Let's start with problem 10 and solve it step by step.

Problem 10:

We are dealing with a binomial distribution where:

• $n = 5$ (number of trials)
• $p = 0.40$ (probability of success in each trial)

The binomial formula is: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where $\binom{n}{k}$ is the binomial coefficient, $p$ is the probability of success, and $1 - p$ is the probability of failure.

Part (a) $X = 1$

$P(X = 1) = \binom{5}{1} (0.40)^1 (0.60)^4$ $P(X = 1) = 5 \times 0.40 \times (0.60)^4 = 5 \times 0.40 \times 0.1296 = 0.2592$

Part (b) $X = 2$

$P(X = 2) = \binom{5}{2} (0.40)^2 (0.60)^3$ $P(X = 2) = 10 \times (0.40)^2 \times (0.60)^3 = 10 \times 0.16 \times 0.216 = 0.3456$

Problem 12:

We are given a binomial distribution where:

• $n = 5$
• $p = 0.30$

Part (a):

We need to list the probabilities for values of $x$ from 0 to 5.

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

For $x = 0$ to $x = 5$:

$P(X = 0) = \binom{5}{0} (0.30)^0 (0.70)^5 = 1 \times 1 \times 0.16807 = 0.16807$ $P(X = 1) = \binom{5}{1} (0.30)^1 (0.70)^4 = 5 \times 0.30 \times 0.2401 = 0.36015$ $P(X = 2) = \binom{5}{2} (0.30)^2 (0.70)^3 = 10 \times 0.09 \times 0.343 = 0.3087$ $P(X = 3) = \binom{5}{3} (0.30)^3 (0.70)^2 = 10 \times 0.027 \times 0.49 = 0.1323$ $P(X = 4) = \binom{5}{4} (0.30)^4 (0.70)^1 = 5 \times 0.0081 \times 0.70 = 0.02835$ $P(X = 5) = \binom{5}{5} (0.30)^5 (0.70)^0 = 1 \times 0.00243 \times 1 = 0.00243$

So, the probabilities are: $P(X = 0) = 0.16807, \, P(X = 1) = 0.36015, \, P(X = 2) = 0.3087$ $P(X = 3) = 0.1323, \, P(X = 4) = 0.02835, \, P(X = 5) = 0.00243$

Part (b):

The mean $\mu$ and standard deviation $\sigma$ can be computed as follows:

The mean for a binomial distribution is: $\mu = n \times p = 5 \times 0.30 = 1.5$

The variance is: $\sigma^2 = n \times p \times (1 - p) = 5 \times 0.30 \times 0.70 = 1.05$

Thus, the standard deviation is: $\sigma = \sqrt{1.05} \approx 1.0247$

Would you like more details on these calculations or have any questions? Here are some related questions:

1. How is the binomial coefficient $\binom{n}{k}$ calculated in binomial probability problems?
2. What are some common applications of the binomial distribution in real-world scenarios?
3. How can the cumulative probability be calculated using binomial probabilities?
4. What is the difference between the mean and expected value in a binomial distribution?
5. How does the value of $p$ (success probability) affect the shape of the binomial distribution?

Tip: The mean $\mu$ of a binomial distribution tells you the average number of successes over many repeated trials, while the standard deviation $\sigma$ gives an idea of how much variation or spread you can expect around this mean.