Question

A national college researcher reported that 64% of students who graduated from high school in 2012 enrolled in college. Twenty seven high school graduates are sampled.

(a) What is the mean number who enroll in college in a sample of 27 high school graduates? Round the answer to two decimal places.

(b) What is the standard deviation of the number who enroll in college in a sample of 27 high school graduates? Round the answer to four decimal places.

Answer 1

To solve these questions, we can utilize the properties of a binomial distribution. Here, the number of trials \( n \) is 27 (the number of high school graduates sampled), and the probability of success (enrolling in college) \( p \) is 0.64.

### (a) Mean Number Who Enroll in College

The mean \( \mu \) of a binomial distribution is calculated using the formula:

\[
\mu = n \cdot p
\]

Substituting the values:

\[
\mu = 27 \cdot 0.64 = 17.28
\]

Thus, the mean number of students who enroll in college is **17.28** (rounded to two decimal places).

### (b) Standard Deviation of the Number Who Enroll in College

The standard deviation \( \sigma \) of a binomial distribution is calculated using the formula:

\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

First, we need to calculate \( (1 - p) \):

\[
1 - p = 1 - 0.64 = 0.36
\]

Now substituting the values into the formula for standard deviation:

\[
\sigma = \sqrt{27 \cdot 0.64 \cdot 0.36}
\]

Calculating the inside of the square root first:

\[
27 \cdot 0.64 = 17.28
\]
\[
17.28 \cdot 0.36 = 6.2208
\]

Now, taking the square root:

\[
\sigma = \sqrt{6.2208} \approx 2.4942
\]

Thus, the standard deviation of the number of students who enroll in college is approximately **2.4942** (rounded to four decimal places).

### Summary:
- (a) Mean: **17.28**
- (b) Standard Deviation: **2.4942**