Several anthropology students are unprepared for a surprise multiple-choice test with 12 questions, and all of their answers are guesses. Each question has six possible answers, and only one of them is correct.

Find the mean for the number of correct answers for such students.
(Round answer to one decimal place.)
μ=μ=

Find the standard deviation for the number of correct answers for such students.
(Round answer to two decimal places.)
σ=σ=

Give the range for the usual number of correct answers.
(Enter answer as an interval using square-brackets only with whole numbers.)
usual values =


Would it be unusual for a student to pass by guessing and getting at least 8 correct answers?

    yes, it would be unusual

Let's break this down step by step.

### **1. Mean number of correct answers (μ)**

This is a binomial distribution problem. In this case, the number of trials \(n = 12\) (because there are 12 questions), and the probability of success on each trial (getting the right answer) is \(p = \frac{1}{6}\), since there are 6 possible answers and only one is correct.

The formula for the **mean** (μ) of a binomial distribution is:
\[
μ = n \times p
\]
Substitute the values:
\[
μ = 12 \times \frac{1}{6} = 2
\]
So, the **mean number of correct answers** is **2**.

### **2. Standard deviation (σ)**

The formula for the **standard deviation** (σ) of a binomial distribution is:
\[
σ = \sqrt{n \times p \times (1 - p)}
\]
Substitute the values:
\[
σ = \sqrt{12 \times \frac{1}{6} \times \left(1 - \frac{1}{6}\right)} = \sqrt{12 \times \frac{1}{6} \times \frac{5}{6}}
\]
\[
σ = \sqrt{\frac{60}{36}} = \sqrt{\frac{5}{3}} \approx 1.291
\]
So, the **standard deviation** is approximately **1.29** (rounded to two decimal places).

### **3. Range for usual number of correct answers**

For the usual values, we use the rule that the "usual" values are within 2 standard deviations of the mean. This means the interval is:
\[
\mu - 2\sigma \quad \text{to} \quad \mu + 2\sigma
\]
Substitute the values:
\[
2 - 2(1.29) \quad \text{to} \quad 2 + 2(1.29)
\]
\[
2 - 2.58 \quad \text{to} \quad 2 + 2.58
\]
\[
-0.58 \quad \text{to} \quad 4.58
\]
Since the number of correct answers must be a whole number, the range of usual values is:
\[
[0, 4]
\]
Thus, the **usual values** for the number of correct answers are **[0, 4]**.

### **4. Would it be unusual to get at least 8 correct answers?**

To determine whether it's unusual to get at least 8 correct answers, we need to look at how likely this is. The usual number of correct answers falls within the range [0, 4], and 8 is far above this range.

The probability of getting 8 or more correct answers in a binomial distribution with \(n = 12\) and \(p = \frac{1}{6}\) is very small. Therefore, getting 8 correct answers is **unusual**.

### **Summary:**
- Mean number of correct answers: **2**
- Standard deviation: **1.29**
- Usual values: **[0, 4]**
- It would be unusual to get at least 8 correct answers by guessing.

Do you need more details on any part of the explanation or have further questions?

---

Here are some related questions to expand your understanding:

1. How do you calculate the probability of getting exactly 3 correct answers using a binomial distribution?
2. What would the mean and standard deviation be if the number of questions were doubled to 24?
3. How does the binomial distribution change if the probability of guessing correctly changes from 1/6 to 1/4?
4. What are the implications of changing the number of answer choices (e.g., to 4 or 5 choices) on the mean and standard deviation?
5. How can you use the normal approximation to the binomial distribution in this case?

Tip: Always remember that binomial distributions have a fixed number of trials and two outcomes (success or failure), making them useful for modeling situations like this.

Several anthropology students are unprepared for a surprise multiple-choice test with 12 questions, and all of their answers are guesses. Each question has six possible answers, and only one of them is correct. Find the mean for the number of correct answers for such students. (Round answer to one decimal place.) μ=μ= Find the standard deviation for the number of correct answers for such students. (Round answer to two decimal places.) σ=σ= Give the range for the usual number of correct answers. (Enter answer as an interval using square-brackets only with whole numbers.) usual values = Would it be unusual for a student to pass by guessing and getting at least 8 correct answers?

Learn how to calculate the mean, standard deviation, and usual values for a binomial distribution problem involving 12 multiple-choice questions with a 1/6 probability of guessing correctly. This step-by-step guide is perfect for high school students.

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