2. Assume a binomial distribution where n = 3 p = 0.60.
a. Refer to the table given by your lecturer and list the probabilities for values of X from 0  to 3.
b. Determine the mean and standard deviation of the distribution.

Given the problem with a binomial distribution where \( n = 3 \) and \( p = 0.60 \), we can solve it as follows:

### Part (a): List the Probabilities for Values of \( X \) from 0 to 3

The probability mass function (PMF) for a binomial distribution is given by:

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

Where:
- \( n = 3 \) (number of trials)
- \( p = 0.60 \) (probability of success)
- \( k \) is the number of successes (values of \( X \) from 0 to 3)

Let's calculate the probabilities:

1. **\( P(X = 0) \)**:
   \[
   P(X = 0) = \binom{3}{0} (0.60)^0 (1-0.60)^3 = 1 \times 1 \times 0.4^3 = 0.064
   \]

2. **\( P(X = 1) \)**:
   \[
   P(X = 1) = \binom{3}{1} (0.60)^1 (1-0.60)^2 = 3 \times 0.60 \times 0.4^2 = 0.288
   \]

3. **\( P(X = 2) \)**:
   \[
   P(X = 2) = \binom{3}{2} (0.60)^2 (1-0.60)^1 = 3 \times 0.60^2 \times 0.4 = 0.432
   \]

4. **\( P(X = 3) \)**:
   \[
   P(X = 3) = \binom{3}{3} (0.60)^3 (1-0.60)^0 = 1 \times 0.60^3 \times 1 = 0.216
   \]

So, the probabilities are:
- \( P(X = 0) = 0.064 \)
- \( P(X = 1) = 0.288 \)
- \( P(X = 2) = 0.432 \)
- \( P(X = 3) = 0.216 \)

### Part (b): Determine the Mean and Standard Deviation of the Distribution

The mean (\(\mu\)) and standard deviation (\(\sigma\)) of a binomial distribution are given by the formulas:

\[
\mu = np
\]
\[
\sigma = \sqrt{np(1-p)}
\]

Given:
- \( n = 3 \)
- \( p = 0.60 \)

1. **Mean**:
   \[
   \mu = 3 \times 0.60 = 1.8
   \]

2. **Standard Deviation**:
   \[
   \sigma = \sqrt{3 \times 0.60 \times (1-0.60)} = \sqrt{3 \times 0.60 \times 0.40} = \sqrt{0.72} \approx 0.8485
   \]

### Summary:
- **Probabilities**:
  - \( P(X = 0) = 0.064 \)
  - \( P(X = 1) = 0.288 \)
  - \( P(X = 2) = 0.432 \)
  - \( P(X = 3) = 0.216 \)
- **Mean**: \( \mu = 1.8 \)
- **Standard Deviation**: \( \sigma \approx 0.8485 \)

Would you like any further details or have any questions? 

Here are 5 related questions for further exploration:
1. How would the probabilities change if the number of trials \( n \) was increased to 5?
2. What is the probability of getting exactly 2 successes when \( n = 4 \) and \( p = 0.60 \)?
3. How does changing the probability \( p \) to 0.70 affect the mean and standard deviation?
4. What is the cumulative probability \( P(X \leq 2) \) for the given distribution?
5. How would you interpret the standard deviation in the context of this binomial distribution?

**Tip:** Understanding the binomial distribution is crucial as it applies to many real-world situations where outcomes are binary, like pass/fail scenarios.

Explore the binomial distribution problem with n = 3 and p = 0.60. Calculate probabilities for X = 0, 1, 2, 3, and find the mean and standard deviation. Understand binomial probability concepts in this detailed solution.

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