A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is $$\frac{x}{121}$$%. The value of $$x$$ is ______.

Given readings: 1.22 mm, 1.23 mm, 1.19 mm, and 1.20 mm. $$\bar{x} = \frac{1.22 + 1.23 + 1.19 + 1.20}{4} = \frac{4.84}{4} = 1.21 \text{ mm}$$

$$|\Delta x_1| = |1.22 - 1.21| = 0.01 \text{ mm}$$

$$|\Delta x_2| = |1.23 - 1.21| = 0.02 \text{ mm}$$

$$|\Delta x_3| = |1.19 - 1.21| = 0.02 \text{ mm}$$

$$|\Delta x_4| = |1.20 - 1.21| = 0.01 \text{ mm}$$

$$\overline{\Delta x} = \frac{0.01 + 0.02 + 0.02 + 0.01}{4} = \frac{0.06}{4} = 0.015 \text{ mm}$$

$$\text{Percentage error} = \frac{\overline{\Delta x}}{\bar{x}} \times 100 = \frac{0.015}{1.21} \times 100$$ $$= \frac{1.5}{1.21} = \frac{150}{121}\%$$

Comparing with the given form $$\frac{x}{121}\%$$:

$$\frac{x}{121} = \frac{150}{121}$$

Hence, x = 150.