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.

Do you try to pad an insurance claim to cover your deductible? About 38% of all U.S. adults will try to pad their insurance claims! Suppose that you are the director of an insurance adjustment office. Your office has just received 124 insurance claims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.) half or more of the claims have been padded

To solve this problem, we need to determine if it’s appropriate to use the normal approximation to the binomial distribution, and if so, use it to estimate the requested probability.

### Step 1: Check if it's appropriate to use the normal approximation

The **normal approximation to the binomial distribution** is appropriate when both \( np \geq 5 \) and \( n(1-p) \geq 5 \), where:
- \( n \) is the number of trials (in this case, the number of insurance claims),
- \( p \) is the probability of success (in this case, the probability of someone padding their insurance claim).

Here, we have:
- \( n = 124 \) (number of claims),
- \( p = 0.38 \) (probability of padding a claim).

Now let's check the conditions:

- \( np = 124 \times 0.38 = 47.12 \)
- \( n(1 - p) = 124 \times 0.62 = 76.88 \)

Since both values are greater than 5, it is appropriate to use the normal approximation.

### Step 2: Apply the normal approximation

The binomial distribution \( X \sim \text{Binomial}(n = 124, p = 0.38) \) can be approximated by a normal distribution with:
- Mean: \( \mu = np = 47.12 \),
- Standard deviation: \( \sigma = \sqrt{np(1 - p)} = \sqrt{124 \times 0.38 \times 0.62} \approx 5.4446 \).

We are asked to find the probability that **half or more** of the claims are padded. This corresponds to finding \( P(X \geq 62) \), where 62 is half of 124.

### Step 3: Use the continuity correction

For a better approximation, we apply the continuity correction by subtracting 0.5, so we need to find \( P(X \geq 61.5) \).

### Step 4: Standardize the value

Now, we standardize the value 61.5 to a z-score using the formula:

\[
z = \frac{x - \mu}{\sigma}
\]

Substitute the values:

\[
z = \frac{61.5 - 47.12}{5.4446} \approx 2.6428
\]

### Step 5: Find the corresponding probability

Using the standard normal distribution table, a z-score of 2.6428 corresponds to a cumulative probability of approximately 0.9959. To find \( P(X \geq 61.5) \), we subtract this value from 1:

\[
P(X \geq 61.5) = 1 - 0.9959 = 0.0041
\]

### Final Answer:

The probability that half or more of the claims have been padded is approximately **0.0041** (or 0.41%).

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Do you have any further questions or would you like more details on any step? Here are some related questions for deeper understanding:

1. What is the continuity correction, and why do we apply it?
2. How would the solution change if the probability of padding claims was different?
3. What are the key conditions for using the normal approximation to the binomial distribution?
4. How do you interpret the z-score in the context of probability?
5. How would this problem be solved without the normal approximation?

**Tip:** When using the normal approximation for binomial distributions, always check if the conditions \( np \geq 5 \) and \( n(1-p) \geq 5 \) are satisfied before proceeding!

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.

Do you try to pad an insurance claim to cover your deductible? About 38% of all U.S. adults will try to pad their insurance claims! Suppose that you are the director of an insurance adjustment office. Your office has just received 124 insurance claims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.) half or more of the claims have been padded.

Learn how to apply the normal approximation to estimate the probability that half or more of 124 insurance claims are padded, given that 38% of adults try to pad their claims. This solution covers the binomial distribution, normal approximation, and continuity correction, suitable for advanced high school or college-level students.

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