https://file.aiquickdraw.com/mathbot/file/1728773154488yyt9ppc6.jpg

Let's go through the steps for solving the remaining parts of this binomial distribution problem.

### Given:
- Number of trials, \( n = 867 \)
- Probability of success, \( p = 28\% = 0.28 \)
- The mean, \( \mu = 242.8 \) (which is correct)

#### 1. **Standard Deviation (\( \sigma \)):**
The formula for the standard deviation of a binomial distribution is:

\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Substituting the given values:

\[
\sigma = \sqrt{867 \cdot 0.28 \cdot (1 - 0.28)}
\]
\[
\sigma = \sqrt{867 \cdot 0.28 \cdot 0.72}
\]
\[
\sigma = \sqrt{867 \cdot 0.2016}
\]
\[
\sigma = \sqrt{174.7872}
\]
\[
\sigma \approx 13.22
\]

So, the standard deviation \( \sigma \) is approximately **13.22**.

#### 2. **Range Rule of Thumb (Minimum and Maximum Usual Values):**

Using the range rule of thumb, the minimum and maximum usual values are calculated as:

- **Minimum usual value**: \( \mu - 2\sigma \)
- **Maximum usual value**: \( \mu + 2\sigma \)

Substitute \( \mu = 242.8 \) and \( \sigma = 13.22 \):

- **Minimum usual value**:
\[
\mu - 2\sigma = 242.8 - 2 \cdot 13.22 = 242.8 - 26.44 = 216.36 \approx 216
\]
- **Maximum usual value**:
\[
\mu + 2\sigma = 242.8 + 2 \cdot 13.22 = 242.8 + 26.44 = 269.24 \approx 269
\]

Thus, the interval for usual values is \( [216, 269] \).

### Final Answers:
1. **Standard deviation (\( \sigma \))**: \( 13.22 \)
2. **Usual values interval**: \( [216, 269] \)

Would you like any further details or clarifications?

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Here are 5 related questions to deepen your understanding:
1. How does the probability of success (\( p \)) affect the mean and standard deviation of a binomial distribution?
2. What happens to the range of usual values if the standard deviation increases?
3. Can you explain why the formula for standard deviation includes both \( p \) and \( (1 - p) \)?
4. How would the binomial distribution change if the number of trials \( n \) were halved?
5. How would you use this binomial distribution to calculate the probability of exactly 250 successes?

**Tip**: Always verify that the probability and number of trials are within reasonable bounds for a binomial distribution—extreme values might require alternative statistical approaches.

Assume that a procedure yields a binomial distribution with n = 867 trials and the probability of success for one trial is p = 28%. Find the mean, standard deviation, and usual values for this distribution.

Solve a binomial distribution problem with 867 trials and a probability of success of 28%. Learn how to calculate the mean, standard deviation, and usual value range using formulas. Ideal for high school students in grades 9-12.

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