Given `Sigma_(i=1)^(20) a_i=100, Sigma_(i-1)^(20) a_i^2=600, Sigma_(i-1)^(20) b_i=140, Sigma_(i-1)^(20)b_i^2=1000`, where `a_i,b_i` denotes length and weight of an observations. Then which is more varying ?

Options

A. (a)Length
B. (b)Weight
C. (c)Equal C.V.
D. (d)None of these

Answer

To determine which of the two variables \( a_i \) (length) and \( b_i \) (weight) has more variation, we will calculate the variance for both sets of observations using the provided summations.

### Step 1: Calculate the variance for \( a_i \)

The formula for variance \( \sigma^2 \) is given by:

\[ \sigma^2 = \frac{\Sigma x_i^2}{n} - \left( \frac{\Sigma x_i}{n} \right)^2 \]

Where:
- \( \Sigma x_i^2 \) is the sum of the squares of the observations,
- \( \Sigma x_i \) is the sum of the observations,
- \( n \) is the number of observations.

For \( a_i \):
- \( \Sigma_{i=1}^{20} a_i = 100 \)
- \( \Sigma_{i=1}^{20} a_i^2 = 600 \)
- \( n = 20 \)

Substituting these values into the variance formula:

\[ \sigma_a^2 = \frac{600}{20} - \left( \frac{100}{20} \right)^2 \]

Calculating each term:

1. \( \frac{600}{20} = 30 \)
2. \( \frac{100}{20} = 5 \) and \( 5^2 = 25 \)

Now substituting back:

\[ \sigma_a^2 = 30 - 25 = 5 \]

### Step 2: Calculate the variance for \( b_i \)

Now we will calculate the variance for \( b_i \):

For \( b_i \):
- \( \Sigma_{i=1}^{20} b_i = 140 \)
- \( \Sigma_{i=1}^{20} b_i^2 = 1000 \)
- \( n = 20 \)

Using the variance formula:

\[ \sigma_b^2 = \frac{1000}{20} - \left( \frac{140}{20} \right)^2 \]

Calculating each term:

1. \( \frac{1000}{20} = 50 \)
2. \( \frac{140}{20} = 7 \) and \( 7^2 = 49 \)

Now substituting back:

\[ \sigma_b^2 = 50 - 49 = 1 \]

### Step 3: Compare the variances

Now we have:
- Variance of \( a_i \): \( \sigma_a^2 = 5 \)
- Variance of \( b_i \): \( \sigma_b^2 = 1 \)

Since \( \sigma_a^2 > \sigma_b^2 \), we conclude that the lengths \( a_i \) are more varying than the weights \( b_i \).

### Final Answer

The variable \( a_i \) (length) is more varying than \( b_i \) (weight).

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