Question

If both mean and standard deviation of 50 observations $x_1, x_2, x_3 ...x_{50}$ are equal to 8. If the $2)^2, (x_2-2)^2, (x_3-2)^2,...(x_{50} - 2)^2$ is $\lambda$, then $\frac{\lambda}{2}$ is equal to _____

Answer 1

2500

Answer 2

Given:

• Mean, $\bar{x} = 8$
• Standard Deviation, $\sigma = 8 \;\Rightarrow\; \sigma^2 = 64$
• Number of observations, $n = 50$

1. Calculate the sums:

   • Total sum of observations: $$\sum_{i=1}^{50} x_i = n \cdot \bar{x} = 50 \times 8 = 400$$
   • Sum of squared deviations from the mean: $$\sum_{i=1}^{50}(x_i-8)^2 = n \cdot \sigma^2 = 50 \times 64 = 3200.$$

2. Express $(x_i-2)^2$ in terms of $(x_i-8)$:

   $$x_i-2 = (x_i-8) + 6$$

   Therefore,

   $$(x_i-2)^2 = (x_i-8)^2 + 12(x_i-8) + 36.$$

3. Sum over all observations:

   $$\sum_{i=1}^{50}(x_i-2)^2 = \sum_{i=1}^{50}(x_i-8)^2 + 12\sum_{i=1}^{50}(x_i-8) + 50 \times 36.$$

   Notice that:

   $$\sum_{i=1}^{50}(x_i-8) = \left(\sum_{i=1}^{50} x_i\right) - 50 \times 8 = 400 - 400 = 0.$$

   Hence,

   $$\sum_{i=1}^{50}(x_i-2)^2 = 3200 + 0 + 1800 = 5000.$$

   Let $\lambda = 5000$.

4. Find $\frac{\lambda}{2}$:

   $$\frac{\lambda}{2} = \frac{5000}{2} = 2500.$$