Certainly! Let's tackle each part step by step.
### a. Finding the Probabilities
In a binomial experiment with \(\( n \)\) trials and a probability \(\( p \)\) of success on each trial, the probability of obtaining exactly \(\( k \)\) successes is given by the binomial probability mass function (PMF):
\(\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]\)
where
- \(\( \binom{n}{k} \)\) is the binomial coefficient, calculated as
\(\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]\)
- \(\( n \)\) is the number of trials
- \(\( k \)\) is the number of successes
- \(\( p \)\) is the probability of success in each trial
- \(\( n! \)\) represents the factorial of \(\( n \)\)
Given \(\( n = 4 \)\) and \(\( p = 0.2 \)\), we can calculate the probabilities for \(\( k = 0 \)\), \(\( 1 \)\), \(\( 2 \)\), \(\( 3 \)\), and \(\( 4 \)\) as follows:
- For \(\( k = 0 \)\):
\(\[ P(X = 0) = \binom{4}{0} (0.2)^0 (1-0.2)^4 = 1 \times 1 \times 0.8^4 \approx 0.4096 \]\)
- For \(\( k = 1 \)\):
\(\[ P(X = 1) = \binom{4}{1} (0.2)^1 (1-0.2)^3 = 4 \times 0.2 \times 0.8^3 \approx 0.4096 \]\)
- For \(\( k = 2 \)\):
\(\[ P(X = 2) = \binom{4}{2} (0.2)^2 (1-0.2)^2 = 6 \times 0.04 \times 0.64 \approx 0.1536 \]\)
- For \(\( k = 3 \)\):
\(\[ P(X = 3) = \binom{4}{3} (0.2)^3 (1-0.2)^1 = 4 \times 0.008 \times 0.8 \approx 0.0256 \]\)
- For \(\( k = 4 \)\):
\(\[ P(X = 4) = \binom{4}{4} (0.2)^4 (1-0.2)^0 = 1 \times 0.0016 \times 1 \approx 0.0016 \]\)
### b. Constructing the Binomial Distribution and Histogram
The binomial distribution for this experiment can be represented by a histogram where the x-axis represents the number of successes (0 to 4) and the y-axis represents the corresponding probabilities we've just calculated. The heights of the bars in the histogram will match the probabilities.
If you were to draw the histogram, you would have a bar above each number of successes from 0 to 4, with heights approximately equal to 0.4096, 0.4096, 0.1536, 0.0256, and 0.0016, respectively.
### c. Computing the Mean and Standard Deviation
The mean \(\( \mu \)\) of a binomial distribution is given by
\(\[ \mu = n \cdot p \]\)
For this experiment,
\(\[ \mu = 4 \cdot 0.2 = 0.8 \]\)
The standard deviation \(\( \sigma \)\) of a binomial distribution is given by
\(\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]\)
So for this experiment,
\(\[ \sigma = \sqrt{4 \cdot 0.2 \cdot (1-0.2)} = \sqrt{4 \cdot 0.2 \cdot 0.8} = \sqrt{0.64} \approx 0.8 \]\)
Therefore, the mean of the random variable associated with this experiment is 0.8 (or 4 successes) and the standard deviation is approximately 0.8.
