In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ocean fishing for billfish is very popular in a certain region of Mexico. In this region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 21 strikes. Find the following probabilities. (Round your answers to four decimal places.)

12 or fewer fish were caught

To determine whether it's appropriate to use the normal approximation to the binomial distribution, we need to check the following two conditions:

1. **\( np \geq 5 \)** (where \( n \) is the number of trials and \( p \) is the probability of success)
2. **\( n(1-p) \geq 5 \)** (where \( 1-p \) is the probability of failure)

Given:
- \( n = 21 \) (number of trials, or the total number of strikes)
- \( p = 0.48 \) (probability of success, or the probability of catching a fish)

Now, let's check the conditions:

1. **\( np = 21 \times 0.48 = 10.08 \geq 5 \)**  
2. **\( n(1-p) = 21 \times (1 - 0.48) = 21 \times 0.52 = 10.92 \geq 5 \)**

Both conditions are satisfied, so it's appropriate to use the normal approximation.

### Step 1: Define the problem in terms of the binomial distribution
- The binomial distribution has a mean \( \mu = np = 21 \times 0.48 = 10.08 \)
- The standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{21 \times 0.48 \times 0.52} = \sqrt{5.2416} \approx 2.2882 \)

### Step 2: Apply the continuity correction
To use the normal distribution as an approximation to the binomial distribution, we need to apply a continuity correction when finding the probability of 12 or fewer fish. We approximate \( P(X \leq 12) \) by calculating \( P(X \leq 12.5) \).

### Step 3: Standardize the variable
We now convert \( X = 12.5 \) to a z-score:

\[
z = \frac{X - \mu}{\sigma} = \frac{12.5 - 10.08}{2.2882} = \frac{2.42}{2.2882} \approx 1.0576
\]

### Step 4: Use the standard normal distribution
Using a standard normal distribution table or calculator, we find the probability corresponding to \( z = 1.0576 \).

\[
P(Z \leq 1.0576) \approx 0.8553
\]

### Final Answer:
The probability that 12 or fewer fish were caught is approximately \( 0.8553 \).

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Would you like more details or clarifications?

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Here are five related questions to explore this concept further:

1. How does the continuity correction affect the approximation in a binomial distribution?
2. What would happen if the sample size \( n \) were smaller in this problem?
3. How is the normal distribution derived from the binomial distribution?
4. What are other common applications of the normal approximation to the binomial distribution?
5. How would the probability change if the success rate (probability of catching a fish) were higher?

**Tip**: Always check both conditions \( np \geq 5 \) and \( n(1-p) \geq 5 \) before using the normal approximation to the binomial distribution.

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ocean fishing for billfish is very popular in a certain region of Mexico. In this region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 21 strikes. Find the following probabilities. (Round your answers to four decimal places.)

12 or fewer fish were caught.

Learn how to use the normal approximation to the binomial distribution to estimate the probability of catching 12 or fewer fish. This problem covers key concepts such as the binomial distribution, z-scores, and continuity correction. Suitable for high school students in Grades 11-12.

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