Question 1191315: show that mean is 3p and variance is 3pq of binomial distribution
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! You're asking to show that for a binomial distribution with *n* = 3 trials, the mean is 3*p* and the variance is 3*p*q*, where *p* is the probability of success on a single trial and *q* = 1 - *p* is the probability of failure.
Here's the proof:
**1. Probability Mass Function:**
The probability mass function for a binomial distribution is given by:
P(X = k) = (nCk) * p^k * q^(n-k)
where nCk is the binomial coefficient "n choose k".
In our case, *n* = 3, so the possible values for *k* (number of successes) are 0, 1, 2, and 3. The probabilities are:
* P(X = 0) = (3C0) * p^0 * q^3 = q^3
* P(X = 1) = (3C1) * p^1 * q^2 = 3pq^2
* P(X = 2) = (3C2) * p^2 * q^1 = 3p^2q
* P(X = 3) = (3C3) * p^3 * q^0 = p^3
**2. Mean (Expected Value):**
The mean (or expected value) of a discrete random variable is given by:
E(X) = Σ [x * P(X = x)] (summed over all possible values of x)
For our binomial distribution:
E(X) = 0 * q^3 + 1 * 3pq^2 + 2 * 3p^2q + 3 * p^3
E(X) = 0 + 3pq^2 + 6p^2q + 3p^3
E(X) = 3p(q^2 + 2pq + p^2)
E(X) = 3p(p + q)^2
Since p + q = 1:
E(X) = 3p * 1^2
E(X) = 3p
Therefore, the mean of the binomial distribution with n=3 is 3p.
**3. Variance:**
The variance of a discrete random variable is given by:
Var(X) = E(X^2) - [E(X)]^2
First, we need to calculate E(X^2):
E(X^2) = Σ [x^2 * P(X = x)]
E(X^2) = 0^2 * q^3 + 1^2 * 3pq^2 + 2^2 * 3p^2q + 3^2 * p^3
E(X^2) = 0 + 3pq^2 + 12p^2q + 9p^3
Now, we can calculate the variance:
Var(X) = E(X^2) - [E(X)]^2
Var(X) = (3pq^2 + 12p^2q + 9p^3) - (3p)^2
Var(X) = 3pq^2 + 12p^2q + 9p^3 - 9p^2
Var(X) = 3p(q^2 + 4pq + 3p^2 - 3p)
Var(X) = 3p[q^2 + 2pq + p^2 + 2pq + 2p^2 - 3p]
Var(X) = 3p[(q+p)^2 + 2p(q+p) - 3p]
Var(X) = 3p[1 + 2p - 3p]
Var(X) = 3p(1-p)
Var(X) = 3pq
Therefore, the variance of the binomial distribution with n=3 is 3pq.
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