A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.34 mm, 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:

$$\text{Least Count} = 0.01\text{ mm}$$

Since the measurement cannot be more precise than the instrument's capability, both the average value and the uncertainty (standard deviation) must be rounded off to two decimal places:

$$\text{Average diameter} = 5.5375\text{ mm} \approx 5.54\text{ mm}$$

$$\text{Uncertainty} = 0.07395\text{ mm} \approx 0.07\text{ mm}$$

$$\text{Diameter} = (5.54 \pm 0.07)\text{ mm}$$