### Answer:

Mean of the Binomial Distribution, $\mu$: 4.55

Standard Deviation of the Binomial Distribution, $\sigma$: 1.2619429464124

P(0) = 0.00064339296875

P(1) = 0.00836410859375

P(2) = 0.04660003359375

P(3) = 0.14423819921875

P(4) = 0.26787094140625

P(5) = 0.29848476328125

P(6) = 0.18477628203125

P(7) = 0.04902227890625

### Solution:

The mean of the binomial distribution is interpreted as the mean number of successes for the distribution. To find the mean, use the formula $$ \mu = n \cdot p $$ where n is the number of trials and p is the probability of success on a single trial. Substituting values for this problem, we have $$ \mu = 7 \cdot 0.65 $$ Multiplying the expression we have $$ \mu = 4.55 $$

The standard deviation of the binomial distribution is interpreted as the standard deviation of the number of successes for the distribution. To find the standard deviation, use the formula $$ \sigma = \sqrt{n \cdot p \cdot (1 - p)} $$ where n is the umber of trials and p is the probability of success on a single trial. Substituting values fo this problem, we have $$ \sigma = \sqrt{7 \cdot 0.65 \cdot (1 - 0.65)} $$ Evaluating the expression on the right, we have $$ \sigma = \sqrt{1.5925)} $$
$$ \sigma = 1.2619429464124 $$

To complete a binomial distribution table, first identify all of the possible values of X. Since there are 7 trials, the values of X range from X = 0 to X = 7.

Next, find each individual binomial probability for each value of X. In this problem, we will be finding 8 probabilities. The sum of all these probabilities will be 1.

### $P(0)$ Probability of exactly 0 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 0 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 0 $.
$$ P(0) = \frac{7!}{0!(7-0)!} \cdot 0.65^0 \cdot (1-0.65)^{7-0} $$
Remember, 0! is 1. Evaluating the expression, we have
$$ P(0) = 0.00064339296875 $$

### $P(1)$ Probability of exactly 1 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 1 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 1 $.
$$ P(1) = \frac{7!}{1!(7-1)!} \cdot 0.65^1 \cdot (1-0.65)^{7-1} $$
Evaluating the expression, we have
$$ P(1) = 0.00836410859375 $$

### $P(2)$ Probability of exactly 2 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 2 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 2 $.
$$ P(2) = \frac{7!}{2!(7-2)!} \cdot 0.65^2 \cdot (1-0.65)^{7-2} $$
Evaluating the expression, we have
$$ P(2) = 0.04660003359375 $$

### $P(3)$ Probability of exactly 3 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 3 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 3 $.
$$ P(3) = \frac{7!}{3!(7-3)!} \cdot 0.65^3 \cdot (1-0.65)^{7-3} $$
Evaluating the expression, we have
$$ P(3) = 0.14423819921875 $$

### $P(4)$ Probability of exactly 4 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 4 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 4 $.
$$ P(4) = \frac{7!}{4!(7-4)!} \cdot 0.65^4 \cdot (1-0.65)^{7-4} $$
Evaluating the expression, we have
$$ P(4) = 0.26787094140625 $$

### $P(5)$ Probability of exactly 5 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 5 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 5 $.
$$ P(5) = \frac{7!}{5!(7-5)!} \cdot 0.65^5 \cdot (1-0.65)^{7-5} $$
Evaluating the expression, we have
$$ P(5) = 0.29848476328125 $$

### $P(6)$ Probability of exactly 6 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 6 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 6 $.
$$ P(6) = \frac{7!}{6!(7-6)!} \cdot 0.65^6 \cdot (1-0.65)^{7-6} $$
Evaluating the expression, we have
$$ P(6) = 0.18477628203125 $$

### $P(7)$ Probability of exactly 7 successes

If using a calculator, you can enter $ \text{trials} = 7 $, $ p = 0.65 $, and $ X = 7 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 7 $, $ p = 0.65 $, and $ X = 7 $.
$$ P(7) = \frac{7!}{7!(7-7)!} \cdot 0.65^7 \cdot (1-0.65)^{7-7} $$
Evaluating the expression, we have
$$ P(7) = 0.04902227890625 $$