Let's go step by step to solve the problem:

Given:

• n = 975 (number of trials)
• p = 0.67 (probability of success for one trial)

Step 1: Find the mean ($\mu$) for the binomial distribution.

The formula for the mean of a binomial distribution is: $\mu = n \times p$ Substituting the values: $\mu = 975 \times 0.67 = 653.25$ So, the mean $\mu$ is approximately 653.3 (rounded to one decimal place).

Step 2: Find the standard deviation ($\sigma$) for the binomial distribution.

The formula for the standard deviation of a binomial distribution is: $\sigma = \sqrt{n \times p \times (1 - p)}$ Substituting the values: $\sigma = \sqrt{975 \times 0.67 \times (1 - 0.67)} = \sqrt{975 \times 0.67 \times 0.33}$ $\sigma = \sqrt{224.38575} \approx 14.98$ So, the standard deviation $\sigma$ is approximately 15.00 (rounded to two decimal places).

Step 3: Use the range rule of thumb to find the minimum and maximum usual values.

The range rule of thumb says that the usual values are within 2 standard deviations of the mean, i.e., between: $\mu - 2\sigma \quad \text{and} \quad \mu + 2\sigma$

For the minimum usual value: $\mu - 2\sigma = 653.3 - 2 \times 15.00 = 653.3 - 30.00 = 623.3$

For the maximum usual value: $\mu + 2\sigma = 653.3 + 2 \times 15.00 = 653.3 + 30.00 = 683.3$

So, the usual values are between 623 and 683 (rounded to whole numbers).

Final Answers:

• Mean (μ) = 653.3
• Standard deviation (σ) = 15.00
• Usual values = [623, 683]

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Would you like more details on any of the steps? Here are some related questions to explore:

1. How do you calculate the variance for a binomial distribution?
2. What does the standard deviation tell you about the spread of the distribution?
3. What happens if the probability of success (p) is close to 0 or 1 in a binomial distribution?
4. How can you use the binomial distribution to calculate probabilities for certain outcomes?
5. How does the binomial distribution relate to the normal distribution when n is large?

Tip: In large trials like this one, the binomial distribution often approximates the normal distribution well, especially when both $np$ and $n(1-p)$ are greater than 5.