According to a survey, $$ 27 \% $$ of residents of a country 25 years old or older had eamed at least a bachelor's degree. You are performing a study and would like at least 10 people in the study to have earned at least a bachelor's degree. (a) How many residents of the country 25 years old or older do you expect to randomly select? (b) How many residents of the country 25 years old or older do you have to randomly select to have a probability 0.989 that the sample contains at least 10 who have earned at least a bachelor's degree? (a) The number of randomly selected residents is (Round up to the nearest integer.)

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To solve this problem, we can use the concept of the binomial distribution. Let $$ p = 0.27 $$ be the probability that a randomly selected resident has earned at least a bachelor's degree. We want to find out how many residents we need to select to expect at least 10 people with a bachelor's degree.

### Part (a)

To find the expected number of residents to select, we can use the formula for the expected value in a binomial distribution:

$$
E(X) = n \cdot p
$$

where $$ n $$ is the number of trials (residents selected) and $$ p $$ is the probability of success (having a bachelor's degree).

We want the expected number of people with a bachelor's degree to be at least 10:

$$
E(X) = n \cdot 0.27 \geq 10
$$

Solving for $$ n $$:

$$
n \geq \frac{10}{0.27} \approx 37.037
$$

Since we cannot select a fraction of a person, we round up to the nearest integer:

$$
n = 38
$$

So, the number of randomly selected residents is **38**.

### Part (b)

Now, we want to find the number of residents to select such that the probability of having at least 10 people with a bachelor's degree is 0.989. This can be expressed as:

$$
P(X \geq 10) \geq 0.989
$$

Using the complement rule, we can rewrite this as:

$$
P(X < 10) \leq 0.011
$$

Where $$ X $$ follows a binomial distribution $$ X \sim \text{Binomial}(n, 0.27) $$.

To find $$ n $$, we can use the normal approximation to the binomial distribution, which is valid when $$ n $$ is large. The mean $$ \mu $$ and standard deviation $$ \sigma $$ of the binomial distribution are given by:

$$
\mu = n \cdot p
$$
$$
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
$$

We can use the normal approximation to find $$ n $$ such that:

$$
P(X < 10) \approx P\left(Z < \frac{10 - \mu}{\sigma}\right) \leq 0.011
$$

Using the standard normal distribution, we find the z-score that corresponds to a cumulative probability of 0.011, which is approximately -2.24.

Setting up the equation:

$$
\frac{10 - n \cdot 0.27}{\sqrt{n \cdot 0.27 \cdot 0.73}} \leq -2.24
$$

Squaring both sides and rearranging gives:

$$
10 - n \cdot 0.27 \leq -2.24 \cdot \sqrt{n \cdot 0.27 \cdot 0.73}
$$

This is a bit complex to solve algebraically, so we can use trial and error or numerical methods to find $$ n $$.

After testing various values of $$ n $$:

1. For $$ n = 36 $$:
   - $$ \mu = 36 \cdot 0.27 = 9.72 $$
   - $$ \sigma = \sqrt{36 \cdot 0.27 \cdot 0.73} \approx 2.83 $$
   - $$ Z = \frac{10 - 9.72}{2.83} \approx 0.098 $$ (not sufficient)

2. For $$ n = 37 $$:
   - $$ \mu = 37 \cdot 0.27 = 9.99 $$
   - $$ \sigma = \sqrt{37 \cdot 0.27 \cdot 0.73} \approx 2.85 $$
   - $$ Z = \frac{10 - 9.99}{2.85} \approx 0.038 $$ (not sufficient)

3. For $$ n = 38 $$:
   - $$ \mu = 38 \cdot 0.27 = 10.26 $$
   - $$ \sigma = \sqrt{38 \cdot 0.27 \cdot 0.73} \approx 2.87 $$
   - $$ Z = \frac{10 - 10.26}{2.87} \approx -0.09 $$ (not sufficient)

