To solve this problem, let's go over each part of the question:

Given Data

• Probability of a person donating blood ($p$) = 0.21
• Number of trials ($n$) = 4

The problem involves finding the mean and variance of a binomial distribution, which we can calculate using the following formulas:

Part (a): Mean of the Binomial Distribution

The mean ($\mu$) of a binomial distribution is calculated using: $\mu = n \times p$ Substituting the values: $\mu = 4 \times 0.21 = 0.84$ So, the mean $\mu$ is 0.84.

Part (b): Variance of the Binomial Distribution

The variance ($\sigma^2$) of a binomial distribution is calculated using: $\sigma^2 = n \times p \times (1 - p)$ Substituting the values: $\sigma^2 = 4 \times 0.21 \times (1 - 0.21)$ $\sigma^2 = 4 \times 0.21 \times 0.79$ $\sigma^2 = 0.6636$ Rounded to the nearest hundredth, the variance $\sigma^2$ is 0.66.

Summary of Answers

• Mean $\mu$ = 0.84
• Variance $\sigma^2$ = 0.66

Would you like further details on the binomial distribution or have any questions?

Related Questions

1. What is the standard deviation of this binomial distribution?
2. How would the mean and variance change if you sampled 10 people instead of 4?
3. Can you calculate the probability of exactly 2 people donating blood in this scenario?
4. How does changing the probability $p$ affect the mean and variance?
5. What is the probability that no one in the sample donates blood?

Tip

For any binomial distribution, remember that the mean and variance are directly influenced by both the number of trials and the probability of success.