A probability distribution of the number of claims per person per year for an auto insurance company are shown in the following table.


Insurance claims per year	Percentage
1	0.1
2	0.2
3	0.1
4	0.15
5	0.4
6	0.05


Find the percentage of claims within one standard deviation of the mean. Give your answer as a decimal. 
   ≈ 0.35

To find the percentage of claims within one standard deviation of the mean for the given probability distribution, we need to calculate:

1. **Mean** (Expected Value \( \mu \)),
2. **Standard Deviation** \( \sigma \),
3. Determine the range for \( \mu - \sigma \) to \( \mu + \sigma \),
4. Find the total percentage of claims that fall within this range.

### Step 1: Calculate the Mean (\( \mu \))
The mean \( \mu \) is given by:

\[
\mu = \sum x_i P(x_i)
\]

Where \( x_i \) are the insurance claims per year and \( P(x_i) \) are the corresponding percentages (probabilities).

\[
\mu = 1(0.1) + 2(0.2) + 3(0.1) + 4(0.15) + 5(0.4) + 6(0.05)
\]

Let's calculate the mean.

\[
\mu = 0.1 + 0.4 + 0.3 + 0.6 + 2 + 0.3 = 3.7
\]

### Step 2: Calculate the Standard Deviation (\( \sigma \))
The standard deviation \( \sigma \) is given by:

\[
\sigma = \sqrt{\sum (x_i - \mu)^2 P(x_i)}
\]

Substitute the values of \( x_i \) and \( P(x_i) \), and \( \mu = 3.7 \).

\[
\sigma^2 = (1 - 3.7)^2(0.1) + (2 - 3.7)^2(0.2) + (3 - 3.7)^2(0.1) + (4 - 3.7)^2(0.15) + (5 - 3.7)^2(0.4) + (6 - 3.7)^2(0.05)
\]

Let's calculate this step by step.

\[
\sigma^2 = (6.76)(0.1) + (2.89)(0.2) + (0.49)(0.1) + (0.09)(0.15) + (1.69)(0.4) + (5.29)(0.05)
\]

\[
\sigma^2 = 0.676 + 0.578 + 0.049 + 0.0135 + 0.676 + 0.2645 = 2.257
\]

\[
\sigma = \sqrt{2.257} \approx 1.5
\]

### Step 3: Range for \( \mu - \sigma \) to \( \mu + \sigma \)
The range is:

\[
\mu - \sigma = 3.7 - 1.5 = 2.2
\]
\[
\mu + \sigma = 3.7 + 1.5 = 5.2
\]

So, we are looking for the percentage of claims that fall between 2.2 and 5.2.

### Step 4: Calculate the Percentage of Claims within One Standard Deviation

From the table, the relevant claims within the range 2.2 to 5.2 are those corresponding to 3, 4, and 5 claims per year. Summing the probabilities:

\[
P(3) = 0.1, \quad P(4) = 0.15, \quad P(5) = 0.4
\]

\[
\text{Total percentage} = 0.1 + 0.15 + 0.4 = 0.65
\]

Thus, the percentage of claims within one standard deviation of the mean is **0.65**.

### Conclusion:
The percentage of claims within one standard deviation of the mean is \( \approx 0.65 \), not \( 0.35 \).

Would you like further details or have any questions?

Here are 5 related questions:
1. How is variance related to standard deviation?
2. What does the mean signify in a probability distribution?
3. How can we apply Chebyshev’s inequality in this context?
4. Why is the empirical rule less relevant to this distribution?
5. What other moments (e.g., skewness) could describe this distribution?

**Tip:** The standard deviation helps determine how spread out the data points are around the mean in a probability distribution.

A probability distribution of the number of claims per person per year for an auto insurance company are shown in the following table. Find the percentage of claims within one standard deviation of the mean. Give your answer as a decimal. ≈ 0.35

Learn how to calculate the percentage of auto insurance claims that fall within one standard deviation of the mean. This detailed step-by-step solution covers key concepts such as mean, standard deviation, and probability distribution. Suitable for high school students in Grades 11-12.

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