Construct a Binomial probability distrubtion table using n=5 and p=0.23 
. Next, record the mean and standard deviation of the distribution below. Where necessary, round all numbers to four decimal places.

To construct a binomial probability distribution table with \(n = 5\) (number of trials) and \(p = 0.23\) (probability of success), we can calculate the probability for each number of successes \(x\), where \(x\) can range from 0 to 5. The formula for the binomial probability mass function is:

\[
P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}
\]
where \(\binom{n}{x}\) is the binomial coefficient, calculated as:

\[
\binom{n}{x} = \frac{n!}{x!(n-x)!}
\]

### Step 1: Calculate the binomial probabilities for each \(x\) from 0 to 5.

Given:
- \(n = 5\)
- \(p = 0.23\)
- \(q = 1 - p = 0.77\)

We calculate each probability as follows:

\[
P(X = x) = \binom{5}{x} (0.23)^x (0.77)^{5-x}
\]

| \(x\) (Number of Successes) | \(\binom{5}{x}\) | Probability \(P(X = x)\) |
|-----------------------------|-------------------|--------------------------|
| 0                           | \(\binom{5}{0} = 1\)     | \(P(X = 0) = 1 \times (0.23)^0 \times (0.77)^5 = 0.2861\) |
| 1                           | \(\binom{5}{1} = 5\)     | \(P(X = 1) = 5 \times (0.23)^1 \times (0.77)^4 = 0.4268\) |
| 2                           | \(\binom{5}{2} = 10\)    | \(P(X = 2) = 10 \times (0.23)^2 \times (0.77)^3 = 0.2181\) |
| 3                           | \(\binom{5}{3} = 10\)    | \(P(X = 3) = 10 \times (0.23)^3 \times (0.77)^2 = 0.0585\) |
| 4                           | \(\binom{5}{4} = 5\)     | \(P(X = 4) = 5 \times (0.23)^4 \times (0.77)^1 = 0.0082\) |
| 5                           | \(\binom{5}{5} = 1\)     | \(P(X = 5) = 1 \times (0.23)^5 \times (0.77)^0 = 0.0005\) |

### Step 2: Calculate the mean and standard deviation.

- **Mean** (\(\mu\)) of a binomial distribution is given by:

\[
\mu = n \times p = 5 \times 0.23 = 1.15
\]

- **Standard Deviation** (\(\sigma\)) of a binomial distribution is given by:

\[
\sigma = \sqrt{n \times p \times (1-p)} = \sqrt{5 \times 0.23 \times 0.77} = \sqrt{0.8855} = 0.9405
\]

### Final Summary:
- **Mean**: \(\mu = 1.15\)
- **Standard Deviation**: \(\sigma = 0.9405\)

Would you like further clarification on these calculations or more details on any step?

Here are five related questions to deepen your understanding:
1. How would the probabilities change if \(p\) were increased to 0.5?
2. What happens to the standard deviation if the number of trials, \(n\), increases?
3. How would you calculate cumulative probabilities for this binomial distribution?
4. What is the probability of getting exactly 3 successes for \(p = 0.4\) and \(n = 7\)?
5. How does the shape of a binomial distribution change as \(p\) approaches 1?

**Tip**: For large values of \(n\) and probabilities near 0.5, the binomial distribution begins to resemble the normal distribution.

Construct a Binomial probability distribution table using n=5 and p=0.23. Next, record the mean and standard deviation of the distribution below. Where necessary, round all numbers to four decimal places.

Learn how to construct a binomial probability distribution table for n=5 and p=0.23. Discover how to calculate the mean (1.15) and standard deviation (0.9405) for this distribution, with detailed step-by-step explanations. Suitable for students in Grades 10-12 studying statistics and probability.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation