To solve the problems, we can use the properties of a binomial distribution, since the problem involves a fixed number of trials (high school graduates), each with two possible outcomes (enrolling in college or not).
Given:
- The probability of success (enrolling in college) \( p = 0.66 \)
- The number of trials (students sampled) \( n = 32 \)
(a) Mean Number Who Enroll in College
The mean \( \mu \) in a binomial distribution is calculated using the formula:
\[ \mu = n \times p \]
Substituting the values:
\[ \mu = 32 \times 0.66 = 21.12 \]
So, the mean number who enroll in college in a sample of 32 high school graduates is 21.12.
(b) Standard Deviation of the Number Who Enroll in College
The standard deviation \( \sigma \) for a binomial distribution is calculated using the formula:
\[ \sigma = \sqrt{n \times p \times (1 - p)} \]
Where \( (1 - p) \) is the probability of failure (not enrolling in college).
Calculating \( 1 - p \):
\[ 1 - p = 1 - 0.66 = 0.34 \]
Now substituting the values into the standard deviation formula:
\[ \sigma = \sqrt{32 \times 0.66 \times 0.34} \]
Calculating inside the square root:
\[ \sigma = \sqrt{32 \times 0.66 \times 0.34} = \sqrt{32 \times 0.2244} = \sqrt{7.1552} \approx 2.6734 \]
Thus, the standard deviation of the number who enroll in college in a sample of 32 high school graduates is approximately 2.6734 (rounded to four decimal places).
Summary
- (a) Mean: 21.12
- (b) Standard Deviation: 2.6734