We have a binomial distribution scenario because the problem involves a fixed number of Bernoulli trials (10) with a given probability of success (0.65). Let's break down the problem:

(A) Mean and Standard Deviation

For a binomial distribution, the mean (μ) and standard deviation (σ) are given by:

• Mean (μ): $\mu = n \cdot p$
• Standard Deviation (σ): $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Where:

• $n = 10$ (number of trials)
• $p = 0.65$ (probability of success)

Calculating the Mean:

$\mu = 10 \cdot 0.65 = 6.5$

Calculating the Standard Deviation:

$\sigma = \sqrt{10 \cdot 0.65 \cdot (1 - 0.65)} = \sqrt{10 \cdot 0.65 \cdot 0.35} \approx \sqrt{2.275} \approx 1.51$

So:

• Mean (μ): 6.5
• Standard Deviation (σ): 1.51 (approximately)

(B) Probability Within 1 Standard Deviation of the Mean

We want to find the probability that the number of successes lies within 1 standard deviation of the mean. This corresponds to the range:

$\mu - \sigma \leq X \leq \mu + \sigma$

Substituting the values:

$6.5 - 1.51 \leq X \leq 6.5 + 1.51$

This gives us the interval:

$4.99 \leq X \leq 8.01$

Since $X$ represents the number of successes, it must be an integer. Thus, we consider $X = 5, 6, 7, 8$.

Now, we need to calculate the probability for each of these values using the binomial probability formula:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}$

For each $X = 5, 6, 7, 8$, we'll compute the corresponding probability and sum them to get the overall probability.

Let me calculate this for you.The probability that the number of successes lies within 1 standard deviation of the mean is approximately 0.819 (or 81.9%).

Summary:

• (A) Mean: 6.5
• (A) Standard Deviation: 1.51
• (B) Probability within 1 Standard Deviation of the Mean: 0.819 (or 81.9%)

Would you like more details or have any other questions?

Here are some related questions to explore:

1. How do you calculate the binomial probability for any specific number of successes?
2. What happens to the mean and standard deviation if the number of trials increases?
3. How does changing the probability $p$ affect the binomial distribution?
4. What is the shape of the binomial distribution for different values of $n$ and $p$?
5. How does the normal approximation relate to the binomial distribution in this context?

Tip: In binomial distributions, as the number of trials $n$ increases and $p$ is not too close to 0 or 1, the distribution becomes more symmetric and can be approximated using a normal distribution.