This calculator helps you compute the mean, variance, and standard deviation of a binomial distribution from your input for the number of trials and the probability of success. The mean and standard deviation of the binomial distribution are useful when using the normal distribution as an approximation to the binomial distribution. You can use the Normal Distribution Calculator with the mean and standard deviation calculated here for that approximation. You’ll find the normal distribution is a good approximation for the binomial distribution when np and n(1 – p) are both greater than 5.
If you are looking for help computing the probabilities that make up a binomial distribution, have a look at Binomial Probability Calculator and Binomial Distribution Calculator With a Step By Step Solution. These calculators give you step by step solutions to the calculations so you can learn from them and do them yourself on an assessment.
Binomial Mean Variance Standard Deviation Calculator
Answer:
Mean of the binomial distribution, $\mu$: 2.45
Variance of the binomial distribution, $\sigma^2$: 1.5925
Standard deviation of the binomial distribution, $\sigma$: 1.2619429464124
P(X), where X is the number of successes
P(0) = 0.04902227890625
P(1) = 0.18477628203125
P(2) = 0.29848476328125
P(3) = 0.26787094140625
P(4) = 0.14423819921875
P(5) = 0.04660003359375
P(6) = 0.00836410859375
P(7) = 0.00064339296875
Solution:
The mean of a binomial distribution is a measure that describes the center of the distribution, and the variance of a binomial distribution is a measure that describes the spread of values in the distribution. Furthermore, we can compute the standard deviation of a binomial distribution by taking the square root of the variance. This calculator helps you find mean, variance, and standard deviation of a binomial distribution.
First, to find the mean of this binomial distribution, we could use the formula for finding the mean for any discrete distribution:
$$ \mu = {\sum}[X \cdot P(X)] $$
But for a binomial distribution, this simplifies to $$ \mu = n \cdot p $$ n is the number of trials and p is the probability of success for a single trial. Using the numbers entered into the calculator above, we have $$ \mu = 7 \cdot 0.35 $$ Evaluating the expression we have $$ \mu = 2.45 $$
To find the variance for this binomial distribution, use the formula $$ \sigma^2 = n \cdot p \cdot (1 - p) $$
Here, n is the number of trials and p is the probability of success on a single trial. Plugging in the values for this particular problem, the variance is computed as $$ \sigma^2 = 7 \cdot 0.35 \cdot (1 - 0.35) $$ Evaluating this expression, we have
$$ \sigma^2 = 1.5925 $$
Last, we can find the standard deviation of this binomial distribution by taking the square root of the variance.
$$ \sigma = \sqrt{(n \cdot p \cdot (1 - p))} $$
Substituting in 7 for n and 0.35 for p, we have
$$ \sigma = \sqrt{(7 \cdot 0.35 \cdot (1 - 0.35))} $$
Evaluating this expression, we have
$$ \sigma = \sqrt{1.5925} $$
$$ \sigma = 1.2619429464124 $$
