The question asks us to find the number of observations, denoted by 'n', given the correlation coefficient (r), the sum of products of deviations from mean ($\sum xy$), the standard deviation of y ($\sigma_y$), and the sum of squared deviations of x ($\sum x^2$).

We are given:

• Correlation coefficient, $r = 0.5$
• Sum of products of deviations, $\sum xy = 120$
• Standard deviation of y, $\sigma_y = 8$
• Sum of squared deviations of x, $\sum x^2 = 90$

The formula for the correlation coefficient (r) using deviations from the mean and standard deviations is:

$$r = \frac{\sum xy}{n \sigma_x \sigma_y}$$

We know that the standard deviation $\sigma_x$ is related to the sum of squared deviations $\sum x^2$ by the formula:

$$\sigma_x = \sqrt{\frac{\sum x^2}{n}}$$

Substituting the expression for $\sigma_x$ into the formula for r:

$$r = \frac{\sum xy}{n \sqrt{\frac{\sum x^2}{n}} \sigma_y}$$

$$r = \frac{\sum xy}{\sqrt{n^2 \cdot \frac{\sum x^2}{n}} \sigma_y}$$

$$r = \frac{\sum xy}{\sqrt{n \sum x^2} \sigma_y}$$

Now, let's substitute the given values into this formula:

$$0.5 = \frac{120}{\sqrt{n \times 90} \times 8}$$

Simplify the denominator:

$$0.5 = \frac{120}{8 \sqrt{90n}}$$

$$0.5 = \frac{15}{\sqrt{90n}}$$

Now, we need to solve for n. Multiply both sides by $\sqrt{90n}$:

$$0.5 \times \sqrt{90n} = 15$$

Divide both sides by 0.5:

$$\sqrt{90n} = \frac{15}{0.5}$$

$$\sqrt{90n} = 30$$

To eliminate the square root, square both sides of the equation:

$$(\sqrt{90n})^2 = 30^2$$

$$90n = 900$$

Finally, solve for n by dividing both sides by 90:

$$n = \frac{900}{90}$$

$$n = 10$$

Thus, the value of 'n' is 10.