Exercise 8.1: Measures of Dispersion - Problem Questions with ... - Free Printable

You've provided a complete solution for calculating the standard deviation of grouped data using the assumed mean method, with assumed mean \( A = 60 \). Let’s walk through and explain each step to ensure clarity.

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📊 Given Data:

| \( x_i \) | \( f_i \) | \( d_i = x_i - 60 \) | \( f_i d_i \) | \( f_i d_i^2 \) |
|-----------|-----------|----------------------|---------------|------------------|
| 45 | 5 | -15 | -75 | 1125 |
| 50 | 13 | -10 | -130 | 1300 |
| 55 | 4 | -5 | -20 | 100 |
| 60 | 9 | 0 | 0 | 0 |
| 65 | 5 | 5 | 25 | 125 |
| 70 | 4 | 10 | 40 | 400 |
| Total | N = 40| | Σfᵢdᵢ = -160 | Σfᵢdᵢ² = 3050 |

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## ✔ Step-by-Step Explanation:

🔹 Step 1: Understand the Assumed Mean Method

Instead of calculating deviations from the actual mean (which we don’t know yet), we pick a convenient value — here, A = 60 — as the assumed mean. We compute deviations \( d_i = x_i - A \), then use these to find the standard deviation.

This method simplifies calculations, especially when dealing with large numbers or grouped data.

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🔹 Step 2: Formula for Standard Deviation (Assumed Mean Method)

The formula used is:

\[
\sigma = \sqrt{ \frac{\sum f_i d_i^2}{N} - \left( \frac{\sum f_i d_i}{N} \right)^2 }
\]

This is derived from the general formula for variance:

\[
\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}
\]

But since we’re using an assumed mean \( A \), we adjust using:

\[
\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left( \frac{\sum f_i d_i}{N} \right)^2
\]

Note: The term \( \left( \frac{\sum f_i d_i}{N} \right) \) is actually the correction factor — it’s the difference between the assumed mean and the actual mean.

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🔹 Step 3: Plug in the Values

From the table:

- \( \sum f_i d_i^2 = 3050 \)
- \( \sum f_i d_i = -160 \)
- \( N = 40 \)

So,

\[
\sigma = \sqrt{ \frac{3050}{40} - \left( \frac{-160}{40} \right)^2 }
\]

Compute each part:

- \( \frac{3050}{40} = 76.25 \)
- \( \frac{-160}{40} = -4 \), so \( (-4)^2 = 16 \)

Then,

\[
\sigma = \sqrt{76.25 - 16} = \sqrt{60.25}
\]

\[
\sqrt{60.25} = 7.76 \quad \text{(since } 7.76^2 = 60.2176 \approx 60.25\text{)}
\]

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## ✔ Final Answer:

> Standard deviation \( \sigma \approx 7.76 \)

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## 💡 Why This Works

Even though we didn’t calculate the actual mean \( \bar{x} \), the formula accounts for it implicitly. The correction term \( \left( \frac{\sum f_i d_i}{N} \right) \) adjusts for the fact that we used an assumed mean instead of the true mean.

In fact, you can also compute the actual mean:

\[
\bar{x} = A + \frac{\sum f_i d_i}{N} = 60 + \frac{-160}{40} = 60 - 4 = 56
\]

So the true mean is 56, and the deviations were calculated from 60 — hence the need for the correction.

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## 🧠 Summary

- Used assumed mean method for easier computation.
- Calculated deviations \( d_i = x_i - 60 \).
- Computed \( \sum f_i d_i \) and \( \sum f_i d_i^2 \).
- Applied the formula:
\[
\sigma = \sqrt{ \frac{\sum f_i d_i^2}{N} - \left( \frac{\sum f_i d_i}{N} \right)^2 }
\]
- Got result: σ ≈ 7.76

This is a correct and well-explained solution!

✔ Answer: Standard deviation \( \sigma \approx 7.76 \)