Five coins are tossed 320 times find the frequencies of the distribution of number of heads and tabulate the result
Answer
To find the frequencies of the distribution of the number of heads when five coins are tossed 320 times, we need to calculate the number of occurrences for each possible outcome: 0, 1, 2, 3, 4, and 5 heads.
Here's a table to tabulate the results:
| Number of Heads | Frequency |
|-----------------|-----------|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
To calculate the frequencies, we can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials (320 in this case)
- k is the number of successful events (number of heads)
- C(n, k) is the number of combinations of n items taken k at a time (binomial coefficient)
- p is the probability of a single successful event (0.5 for a fair coin)
Let's calculate the frequencies for each outcome:
For 0 heads:
- k = 0
- P(X = 0) = C(320, 0) * 0.5^0 * (1-0.5)^(320-0)
- P(X = 0) = 1 * 1 * 0.5^320
- P(X = 0) ≈ 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009
Since the probability is extremely close to zero, we can assume it to be zero for practical purposes. Therefore, the frequency for 0 heads is 0.
For 1 head:
- k = 1
- P(X = 1) = C(320, 1) * 0.5^1 * (1-0.5)^(320-1)
- P(X = 1) = 320 * 0.5 * 0.5^319
- P(X = 1) ≈ 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003
Again, since the probability is extremely close to zero, we can assume it to be zero for practical purposes. Therefore, the frequency for 1 head is 0.
For 2 heads:
- k = 2
- P(X = 2) = C(320, 2) * 0.5^2 * (1-0.5)^(320-2)
- P(X = 2) = (320*319/2) * 0.25 * 0.5^318
- P(X = 2) ≈ 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002
Again, the probability is extremely close to zero, so for practical purposes, the frequency for 2 heads is 0.
For 3 heads:
- k = 3
- P(X = 3) = C(320, 3) * 0.5^3 * (1-0.5)^(320-3)
- P(X = 3) = (320*319*318/6) * 0.125 * 0.5^317
- P(X = 3) ≈ 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
The frequency for 3 heads can also be assumed to be zero for practical purposes.
For 4 heads:
- k = 4
- P(X = 4) = C(320, 4) * 0.5^4 * (1-0.5)^(320-4)
- P(X = 4) = (320*319*318*317/24) * 0.0625 * 0.5^316
- P(X = 4) ≈ 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009
Again, we can assume the frequency for 4 heads to be zero.
For 5 heads:
- k = 5
- P(X = 5) = C(320, 5) * 0.5^5 * (1-0.5)^(320-5)
- P(X = 5) = (320*319*318*317*316/120) * 0.03125 * 0.5^315
- P(X = 5) ≈ 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000009
We can assume the frequency for 5 heads to be approximately 1, as the probability is extremely close to 1.
Updated frequency table:
| Number of Heads | Frequency |
|-----------------|-----------|
| 0 | 0 |
| 1 | 0 |
| 2 | 0 |
| 3 | 0 |
| 4 | 0 |
| 5 | 1 |
Therefore, the frequencies of the distribution of the number of heads when five coins are tossed 320 times are 0 for all outcomes except 5, which has a frequency of 1.
To find the frequencies of the distribution of the number of heads when tossing five coins 320 times, you need to simulate the coin toss experiment and record the number of heads obtained in each trial. Then you can tabulate the results to obtain the frequency distribution.
Here's a step-by-step approach to simulate the experiment and calculate the frequencies:
1. Initialize a list or an array to store the number of heads in each trial.
2. Repeat the following steps 320 times:
a. Toss five coins simultaneously.
b. Count the number of heads obtained.
c. Add the count to the list or array from step 1.
3. Calculate the frequency of each possible number of heads (0, 1, 2, 3, 4, and 5) by counting how many times each number appears in the list or array from step 1.
4. Tabulate the results in a table, displaying the possible number of heads on one column and their corresponding frequencies in the other column.
Let's go through an example:
1. Initialize a list: `heads_list = []`.
2. Repeat the following steps 320 times:
a. Toss five coins simultaneously.
b. Count the number of heads obtained.
c. Add the count to the `heads_list`.
For instance, if you get 3 heads in a trial, add 3 to the `heads_list`.
3. Calculate the frequency of each possible number of heads:
- Count the number of times each number appears in the `heads_list`.
For example, if the 'heads_list' is [1, 2, 1, 3, 3, 4, 0, ...], you would get:
- Number of 0 heads: 1
- Number of 1 heads: 2
- Number of 2 heads: 1
- Number of 3 heads: 2
- Number of 4 heads: 1
- Number of 5 heads: 0
4. Tabulate the results:
| Number of Heads | Frequency |
|-----------------|-----------|
| 0 | 1 |
| 1 | 2 |
| 2 | 1 |
| 3 | 2 |
| 4 | 1 |
| 5 | 0 |
This table shows the frequencies of each possible number of heads obtained when tossing five coins 320 times.