10 coins are tossed at random. Obtain the mean and variance of the number of heads obtained.

Answer

To solve the problem of finding the mean and variance of the number of heads obtained when tossing 10 coins, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of heads obtained when tossing 10 coins.

### Step 2: Identify the Parameters
- The number of trials (n) is 10 (since we are tossing 10 coins).
- The probability of success (getting a head) for each coin toss is \( p = \frac{1}{2} \).
- The probability of failure (getting a tail) is \( q = 1 - p = \frac{1}{2} \).

### Step 3: Calculate the Mean
The mean (expected value) of a binomial distribution is given by the formula:
\[ E(X) = n \cdot p \]
Substituting the values:
\[ E(X) = 10 \cdot \frac{1}{2} = 5 \]

### Step 4: Calculate the Variance
The variance of a binomial distribution is given by the formula:
\[ Var(X) = n \cdot p \cdot q \]
Substituting the values:
\[ Var(X) = 10 \cdot \frac{1}{2} \cdot \frac{1}{2} = 10 \cdot \frac{1}{4} = 2.5 \]

### Final Result
Thus, the mean number of heads obtained is \( 5 \) and the variance is \( 2.5 \).

---