NORMAL DISTRIBUTION

Statistics always deals with variability. There will be no statistics without variability. In our daily life, variability is ubiquitous. Today, we are going to study a kind of variability widely found in measurements. Let's see the case of manufacturing bottled water. Suppose that the labels on the bottles say that the volumes are 220 cc. Does it mean that all the bottles contain exactly 220 cc of water? In fact, if we measure the volumes accurately with a sophisticated measurement device, not all of them has the volume of 220 cc, but the volumes are around 220 cc. Some of them have the volume of 219 cc, some have the volume of 222 cc, 218, 219.5, etc. Statistically speaking, the volumes of water in the bottles have a mean, and since the volumes vary, they have a standard deviation.

Now, can we estimate the percentage of bottled water whose volume is less than 217 cc? This is what this post addresses. If we know that the volume is normally distributed with a certain mean and standard deviation, we will be able to answer it. What is a normal distribution?

Normal distribution (or Gaussian Distribution) is a continuous probability distribution with the following probability density function (pdf):

$n(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{1}{2} (\frac{x - \mu}{\sigma})^2}$

So, if the volumes are normally distributed with the mean of 220 cc and standard deviation of 0.5 cc, then the pdf is: $n(x) = \frac{1}{0.5 \sqrt{2 \pi}} e^{- \frac{1}{2} (\frac{x - 220}{0.5})^2}$

(to be continued)

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CONSTRUCTING THE FREQUENCY DISTRIBUTION TABLE

The frequency distribution table is a table that divides data into groups (classes) and shows how many data values occur in each group/class. Below is an example of frequency distribution table. Now we are learning how to create a frequency distribution table. Suppose we have a collection of ungrouped data on last year’s advertising expenditures of 40 logistics companies, recorded in millions Rupiahs. To construct a frequency distribution table of the ungrouped data, apply the following steps. Step 1: Find the range of the data The range (R) is defined as the difference between the largest data and the smallest data. In this case, R = 307 - 242 = 65. Step 2: Determine the number of categories/classes (k) Applying Sturges rule (k = 1 + 3,322 log n, where n = the number of data), we have: $k = 1 + 3.322 \: log \: 40 \approx 6.32$ As the value of k must be a natural number, 6.32 is rounded up to 7, so k = 7. Step 3: Determine the class width (c) To find c, use $...

THE QUARTILES AND MEDIAN OF GROUPED DATA

In this post, we will learn how to determine the quartiles when some quantitative data are presented in a frequency distribution table. For example, we have the following data, showing Flesch Readability Score of 80 monthly bulletin articles published by Britt and Co. Ltd. Find the quartiles of these readability scores. To answer this, first augment the table with a new column to the right of the frequency column, namely Data Numbers column. There are 5 data in the first class, so the class contains data no. 1 to no. 5. There are 7 data in the second class, so the class contains data no. 6 to no. 12. There are 13 data in the third class, so the class contains data no. 13 to no. 25. Continuing this way, we get the following: In this case, finding the first quartile means finding the 20 th data, after the data have been ordered from the smallest to the highest (20 = ¼ x 80). Note that the 20 th data is in the third class (20 is ...

CALCULATING THE MEAN OF GROUPED DATA

Sometimes quantitative data are presented in the form of a frequency distribution table (FDT). A typical FDT is as follows. Suppose that the table above presents the duration of 16 cell phone conversations between pairs of teens. There are 2 conversations with duration from 30 seconds to 44 seconds, 3 conversations with duration from 45 seconds to 59 seconds, etc. How do we calculate the mean of the data? Step 1: Determine the midpoint of each class If M i denotes the midpoint of class i, $M_{i} = \frac{LB_{i}+UB_{i}}{2}$ where LB i = lower bound of class i and UB i = upper bound of class i. The lower bounds of class 1, 2, 3, 4, 5 are 30, 45, 60, 75, 90, respectively and the upper bounds are 44, 59, 74, 89, 104, respectively. Then, $M_{1} = \frac{30+44}{2} = 37$. Similarly, $M_{2} = \frac{45+59}{2} = 52$. Continuing this way, we have the following table. Step 2: Multiply each class frequency f i by the corresponding class midpoint ...

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