Answer :
Certainly! Let's tackle each part step by step.
### a. Finding the Probabilities
In a binomial experiment with [tex]\( n \)[/tex] trials and a probability [tex]\( p \)[/tex] of success on each trial, the probability of obtaining exactly [tex]\( k \)[/tex] successes is given by the binomial probability mass function (PMF):
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
- [tex]\( n \)[/tex] is the number of trials
- [tex]\( k \)[/tex] is the number of successes
- [tex]\( p \)[/tex] is the probability of success in each trial
- [tex]\( n! \)[/tex] represents the factorial of [tex]\( n \)[/tex]
Given [tex]\( n = 4 \)[/tex] and [tex]\( p = 0.2 \)[/tex], we can calculate the probabilities for [tex]\( k = 0 \)[/tex], [tex]\( 1 \)[/tex], [tex]\( 2 \)[/tex], [tex]\( 3 \)[/tex], and [tex]\( 4 \)[/tex] as follows:
- For [tex]\( k = 0 \)[/tex]:
[tex]\[ P(X = 0) = \binom{4}{0} (0.2)^0 (1-0.2)^4 = 1 \times 1 \times 0.8^4 \approx 0.4096 \][/tex]
- For [tex]\( k = 1 \)[/tex]:
[tex]\[ P(X = 1) = \binom{4}{1} (0.2)^1 (1-0.2)^3 = 4 \times 0.2 \times 0.8^3 \approx 0.4096 \][/tex]
- For [tex]\( k = 2 \)[/tex]:
[tex]\[ P(X = 2) = \binom{4}{2} (0.2)^2 (1-0.2)^2 = 6 \times 0.04 \times 0.64 \approx 0.1536 \][/tex]
- For [tex]\( k = 3 \)[/tex]:
[tex]\[ P(X = 3) = \binom{4}{3} (0.2)^3 (1-0.2)^1 = 4 \times 0.008 \times 0.8 \approx 0.0256 \][/tex]
- For [tex]\( k = 4 \)[/tex]:
[tex]\[ P(X = 4) = \binom{4}{4} (0.2)^4 (1-0.2)^0 = 1 \times 0.0016 \times 1 \approx 0.0016 \][/tex]
### 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 [tex]\( \mu \)[/tex] of a binomial distribution is given by
[tex]\[ \mu = n \cdot p \][/tex]
For this experiment,
[tex]\[ \mu = 4 \cdot 0.2 = 0.8 \][/tex]
The standard deviation [tex]\( \sigma \)[/tex] of a binomial distribution is given by
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \][/tex]
So for this experiment,
[tex]\[ \sigma = \sqrt{4 \cdot 0.2 \cdot (1-0.2)} = \sqrt{4 \cdot 0.2 \cdot 0.8} = \sqrt{0.64} \approx 0.8 \][/tex]
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.
### a. Finding the Probabilities
In a binomial experiment with [tex]\( n \)[/tex] trials and a probability [tex]\( p \)[/tex] of success on each trial, the probability of obtaining exactly [tex]\( k \)[/tex] successes is given by the binomial probability mass function (PMF):
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
- [tex]\( n \)[/tex] is the number of trials
- [tex]\( k \)[/tex] is the number of successes
- [tex]\( p \)[/tex] is the probability of success in each trial
- [tex]\( n! \)[/tex] represents the factorial of [tex]\( n \)[/tex]
Given [tex]\( n = 4 \)[/tex] and [tex]\( p = 0.2 \)[/tex], we can calculate the probabilities for [tex]\( k = 0 \)[/tex], [tex]\( 1 \)[/tex], [tex]\( 2 \)[/tex], [tex]\( 3 \)[/tex], and [tex]\( 4 \)[/tex] as follows:
- For [tex]\( k = 0 \)[/tex]:
[tex]\[ P(X = 0) = \binom{4}{0} (0.2)^0 (1-0.2)^4 = 1 \times 1 \times 0.8^4 \approx 0.4096 \][/tex]
- For [tex]\( k = 1 \)[/tex]:
[tex]\[ P(X = 1) = \binom{4}{1} (0.2)^1 (1-0.2)^3 = 4 \times 0.2 \times 0.8^3 \approx 0.4096 \][/tex]
- For [tex]\( k = 2 \)[/tex]:
[tex]\[ P(X = 2) = \binom{4}{2} (0.2)^2 (1-0.2)^2 = 6 \times 0.04 \times 0.64 \approx 0.1536 \][/tex]
- For [tex]\( k = 3 \)[/tex]:
[tex]\[ P(X = 3) = \binom{4}{3} (0.2)^3 (1-0.2)^1 = 4 \times 0.008 \times 0.8 \approx 0.0256 \][/tex]
- For [tex]\( k = 4 \)[/tex]:
[tex]\[ P(X = 4) = \binom{4}{4} (0.2)^4 (1-0.2)^0 = 1 \times 0.0016 \times 1 \approx 0.0016 \][/tex]
### 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 [tex]\( \mu \)[/tex] of a binomial distribution is given by
[tex]\[ \mu = n \cdot p \][/tex]
For this experiment,
[tex]\[ \mu = 4 \cdot 0.2 = 0.8 \][/tex]
The standard deviation [tex]\( \sigma \)[/tex] of a binomial distribution is given by
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \][/tex]
So for this experiment,
[tex]\[ \sigma = \sqrt{4 \cdot 0.2 \cdot (1-0.2)} = \sqrt{4 \cdot 0.2 \cdot 0.8} = \sqrt{0.64} \approx 0.8 \][/tex]
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.
