Mean of the Binomial Distribution, $\mu$: 3.9
Standard Deviation of the Binomial Distribution, $\sigma$: 1.1683321445548

P(0) = 0.001838265625
P(1) = 0.02048353125
P(2) = 0.095102109375
P(3) = 0.2354909375
P(4) = 0.328005234375
P(5) = 0.24366103125
P(6) = 0.075418890625

### 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 $$\mu = 6 \cdot 0.65$$ Multiplying the expression we have $$\mu = 3.9$$

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 $$\sigma = \sqrt{6 \cdot 0.65 \cdot (1 - 0.65)}$$ Evaluating the expression on the right, we have $$\sigma = \sqrt{1.365)}$$ $$\sigma = 1.1683321445548$$

To complete a binomial distribution table, first identify all of the possible values of X. Since there are 6 trials, the values of X range from X = 0 to X = 6.

Next, find each individual binomial probability for each value of X. In this problem, we will be finding 7 probabilities. The sum of all these probabilities will be 1.

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 0$. $$P(0) = \frac{6!}{0!(6-0)!} \cdot 0.65^0 \cdot (1-0.65)^{6-0}$$ Remember, 0! is 1. Evaluating the expression, we have $$P(0) = 0.001838265625$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 1$. $$P(1) = \frac{6!}{1!(6-1)!} \cdot 0.65^1 \cdot (1-0.65)^{6-1}$$ Evaluating the expression, we have $$P(1) = 0.02048353125$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 2$. $$P(2) = \frac{6!}{2!(6-2)!} \cdot 0.65^2 \cdot (1-0.65)^{6-2}$$ Evaluating the expression, we have $$P(2) = 0.095102109375$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 3$. $$P(3) = \frac{6!}{3!(6-3)!} \cdot 0.65^3 \cdot (1-0.65)^{6-3}$$ Evaluating the expression, we have $$P(3) = 0.2354909375$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 4$. $$P(4) = \frac{6!}{4!(6-4)!} \cdot 0.65^4 \cdot (1-0.65)^{6-4}$$ Evaluating the expression, we have $$P(4) = 0.328005234375$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 5$. $$P(5) = \frac{6!}{5!(6-5)!} \cdot 0.65^5 \cdot (1-0.65)^{6-5}$$ Evaluating the expression, we have $$P(5) = 0.24366103125$$

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

If using a calculator, you can enter $\text{trials} = 6$, $p = 0.65$, 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, $n = 6$, $p = 0.65$, and $X = 6$. $$P(6) = \frac{6!}{6!(6-6)!} \cdot 0.65^6 \cdot (1-0.65)^{6-6}$$ Evaluating the expression, we have $$P(6) = 0.075418890625$$