Question

find the mean \\(\bar{x}\\), median \\(\tilde{x}\\), and mode \\(\hat{x}\\).

1. number of students in six classes

\\34 \quad 35 \quad 34 \quad 35 \quad 36 \quad 36\\

1. costs (\\(\mathbb{p}\\)) of the same bottle of fresh milk from different stores

\\185.00 \quad 180.00 \quad 195.50 \quad 198.70 \quad 196.25\\

1. costs (\\(\mathbb{p}\\)) of lunch of george for five school days

\\95.00 \quad 120.00 \quad 99.00 \quad 135.00 \quad 100.00\\

1. number of students in each class in a small school

\\20 \quad 15 \quad 20 \quad 22 \quad 18 \quad 17 \quad 19 \quad 19 \quad 20 \quad 21 \quad 20 \quad 20\\

1. age in years of players in a kiddie basketball league

\\9 \quad 10 \quad 10 \quad 12 \quad 9 \quad 10 \quad 12 \quad 11 \quad 10 \quad 12\\

Explanation:

🆕 New Concept Discovered: Measures of Central Tendency
Finding the middle, average, or most common value in a dataset.

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Step 1: Solve Problem 1

Dataset: \(34, 35, 34, 35, 36, 36\)

• Mean (\(\bar{x}\)): Add all values together and divide by the total number of values (\(n = 6\)).

\[
\bar{x} = \frac{34 + 35 + 34 + 35 + 36 + 36}{6} = \frac{210}{6} = 35
\]

• Median (\(\tilde{x}\)): Arrange the data in ascending order: \(34, 34, 35, 35, 36, 36\). Since there is an even number of values, find the average of the two middle values (3rd and 4th values):

\[
\tilde{x} = \frac{35 + 35}{2} = 35
\]

• Mode (\(\hat{x}\)): The value(s) that appear most frequently. Here, \(34\), \(35\), and \(36\) each appear exactly twice. This dataset is trimodal:

\[
\hat{x} = 34, 35, \text{ and } 36
\]

---

Step 2: Solve Problem 2

Dataset: \(185.00, 180.00, 195.50, 198.70, 196.25\) (\(n = 5\))

• Mean (\(\bar{x}\)):

\[
\bar{x} = \frac{185.00 + 180.00 + 195.50 + 198.70 + 196.25}{5} = \frac{955.45}{5} = 191.09
\]

• Median (\(\tilde{x}\)): Arrange in ascending order: \(180.00, 185.00, 195.50, 196.25, 198.70\). The middle (3rd) value is:

\[
\tilde{x} = 195.50
\]

• Mode (\(\hat{x}\)): No value is repeated:

\[
\hat{x} = \text{No mode}
\]

---

Step 3: Solve Problem 3

Dataset: \(95.00, 120.00, 99.00, 135.00, 100.00\) (\(n = 5\))

• Mean (\(\bar{x}\)):

\[
\bar{x} = \frac{95.00 + 120.00 + 99.00 + 135.00 + 100.00}{5} = \frac{549.00}{5} = 109.80
\]

• Median (\(\tilde{x}\)): Arrange in ascending order: \(95.00, 99.00, 100.00, 120.00, 135.00\). The middle (3rd) value is:

\[
\tilde{x} = 100.00
\]

• Mode (\(\hat{x}\)): No value is repeated:

\[
\hat{x} = \text{No mode}
\]

---

Step 4: Solve Problem 4

Dataset: \(20, 15, 20, 22, 18, 17, 19, 19, 20, 21, 20, 20\) (\(n = 12\))

• Mean (\(\bar{x}\)):

\[
\bar{x} = \frac{20 + 15 + 20 + 22 + 18 + 17 + 19 + 19 + 20 + 21 + 20 + 20}{12} = \frac{231}{12} = 19.25
\]

• Median (\(\tilde{x}\)): Arrange in ascending order: \(15, 17, 18, 19, 19, 20, 20, 20, 20, 20, 21, 22\). The two middle values (6th and 7th) are both \(20\):

\[
\tilde{x} = \frac{20 + 20}{2} = 20
\]

• Mode (\(\hat{x}\)): The value \(20\) appears most frequently (5 times):

\[
\hat{x} = 20
\]

---

Step 5: Solve Problem 5

Dataset: \(9, 10, 10, 12, 9, 10, 12, 11, 10, 12\) (\(n = 10\))

• Mean (\(\bar{x}\)):

\[
\bar{x} = \frac{9 + 10 + 10 + 12 + 9 + 10 + 12 + 11 + 10 + 12}{10} = \frac{105}{10} = 10.5
\]

• Median (\(\tilde{x}\)): Arrange in ascending order: \(9, 9, 10, 10, 10, 10, 11, 12, 12, 12\). The two middle values (5th and 6th) are both \(10\):

\[
\tilde{x} = \frac{10 + 10}{2} = 10
\]

• Mode (\(\hat{x}\)): The value \(10\) appears most frequently (4 times):

\[
\hat{x} = 10
\]

---

Answer:

1. Mean (\(\bar{x}\)): \(35\), Median (\(\tilde{x}\)): \(35\), Mode (\(\hat{x}\mathbb{}\)): \(34, 35, 36\)
2. Mean (\(\bar{x}\)): \(191.09\), Median (\(\tilde{x}\)): \(195.50\), Mode (\(\hat{x}\)): No mode
3. Mean (\(\bar{x}\)): \(109.80\), Median (\(\tilde{x}\)): \(100.00\), Mode (\(\hat{x}\)): No mode
4. Mean (\(\bar{x}\)): \(19.25\), Median (\(\tilde{x}\)): \(20\), Mode (\(\hat{x}\)): \(20\)
5. Mean (\(\bar{x}\)): \(10.5\), Median (\(\tilde{x}\)): \(10\), Mode (\(\hat{x}\)): \(10\)