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In question 8.54, you are given a table of insurance claims and are asked to calculate the **Standard Deviation** and **Variance** based on the provided data. Let's break this down step by step.

### Provided Data:
The table gives the **number of claims** for various claim amounts:

| Claim (Rs.)   | No. of claims |
|---------------|---------------|
| 1 – 1000      | 5             |
| 1001 – 2000   | 30            |
| 2001 – 3000   | 60            |
| 3001 – 4000   | 70            |
| 4001 – 5000   | 80            |

To calculate the **standard deviation** and **variance**, we follow these steps:

### Step 1: Find the Midpoint of Each Claim Range
For each claim range, we take the midpoint, which represents the approximate claim amount for the group.

\[
\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}
\]

The midpoints for each range:

- \( \text{Midpoint for } 1 – 1000 = \frac{1 + 1000}{2} = 500.5 \)
- \( \text{Midpoint for } 1001 – 2000 = \frac{1001 + 2000}{2} = 1500.5 \)
- \( \text{Midpoint for } 2001 – 3000 = \frac{2001 + 3000}{2} = 2500.5 \)
- \( \text{Midpoint for } 3001 – 4000 = \frac{3001 + 4000}{2} = 3500.5 \)
- \( \text{Midpoint for } 4001 – 5000 = \frac{4001 + 5000}{2} = 4500.5 \)

### Step 2: Calculate the Weighted Mean (Expected Value)
The weighted mean (μ) is calculated as:

\[
\mu = \frac{\sum (\text{Midpoint} \times \text{No. of claims})}{\sum \text{No. of claims}}
\]

Substitute the values:

\[
\mu = \frac{(500.5 \times 5) + (1500.5 \times 30) + (2500.5 \times 60) + (3500.5 \times 70) + (4500.5 \times 80)}{5 + 30 + 60 + 70 + 80}
\]

\[
\mu = \frac{2502.5 + 45015 + 150030 + 245035 + 360040}{245}
\]

\[
\mu = \frac{800622.5}{245} = 3267.03
\]

So, the **mean claim amount** is approximately **3267.03 Rs**.

### Step 3: Calculate the Variance
The variance is given by:

\[
\text{Variance} = \frac{\sum (\text{No. of claims} \times (\text{Midpoint} - \mu)^2)}{\sum \text{No. of claims}}
\]

Now, calculate the squared deviations:

\[
\sum (\text{No. of claims} \times (\text{Midpoint} - \mu)^2) = (5 \times (500.5 - 3267.03)^2) + (30 \times (1500.5 - 3267.03)^2) + (60 \times (2500.5 - 3267.03)^2) + (70 \times (3500.5 - 3267.03)^2) + (80 \times (4500.5 - 3267.03)^2)
\]

Simplifying each term:

\[
= 5 \times (2766.53)^2 + 30 \times (1766.53)^2 + 60 \times (766.53)^2 + 70 \times (233.47)^2 + 80 \times (1233.47)^2
\]

\[
= 5 \times 7653128.5 + 30 \times 3124190.3 + 60 \times 587560.9 + 70 \times 54510.6 + 80 \times 1523113.7
\]

\[
= 38265642.5 + 93725709 + 35253654 + 3815742 + 121849096
\]

\[
= 293265844.5
\]

Now divide by the total number of claims (245):

\[
\text{Variance} = \frac{293265844.5}{245} = 1194502.22
\]

So, the **variance** is approximately **1,194,502**.

### Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance:

\[
\text{Standard Deviation} = \sqrt{1,194,502} \approx 1093.84
\]

Thus, the **standard deviation** is approximately **1093 Rs**.

### Final Answer:
- Variance: **1,194,502**
- Standard Deviation: **1093**

Would you like more details or have any questions?

Here are some related questions:
1. How does changing the number of claims affect the standard deviation?
2. What is the relationship between variance and standard deviation?
3. How would the calculations change if the claim ranges were unequal?
4. Can the standard deviation ever be greater than the mean in such cases?
5. How do weighted means differ from simple averages?

**Tip:** Always use weighted mean when working with grouped data to ensure accurate calculations.

Learn how to solve for variance and standard deviation in question 8.54 using grouped insurance claims data. Detailed calculations show the weighted mean, variance, and standard deviation for the given claims range. A perfect example of applying statistics in real-life scenarios.

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