1. Key Concepts and Formulas

• Binomial Distribution Parameters:
  • $n$: Number of trials (a positive integer).
  • $p$: Probability of success in a single trial ($0 < p < 1$).
  • $q$: Probability of failure in a single trial ($q = 1 - p$, so $0 < q < 1$).
• Mean of Binomial Distribution ($\mu$): $\mu = np$
• Variance of Binomial Distribution ($\sigma^2$): $\sigma^2 = npq$
• Relationship between Mean and Variance: For a binomial distribution, $\sigma^2 = \mu q$. Since $0 < q < 1$, it implies $\sigma^2 < \mu$ (unless $q=0$, which means $p=1$ and $\sigma^2=0$).
• Quadratic Equation Roots: If the sum of two quantities $x_1, x_2$ is $S$ and their product is $P$, then $x_1$ and $x_2$ are the roots of the quadratic equation $x^2 - Sx + P = 0$.

2. Step-by-Step Solution

Step 1: Formulating Equations from Given Information

The problem states that the sum and product of the mean ($\mu$) and variance ($\sigma^2$) of a binomial distribution are 82.5 and 1350, respectively.

1. Sum of mean and variance: $\mu + \sigma^2 = 82.5$
2. Product of mean and variance: $\mu \cdot \sigma^2 = 1350$

Step 2: Solving for the Mean and Variance

We can treat $\mu$ and $\sigma^2$ as the roots of a quadratic equation $x^2 - Sx + P = 0$, where $S = 82.5$ and $P = 1350$. Substituting the given values: $x^2 - 82.5x + 1350 = 0$ To solve this quadratic equation, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ or factorization. For factorization, we look for two numbers that sum to 82.5 and multiply to 1350. By inspection, 60 and 22.5 satisfy these conditions ($60 + 22.5 = 82.5$ and $60 \times 22.5 = 1350$). So, the quadratic equation can be factored as: $(x - 60)(x - 22.5) = 0$ The roots are $x = 60$ and $x = 22.5$. These are the possible values for $\mu$ and $\sigma^2$.

Step 3: Assigning Values to Mean and Variance

For a binomial distribution, the mean is always greater than or equal to the variance ($\mu \ge \sigma^2$), because $\sigma^2 = \mu q$ and $0 < q \le 1$. Therefore, we assign the larger value to the mean and the smaller value to the variance:

• Mean ($\mu$) = $np = 60$
• Variance ($\sigma^2$) = $npq = 22.5$

Step 4: Calculating the Probability of Failure ($q$) and Success ($p$)

We have a system of two equations:

1. $np = 60$
2. $npq = 22.5$ To find $q$, divide the second equation by the first equation: $\frac{npq}{np} = \frac{22.5}{60}$ $q = \frac{22.5}{60}$ To simplify the fraction, multiply the numerator and denominator by 10 to remove the decimal: $q = \frac{225}{600}$ Divide both by 75 (or successively by 25 then 3): $q = \frac{225 \div 75}{600 \div 75} = \frac{3}{8}$ Now, we find the probability of success $p$ using the relation $p + q = 1$: $p = 1 - q = 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8}$ Both $p = 5/8$ and $q = 3/8$ are valid probabilities (between 0 and 1).

Step 5: Determining the Number of Trials ($n$)

Using the formula for the mean, $\mu = np$, and the values we found for $\mu$ and $p$: $np = 60$ Substitute $p = \frac{5}{8}$: $n \left(\frac{5}{8}\right) = 60$ To solve for $n$, multiply both sides by the reciprocal of $\frac{5}{8}$, which is $\frac{8}{5}$: $n = 60 \times \frac{8}{5}$ $n = (60 \div 5) \times 8$ $n = 12 \times 8$ $n = 96$ Therefore, the number of trials in the binomial distribution is 96.

3. Common Mistakes & Tips

• Order of Mean and Variance: Always remember that for a binomial distribution, $\mu \ge \sigma^2$. Incorrectly assigning the values from the quadratic equation (e.g., $\mu = 22.5, \sigma^2 = 60$) would lead to $q > 1$, which is impossible for a probability.
• Decimal to Fraction Conversion: Converting decimals to fractions early often simplifies calculations and reduces the chance of arithmetic errors, especially during division and simplification.
• Probability Constraints: Always verify that your calculated $p$ and $q$ values are between 0 and 1 (exclusive for non-degenerate distributions). If not, recheck your calculations.
• Integer $n$: The number of trials $n$ must be a positive integer. If your final $n$ is a fraction or negative, there's an error in your steps.

4. Summary

We began by setting up two equations based on the given sum and product of the mean and variance. These equations led to a quadratic equation, whose roots represented the mean and variance. By applying the property that the mean is always greater than or equal to the variance for a binomial distribution, we correctly identified $\mu = 60$ and $\sigma^2 = 22.5$. We then used these values to calculate the probability of failure ($q = 3/8$) and success ($p = 5/8$). Finally, substituting $p$ and $\mu$ into the mean formula ($\mu = np$) allowed us to determine the number of trials, $n=96$.

5. Final Answer

The final answer is $\boxed{82}$.