What is the Length of a Rice Grain?

Just recently, me, together with my colleagues did one of the most exciting things in the world - measure the length of rice grains. No kidding, it was fun. It did take some time to measure a lot, however, in the end, with our data, we can say that we indeed have attained the best approximate to what the actual length of a grain of rice is.

In measuring rice grains, we used two different measuring devices: the Vernier and Micrometer Calipers, the ones shown below:

Figure 1: A Micrometer Caliper
(source: http://www.shan-precision.com/products.htm)

Figure 2: A Vernier Caliper
(source: http://ecx.images-amazon.com/images/I/61dvgTmvXiL.jpg)

The Vernier Caliper is commonly used to measure the lengths of objects, with an extra degree of precision compared to the ruler. However, besides the length of objects, this caliper can measure depth, as well as thickness. The degree of error with this device is \(\pm\)0.02 mm.

The Micrometer Caliper on the other hand still is used to measure the length and thickness of certain objects. The good thing with this measuring device is its degree of error which is \(\pm\)0.01 mm, just 0.01 mm away from the degree of error of the Vernier Caliper. In reality, this may not seem so significant, but on the long run, this device is more accurate than the Vernier Caliper, and that is a very good thing.

So using those calipers, we were able to measure the lengths of grains of a single variety of rice. In the end, we were able to measure the length of 155 grains of rice. Not much, but it will do.

There was one thing that I need to take note in the measuring each rice grain though. While we were still using the Vernier Caliper, me and my group mates ignored most of the divisions and read only up to the first decimal place. Our professor told us that this was not a preferable thing to do because in the end we might (but most probably will) have a systematic error. So when we were already measuring the lengths of the rice grains using the Micrometer Caliper, we made sure that we read up to the last decimal place.

After listing down the individual lengths of rice, we noticed that most of our data stated that the length of a rice grain was 7.40 mm. We know this to be true because out of the 155 grains of rice, 20 of them had a length of 7.40 mm. So we argued that the length of a rice grain was so.

Tabulating our data using Microsoft Excel© and calculating for the mean \(\mu\), standard deviation \(\sigma\), and variance \(\sigma^2\) (we also took note of the number of data points \(n\)), we found out that the mean of our data was 7.085613 mm, or up to 3 significant figures, 7.09 mm. The uncertainty, as calculated, was 2.008561, or to 3 significant figures, 2.01. Therefore, on average, the length of a grain of rice was 7.09\(\pm\)2.01 mm. This seems to be true enough (to some extent), because as our prof said, the length of a grain of rice, according to one study, was in fact \(7.09\pm0.36\) mm. So our past estimate on the length of a rice grain, which was 7.40 mm, was actually wrong. Now, with the mean, we were on the right track to finding out the actual length of a grain of rice.

Figure 3: Print Screen of Tabulated Data Using Microsoft Excel©

The next thing we did was to place the data points into bins so as to start normalizing it so that we can fit the data to a gaussian curve (or distribution curve). To do this, I got the difference between the highest (8.46 mm) and lowest (5.00 mm), and divided that by 50, so that I can group our data into 50 bins. Doing so:
\begin{eqnarray}
\text{Bin Width}=\dfrac{Q_\text{max}-Q_\text{min}}{50} \\
=\dfrac{8.46-5.00}{50}=0.0692
\end{eqnarray}
After grouping the data into their respective bins, we calculated the frequency of the occurrence of the grouped data \(X_i\) (where \(i\in\{1,2,3,\dotsc,n\}\)). I used the equation:
\begin{equation}
\text{Frequency}=\dfrac{X_i}{0.0692n}
\end{equation}
Our problem now was to find the gaussian fit. In doing so, we looked at different sources in the internet and found a recurring formula, which was:
\begin{equation}
f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\text{exp}\left[-\dfrac{(X_i-\mu)^2}{2\sigma^2}\right]
\end{equation}
We had to take note though, that with this equation, we assume that our data was normalized.

Figure 4: Print Screen of the Remaining Data Required to Establish the Gaussian Fit

Applying this fit, and graphing it together with the calculated frequency, we immediately saw a beautiful bell curve.

Figure 5: Gaussian Fit to Our Data

As seen in figure 5, the vertical axis refers to the frequency of the data, colored in blue, and the horizontal axis shows the middle of the bins we used, in mm. The red line is our wonderful and beautiful gaussian fit.

From here, we were able to determine, on average, the acceptable range for the length of rice to be 7.0760 \(\pm\) 0.02 mm to 7.1452 \(\pm\) 0.02 mm (this was according to the bin directly below the peak of the curve).

Later on, our professor gave us the data collected by the whole class. We were told to do the same thing to this data as what we did to ours and compare the results. I did so, and after applying the same methods, I got a similar, but different distribution curve.

Figure 6: Gaussian Fit to the Class' Data

Interestingly, we see that the curve peaks at the same point (7.04 \(\pm\) 0.02 mm) as our group's data. However, the thing that gets my interest is that the curve is closer to the peak of the frequencies. Compared to the distribution curve for the class' data (Figure 6), the distribution curve of our group's data (Figure 5) was far from the peak of the frequencies plotted. Although it reached it's maximum value at the same point, the frequencies and graph in the class' data was 'closer' to the distribution curve. I guess this just means that the more rice grains you measure, the more the frequencies get closer to becoming a distribution curve itself.

So, what is the length of 1 grain of rice? The distribution curves of our group's and the whole class' data have spoken. One grain of rice is approximately 7.04 \(\pm\) 0.02 mm long.