In this problem, we are dealing with the number of heads that appear in 16 tosses of a fair coin. The variable $X$ represents this number of heads, and it follows a binomial distribution since:

1. Each coin toss is independent.
2. The probability of heads (success) on any given toss is 0.5.

A binomial random variable $X$ can be defined as:

$X \sim \text{Binomial}(n, p)$

Where:

• $n = 16$ is the number of trials (tosses),
• $p = 0.5$ is the probability of success (getting heads).

1. Expected Value $E(X)$

For a binomial distribution, the expected value is given by:

$E(X) = n \cdot p$

Substituting the values:

$E(X) = 16 \cdot 0.5 = 8$

So, the expected number of heads in 16 tosses is 8.

2. Variance $\text{Var}(X)$

For a binomial distribution, the variance is given by:

$\text{Var}(X) = n \cdot p \cdot (1 - p)$

Substituting the values:

$\text{Var}(X) = 16 \cdot 0.5 \cdot (1 - 0.5) = 16 \cdot 0.5 \cdot 0.5 = 4$

So, the variance of $X$ is 4.

Summary:

• $E(X) = 8$
• $\text{Var}(X) = 4$

Would you like more details or further explanations?

Related Questions:

1. How do you calculate the standard deviation of $X$?
2. What is the probability of getting exactly 8 heads in 16 tosses?
3. How does the binomial distribution change if the probability of heads is not 0.5?
4. Can the number of heads be greater than 16? Why or why not?
5. What is the cumulative probability of getting at most 10 heads in 16 tosses?

Tip:

In a binomial distribution, the standard deviation can be found using the square root of the variance, $\sigma = \sqrt{\text{Var}(X)}$.