Question

A binomial experiment has the given number of trials N and the given success probability P.
n= 13, p= 0.8
determine the probability(6) round to 4 decimal places

Answer 1

No the answer is .0058

now find the mean. round to 2 decimal places

Answer 2

using P(6) as 0.0058,
(c)Find the variance and standard deviation. Round the variance to two decimal places and standard deviation to at least three decimal places.

Answer 3

To calculate the probability of getting exactly \( k = 6 \) successes in a binomial experiment with \( n = 13 \) trials and a success probability \( p = 0.8 \), you can use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
- \( p \) is the probability of success
- \( n \) is the number of trials
- \( k \) is the number of successes

Given:
- \( n = 13 \)
- \( p = 0.8 \)
- \( k = 6 \)

1. Calculate \( \binom{13}{6} \):

\[
\binom{13}{6} = \frac{13!}{6!(13-6)!} = \frac{13!}{6!7!} = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 1716
\]

2. Calculate \( p^6 \) and \( (1-p)^{n-k} \):

\[
p^6 = (0.8)^6 = 0.262144
\]

\[
1-p = 0.2 \quad \text{and} \quad (1-p)^{n-k} = (0.2)^{7} = 0.000128
\]

3. Now, plug in the values into the binomial formula:

\[
P(X = 6) = \binom{13}{6} p^6 (1-p)^{7}
\]

\[
P(X = 6) = 1716 \times 0.262144 \times 0.000128
\]

4. Calculate the probability:

\[
P(X = 6) = 1716 \times 0.262144 \times 0.000128 \approx 0.0573
\]

After rounding to four decimal places, the final answer is:

\[
\boxed{0.0573}
\]

Answer 4

I apologize for the confusion earlier. Let's first confirm that the calculation was correct.

### Binomial Probability Calculation

For \( n = 13 \) and \( p = 0.8 \):

The probability of getting exactly \( k = 6 \) successes in a binomial distribution is calculated as:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

1. **Calculate the binomial coefficient:**

\[
\binom{13}{6} = \frac{13!}{6!7!} = 1716
\]

2. **Calculate \( p^k \) and \( (1-p)^{n-k} \):**

\[
p^6 = (0.8)^6 = 0.262144
\]

\[
(1-p)^{7} = (0.2)^7 = 0.000128
\]

3. **Now calculate the probability:**

\[
P(X = 6) = 1716 \times 0.262144 \times 0.000128 \approx 0.0573
\]

If you have found \( P(X = 6) = 0.0058 \) through a different means or calculation, please let me know as I would like to help and clarify.

### Mean of the Binomial Distribution

The mean \( \mu \) of a binomial distribution is calculated using the formula:

\[
\mu = n \times p
\]

For the given values:

\[
\mu = 13 \times 0.8 = 10.4
\]

Thus, rounding to two decimal places, the mean is:

\[
\boxed{10.40}
\]

Answer 5

To find the variance and standard deviation of a binomial distribution, we can use the following formulas:

1. **Variance (\(\sigma^2\))** of a binomial distribution is given by:

\[
\sigma^2 = n \cdot p \cdot (1 - p)
\]

2. **Standard Deviation (\(\sigma\))** is the square root of the variance:

\[
\sigma = \sqrt{\sigma^2}
\]

### Given Values:
- \( n = 13 \)
- \( p = 0.8 \)
- \( 1 - p = 0.2 \)

### Step 1: Calculate the Variance

\[
\sigma^2 = 13 \cdot 0.8 \cdot 0.2
\]

Calculating this gives:

\[
\sigma^2 = 13 \cdot 0.8 \cdot 0.2 = 13 \cdot 0.16 = 2.08
\]

### Step 2: Round the Variance

The variance, rounded to two decimal places, is:

\[
\sigma^2 \approx 2.08
\]

### Step 3: Calculate the Standard Deviation

Now we'll calculate the standard deviation:

\[
\sigma = \sqrt{2.08} \approx 1.440
\]

### Step 4: Round the Standard Deviation

The standard deviation, rounded to at least three decimal places, is:

\[
\sigma \approx 1.440
\]

### Summary Results:

- Variance: \(\boxed{2.08}\)
- Standard Deviation: \(\boxed{1.440}\)