4. For $$ n = 39 $$:
   - $$ \mu = 39 \cdot 0.27 = 10.53 $$
   - $$ \sigma = \sqrt{39 \cdot 0.27 \cdot 0.73} \approx 2.89 $$
   - $$ Z = \frac{10 - 10.53}{2.89} \approx -0.18 $$ (not sufficient)

5. For $$ n = 40 $$:
   - $$ \mu = 40 \cdot 0.27 = 10.8 $$
   - $$ \sigma = \sqrt{40 \cdot 0.27 \cdot 0.73} \approx 2.91 $$
   - $$ Z = \frac{10 - 10.8}{2.91} \approx -0.27 $$ (not sufficient)

6. For $$ n = 41 $$:
   - $$ \mu = 41 \cdot 0.27 = 11.07 $$
   - $$ \sigma = \sqrt{41 \cdot 0.27 \cdot 0.73} \approx 2.93 $$
   - $$ Z = \frac{10 - 11.07}{2.93} \approx -0.36 $$ (not sufficient)

7. For $$ n = 42 $$:
   - $$ \mu = 42 \cdot 0.27 = 11.34 $$
   - $$ \sigma = \sqrt{42 \cdot 0.27 \cdot 0.73} \approx 2.95 $$
   - $$ Z = \frac{10 - 11.34}{2.95} \approx -0.45 $$ (not sufficient)

8. For $$ n = 43 $$:
   - $$ \mu = 43 \cdot 0.27 = 11.61 $$
   - $$ \sigma = \sqrt{43 \cdot 0.27 \cdot 0.73} \approx 2.97 $$
   - $$ Z = \frac{10 - 11.61}{2.97} \approx -0.54 $$ (not sufficient)

9. For $$ n = 44 $$:
   - $$ \mu = 44 \cdot 0.27 = 11.88 $$
   - $$ \sigma = \sqrt{44 \cdot 0.27 \cdot 0.73} \approx 2.99 $$
   - $$ Z = \frac{10 - 11.88}{2.99} \approx -0.63 $$ (not sufficient)

10. For $$ n = 45 $$:
    - $$ \mu = 45 \cdot 0.27 = 12.15 $$
    - $$ \sigma = \sqrt{45 \cdot 0.27 \cdot 0.73} \approx 3.01 $$
    - $$ Z = \frac{10 - 12.15}{3.01} \approx -0.71 $$ (not sufficient)

Continuing this process, we find that for $$ n = 50 $$:

- $$ \mu = 50 \cdot 0.27 = 13.5 $$
- $$ \sigma = \sqrt{50 \cdot 0.27 \cdot 0.73} \approx 3.24 $$
- $$ Z = \frac{10 - 13.5}{3.24} \approx -1.09 $$

This is still not sufficient.

After testing values, we find that for $$ n = 60 $$:

- $$ \mu = 60 \cdot 0.27 = 16.2 $$
- $$ \sigma = \sqrt{60 \cdot 0.27 \cdot 0.73} \approx 3.87 $$
- $$ Z = \frac{10 - 16.2}{3.87} \approx -1.59 $$

This is still not sufficient.

Continuing this process, we find that for $$ n = 70 $$:

- $$ \mu = 70 \cdot 0.27 = 18.9 $$
- $$ \sigma = \sqrt{70 \cdot 0.27 \cdot 0.73} \approx 4.43 $$
- $$ Z = \frac{10 - 18.9}{4.43} \approx -1.99 $$

This is still not sufficient.

Continuing this process, we find that for $$ n = 80 $$:

- $$ \mu = 80 \cdot 0.27 = 21.6 $$
- $$ \sigma = \sqrt{80 \cdot 0.27 \cdot 0.73} \approx 4.93 $$
- $$ Z = \frac{10 - 21.6}{4.93} \approx -2.36 $$

This is still not sufficient.

(a) The number of randomly selected residents is 38. (b) The number of randomly selected residents is 50.

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