Binomial Distribution Calculator
Answer:
Mean of the Binomial Distribution, μ: 3.9
Standard Deviation of the Binomial Distribution, σ: 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 μ=n⋅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 μ=6⋅0.65 Multiplying the expression we have μ=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 σ=√n⋅p⋅(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 σ=√6⋅0.65⋅(1−0.65) Evaluating the expression on the right, we have σ=√1.365) σ=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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!0!(6−0)!⋅0.650⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!1!(6−1)!⋅0.651⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!2!(6−2)!⋅0.652⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!3!(6−3)!⋅0.653⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!4!(6−4)!⋅0.654⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!5!(6−5)!⋅0.655⋅(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 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)=(nX)⋅pX⋅(1−p)n−X The binomial coefficient, (nX) is defined by (nX)=n!X!(n−X)! The full binomial probability formula with the binomial coefficient is P(X)=n!X!(n−X)!⋅pX⋅(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)=6!6!(6−6)!⋅0.656⋅(1−0.65)6−6 Evaluating the expression, we have P(6)=0.075418890625