Question

1 flipping ten coins

1. flip ten coins ten times and record the number of heads in each ten flip trial.

record them here:

1	2	3	4	5	6	7	8	9	10
3	2	4	5	6	5	5	2	3	6

1. calculate the arithmetic mean of your data.

\\( \langle x \
angle = \frac{1}{n} \times \sum_{i=1}^{n} x_i \\)\\( \langle h \
angle = \\)

1. draw a histogram as in fig. 8.1b in the text.
2. calculate the mean of your data from the histogram as demonstrated in example 8.4. show your steps below.

Explanation:

Part 2: Calculate the arithmetic mean

Step 1: Identify the data values

The data values (number of heads in each trial) are: \( 3, 2, 4, 5, 6, 5, 5, 2, 3, 6 \)

Step 2: Sum the data values

Sum \( = 3 + 2 + 4 + 5 + 6 + 5 + 5 + 2 + 3 + 6 \)
\( = (3+2)+(4+5)+(6+5)+(5+2)+(3+6) \)
\( = 5 + 9 + 11 + 7 + 9 \)
\( = 5 + 9 = 14; 14 + 11 = 25; 25 + 7 = 32; 32 + 9 = 41 \)
Wait, let's recalculate the sum correctly:
\( 3 + 2 = 5 \)
\( 5 + 4 = 9 \)
\( 9 + 5 = 14 \)
\( 14 + 6 = 20 \)
\( 20 + 5 = 25 \)
\( 25 + 5 = 30 \)
\( 30 + 2 = 32 \)
\( 32 + 3 = 35 \)
\( 35 + 6 = 41 \)
Wait, no, let's add all numbers one by one:
\( 3, 2, 4, 5, 6, 5, 5, 2, 3, 6 \)
Let's list them: 3, 2, 4, 5, 6, 5, 5, 2, 3, 6.
Count the number of terms: \( n = 10 \)
Sum: \( 3 + 2 = 5; 5 + 4 = 9; 9 + 5 = 14; 14 + 6 = 20; 20 + 5 = 25; 25 + 5 = 30; 30 + 2 = 32; 32 + 3 = 35; 35 + 6 = 41 \). Wait, that's 9 terms? No, wait, there are 10 terms: positions 1 - 10. Let's list all 10:
1: 3
2: 2
3: 4
4: 5
5: 6
6: 5
7: 5
8: 2
9: 3
10: 6
Now sum: \( 3 + 2 + 4 + 5 + 6 + 5 + 5 + 2 + 3 + 6 \)
Let's group like terms:
Number of 2s: 2 (positions 2, 8) → \( 2×2 = 4 \)
Number of 3s: 2 (positions 1, 9) → \( 3×2 = 6 \)
Number of 4s: 1 (position 3) → \( 4×1 = 4 \)
Number of 5s: 3 (positions 4, 6, 7) → \( 5×3 = 15 \)
Number of 6s: 2 (positions 5, 10) → \( 6×2 = 12 \)
Now sum these: \( 4 + 6 + 4 + 15 + 12 = 41 \)? Wait, no: \( 4 + 6 = 10; 10 + 4 = 14; 14 + 15 = 29; 29 + 12 = 41 \). Wait, but 2+2=4, 3+3=6, 4=4, 5+5+5=15, 6+6=12. Total: 4+6=10, +4=14, +15=29, +12=41. But there are 10 terms, so sum is 41? Wait, let's add again: 3+2=5, +4=9, +5=14, +6=20, +5=25, +5=30, +2=32, +3=35, +6=41. Yes, sum is 41.

Step 3: Calculate the mean

The formula for the arithmetic mean \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \), where \( n = 10 \) and \( \sum x_i = 41 \)
So \( \bar{x} = \frac{41}{10} = 4.1 \)

Step 1: Calculate \( x_i \times f_i \) for each \( x_i \)

• For \( x = 2 \): \( 2 \times 2 = 4 \)
• For \( x = 3 \): \( 3 \times 2 = 6 \)
• For \( x = 4 \): \( 4 \times 1 = 4 \)
• For \( x = 5 \): \( 5 \times 3 = 15 \)
• For \( x = 6 \): \( 6 \times 2 = 12 \)

Step 2: Sum the \( x_i \times f_i \) values

Sum \( = 4 + 6 + 4 + 15 + 12 = 41 \) (same as before)

Step 3: Sum the frequencies \( \sum f_i \)

\( \sum f_i = 2 + 2 + 1 + 3 + 2 = 10 \) (total number of trials)

Step 4: Calculate the mean

\( \bar{x} = \frac{41}{10} = 4.1 \)

Answer:

The arithmetic mean \( \langle h
angle = 4.1 \)

Part 4: Calculate the mean from the histogram (assuming we first create the histogram)

First, we need to create the histogram. Let's find the frequency of each number of heads:

• Number of heads = 2: frequency \( f_2 = 2 \) (trials 2, 8)
• Number of heads = 3: frequency \( f_3 = 2 \) (trials 1, 9)
• Number of heads = 4: frequency \( f_4 = 1 \) (trial 3)
• Number of heads = 5: frequency \( f_5 = 3 \) (trials 4, 6, 7)
• Number of heads = 6: frequency \( f_6 = 2 \) (trials 5, 10)

To calculate the mean from the histogram, we use the formula for the mean of a frequency distribution: \( \bar{x} = \frac{\sum (x_i \times f_i)}{\sum f_i} \)