The problem from your screenshot involves calculating the mean and standard deviation for students enrolling in college based on a given percentage. Here's how to solve it:

Part (a): Finding the mean number of students who enroll in college.

• Given: 67% of students enrolled in college.
• Sample size: 28 high school graduates.

The mean ($\mu$) is calculated by multiplying the probability by the sample size: $\mu = p \times n$ Where:

• $p = 0.67$ (the percentage of students enrolling in college, converted to a decimal),
• $n = 28$ (the number of students sampled).

$\mu = 0.67 \times 28 = 18.76$

So, the mean number of students who enroll in college is 18.76.

Part (b): Finding the standard deviation.

The standard deviation ($\sigma$) for a binomial distribution is calculated using the formula: $\sigma = \sqrt{n \times p \times (1 - p)}$ Where:

• $n = 28$,
• $p = 0.67$,
• $1 - p = 0.33$ (the probability of a student not enrolling).

$\sigma = \sqrt{28 \times 0.67 \times 0.33}$ Let me calculate this value for you.The standard deviation ($\sigma$) is approximately 2.4881.

Summary:

• Mean number of students who enroll in college: 18.76.
• Standard deviation: 2.4881 (rounded to four decimal places).

Would you like further details or clarification on any part?

Here are some related questions to explore:

1. How would the mean change if the sample size increased?
2. What happens to the standard deviation if the percentage of students enrolling decreases?
3. How is this calculation affected if the sample size is much larger?
4. Could you explain why we use the square root in the standard deviation formula?
5. How do we interpret the standard deviation in the context of this problem?

Tip: When working with percentages, always convert them to decimals before performing calculations.