? Show your work or explain how you got your answers.

Learn how to calculate the mean and median of baseball throw distances with a step-by-step solution. The problem includes distances like 74, 88, 91, and 102 yards. Explore how adding more data points affects the mean and median, suitable for students in Grades 6-8.

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