**Use the normal distribution to approximate the desired probability.**

Find the probability that in 377 tosses of a single 13-sided die, we will get at least 23 threes.

Round your answer to **4 places after the decimal point**.

To solve this problem using the normal distribution, we'll first model the situation as a binomial distribution and then apply the normal approximation.

### Step 1: Define the binomial distribution
The problem involves 377 tosses of a 13-sided die. Each toss can either result in a "three" or not. The probability of getting a three on a single toss is:

\[
p = \frac{1}{13}
\]

Thus, the probability of not getting a three is:

\[
q = 1 - p = \frac{12}{13}
\]

The number of trials is \(n = 377\), and we are looking for the probability of getting at least 23 threes. This is a binomial distribution with parameters \(n\) and \(p\).

### Step 2: Normal approximation to the binomial distribution
The mean (\(\mu\)) and standard deviation (\(\sigma\)) for the binomial distribution are calculated as:

\[
\mu = n \cdot p = 377 \cdot \frac{1}{13} \approx 29
\]
\[
\sigma = \sqrt{n \cdot p \cdot q} = \sqrt{377 \cdot \frac{1}{13} \cdot \frac{12}{13}} \approx 5.188
\]

### Step 3: Apply continuity correction
To approximate the probability of getting at least 23 threes, we apply a continuity correction. We're interested in:

\[
P(X \geq 23) = P(X > 22.5)
\]

### Step 4: Standardize the value
We now convert this to the standard normal distribution (Z) using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

Substituting \(X = 22.5\), \(\mu = 29\), and \(\sigma = 5.188\):

\[
Z = \frac{22.5 - 29}{5.188} \approx -1.252
\]

### Step 5: Find the probability using the standard normal distribution
Now we find the cumulative probability corresponding to \(Z = -1.252\) from the standard normal table. The value for \(P(Z \leq -1.252)\) is approximately 0.1056.

Since we want the probability of \(X \geq 23\), we compute the complement:

\[
P(X \geq 23) = 1 - P(Z \leq -1.252) = 1 - 0.1056 = 0.8944
\]

Thus, the probability that in 377 tosses of a 13-sided die we will get at least 23 threes is approximately:

\[
\boxed{0.8944}
\]

Let me know if you need further explanation or have questions!

Here are five related questions to expand on this:

1. How would the probability change if the number of tosses increased to 500?
2. What is the expected number of threes in 500 tosses of a 13-sided die?
3. How does the continuity correction affect the approximation in this case?
4. Can the normal approximation be used for any number of trials? Why or why not?
5. How do you calculate the variance for a binomial distribution?

**Tip:** When using the normal approximation for the binomial distribution, always apply a continuity correction to improve accuracy.

Find the probability that in 377 tosses of a single 13-sided die, we will get at least 23 threes. Round your answer to 4 places after the decimal point.

Find the probability that in 377 tosses of a 13-sided die, at least 23 will result in a three. This problem uses normal approximation to binomial distribution and includes concepts like continuity correction, mean, and standard deviation calculations. Learn how to solve it step by step.

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