Binomial Distribution Calculator
Answer:
Mean of the Binomial Distribution, $\mu$: 3.25
Standard Deviation of the Binomial Distribution, $\sigma$: 1.0665364503851
P(0) = 0.0052521875
P(1) = 0.0487703125
P(2) = 0.181146875
P(3) = 0.336415625
P(4) = 0.3123859375
P(5) = 0.1160290625
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 $$ \mu = 5 \cdot 0.65 $$ Multiplying the expression we have $$ \mu = 3.25 $$
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 $$ \sigma = \sqrt{5 \cdot 0.65 \cdot (1 - 0.65)} $$ Evaluating the expression on the right, we have $$ \sigma = \sqrt{1.1375)} $$ $$ \sigma = 1.0665364503851 $$
To complete a binomial distribution table, first identify all of the possible values of X. Since there are 5 trials, the values of X range from X = 0 to X = 5.
Next, find each individual binomial probability for each value of X. In this problem, we will be finding 6 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} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 0 $. $$ P(0) = \frac{5!}{0!(5-0)!} \cdot 0.65^0 \cdot (1-0.65)^{5-0} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.0052521875 $$
$P(1)$ Probability of exactly 1 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 1 $. $$ P(1) = \frac{5!}{1!(5-1)!} \cdot 0.65^1 \cdot (1-0.65)^{5-1} $$ Evaluating the expression, we have $$ P(1) = 0.0487703125 $$
$P(2)$ Probability of exactly 2 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 2 $. $$ P(2) = \frac{5!}{2!(5-2)!} \cdot 0.65^2 \cdot (1-0.65)^{5-2} $$ Evaluating the expression, we have $$ P(2) = 0.181146875 $$
$P(3)$ Probability of exactly 3 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 3 $. $$ P(3) = \frac{5!}{3!(5-3)!} \cdot 0.65^3 \cdot (1-0.65)^{5-3} $$ Evaluating the expression, we have $$ P(3) = 0.336415625 $$
$P(4)$ Probability of exactly 4 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 4 $. $$ P(4) = \frac{5!}{4!(5-4)!} \cdot 0.65^4 \cdot (1-0.65)^{5-4} $$ Evaluating the expression, we have $$ P(4) = 0.3123859375 $$
$P(5)$ Probability of exactly 5 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ 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 = 5 $, $ p = 0.65 $, and $ X = 5 $. $$ P(5) = \frac{5!}{5!(5-5)!} \cdot 0.65^5 \cdot (1-0.65)^{5-5} $$ Evaluating the expression, we have $$ P(5) = 0.1160290625 $$