Animated Solution for Physics - Physics and Measurement: A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings; $5.50\text{ mm}$, $5.55\text{ mm}$, $5.45\text{ mm}$ and $5.65\text{ mm}$. The average of these four readings is $5.5375\text{ mm}$ and the standard deviation of the data is $0.07395\text{ mm}$. The average diameter of the pencil should be therefore recorded as

Imagine you are in the physics lab, carefully measuring the diameter of a pencil using a vernier caliper. You take four readings: $5.50\text{ mm}$, $5.55\text{ mm}$, $5.45\text{ mm}$, and $5.65\text{ mm}$.

When you plug these numbers into a calculator to find the average and the standard deviation (which represents our error), the calculator spits out $5.5375\text{ mm}$ for the average and $0.07395\text{ mm}$ for the error. It looks incredibly precise, doesn't it? But here is the catch: a calculator doesn't know physics. It doesn't know the limitations of the instrument you used. Writing down all those digits is scientifically incorrect because it implies a level of precision that your vernier caliper simply cannot provide.

To report our final answer correctly, we must look at our raw data. Notice how every reading is recorded up to the second decimal place (e.g., $5.50$). This tells us that the least count of our vernier caliper is $0.01\text{ mm}$.

Our final answer must respect this precision. We cannot magically gain more precision just by doing math. The rules of significant figures dictate that our final reported value and its error must align with the least count of the measuring instrument.

Let's fix the error first. The standard deviation, or error, should generally be rounded to the same decimal place as the least count of the instrument.

Since our least count is $0.01\text{ mm}$ (two decimal places), we must round our calculated error of $0.07395\text{ mm}$ to the second decimal place. Looking at the third decimal digit (which is $3$), we see it is less than $5$, so we round down. Wait, let's look closer: $0.07395$ rounded to two decimal places is simply $0.07\text{ mm}$.

Next, we round the average. The golden rule here is that the average must be rounded to the exact same decimal place as the rounded error.

Our error is rounded to the hundredths place ($0.07$). Therefore, our average of $5.5375\text{ mm}$ must also be rounded to the hundredths place. We look at the digit in the thousandths place, which is $7$. Since $7$ is greater than or equal to $5$, we round up the digit in the hundredths place. Thus, $5.53$ becomes $5.54\text{ mm}$.

Putting it all together, the correctly recorded diameter is the rounded average plus or minus the rounded error.

Final Answer: $d=(5.54\pm 0.07)\text{ mm}$

This perfectly respects the precision of our instrument and the rules of significant figures. Always remember: you are the physicist, and you dictate the precision, not the calculator!