Title: Default for 36.2 (for Edinburgh) Exploring Data
Course: HELM Workbooks 2023-2024

Exploratory data analysis

The basics of EDA

The basic principles followed in EDA are:

- To measure the location and spread of a distribution we use statistics which are resistant to departures from normality;
- To summarise shape location and spread we use several statistics rather than just two;
- Visual displays as well as numerical displays are used to summarise information obtained about shape, location and spread.

You can see these principles illustrated below.

Traditionally, the location and spread of a distribution are measured by calculating its mean and standard deviation. The problem with these statistics is that is sensitive to the influence of extreme values. For example, the data set \[1,2,2,3,3,3,4,4,4,5,5,6\] has mean \(\bar{x}=3.5\) and standard deviation \(s_{n-1}=1.45\). These values are quite acceptable since the distribution is symmetrical about its mean of \(3.5\). The symmetry is easily seen simply by inspecting the data although the bar chart below might make the symmetry more obvious.

[Image: It is a data bar chart, in which any coloumn represents how many times (namely the frequency) the numbers of the data se]

The effects of changes in extreme values are easily illustrated by looking at what happens if we take the last number to be \(60\) instead of \(6\). This destroys the symmetry of the distribution and gives mean \(\bar{x}=8\) and standard deviation \(s_{n-1}=16.42\). Clearly, these values do not describe the distribution very well at all; a mean which is higher than \(92\%\) of the members of the distribution can hardly be described as a representative!

The simplest and most common examples of resistant or robust statistics are those based on the idea of rank order. These are the:

- Minimum value
- Lower quartile
- Median
- Upper quartile
- Maximum value

You will find that this five-number summary, especially when used in conjunction with the inter-quartile range, gives an adequate representation of a non-symmetrical distribution.

Notice that the median and quartile values in the above example are unaffected by the change in extreme value for the above example.

The box-and-whisker diagram

In order to visually summarise a data set, we can use a box and whisker plot. To draw a box-and-whisker diagram we first need to evaluate the of the five-number summaries. Consider the Inter-Party Competition data:

9.5 | 11.9 | 20.0 | 33.4 | 40.1 | 50.0 | 12.7 | 21.0 | 33.6 | 40.6
50.0 | 15.5 | 26.4 | 35.4 | 41.1 | 50.0 | 17.7 | 37.9 | 41.3 | 50.0
41.9 | 50.4 | 43.0 | 43.3 | 43.6 | 43.7 | 43.8 | 44.7 | 44.9 | 45.0
45.1 | 45.2 | 45.3 | 45.5 | 46.1 | 46.5 | 46.6 | 47.1 | 48.0 | 48.2
48.5 | 48.4 | 48.6 | 48.7 | 48.8 | 48.9 | 49.4 | 49.5 | 49.6 | 49.8

The five-number summaries are:

- Minimum \(= 9.5\)
- Lower Quartile \(= 40.1\)
- Median \(= 45.05\)
- Upper Quartile \(= 48.6\)
- Maximum \(= 50.4\)

The box-and-whisker diagram for this data is given below:

[Image]

It is constructed as follows:

1. The Box
  - The left-hand vertical is placed at the lower quartile (\(40.1\));
  - The right-hand vertical is placed at the upper quartile (\(48.6\));
  - The vertical in the box is placed at the median (\(45.05\)).
2. The Whiskers
These dashed lines span out from the lower/upper quartiles towards the minimum/maximum value, but their lengths should not exceed \(1.5\) times the inter-quartile range. Notice that the inter-quartile range of the data is \(8.5\).
  - The furthest extend permitted for the upper whisker is \(48.6 + 1.5 \times 8.5 = 61.35\).
  - The upper whisker therefore extends to the largest value in the data that does not exceed this limit, the maximum value \(50.4\).
  - The furtherest extent permitted for the lower whisker is \(40.1 - 1.5 \times 8.5 = 27.35\).
  - The lower whisker therefore extends to the smallest value in the data that does not exceed this limit, which is \(33.4\).
3. The Outlying Values
Mark as a dot any value that lie beyond furthest reaches of the whiskers. In this example, there are no values greater than \(61.35\), but there are \(8\) values which are less than \(37.35\).

Notice that about half of the data values lie in the box and that the tails show up well in the diagram. The diagram shows the left-skew (skewness refers to the tail) present in the data.

Note: There are several similar methods recommended by different authors for drawing box-and-whisker plots and so the methods recommended in statistical texts may vary a little from those given here.

Outliers

Outliers are values which are well outside the range covered by the vast bulk of a data set - a precise definition is impossible although some simple criteria do exist which may be used to detect outliers and accept or reject outliers. The large dots in box-and-whiskers diagrams illustrate the concept of outliers. Outliers can be extremely important since they may be (for example) erroneous data or they may point the way to further investigations of a data set.

For example, one statistic used to measure the state of the industrial development of a nation is the number of miles of railway track built per square mile of land. The box-and-whisker plot below summarises this variable for a total of 26 nations in the year 1972 according to one author.

[Image: This box-and-whisker plot summarises the number of miles of railway track built per square mile of land for a total of 2]

The figure for Cuba literally means that the whole island is covered by tracks which are placed about \(3\)m apart! Clearly, there is an error in the data. In fact, the \(1972\) Statistical Abstract of Latin America gives the figure for Cuba as \(71.75\) miles of railway per square mile of land. Note that the figure is still an outlier but is much more believable.

Criteria for rejecting outliers

As you already know, outliers may be taken to be observations which lie well outside the range of most of a sample. They are important for several reasons:

1. they can have a misleading effect on statistics such as the mean and standard deviation;
2. their occurrence may be due to incorrect observation, measurement or recording. In this case, it is often possible to correct the data;
3. their presence can induce a false skewness in a data set;
4. they may actually be members of a population not under consideration. For example, a study of urban families may involve recording the number of children in a family, say between \(0\) and \(4\) for the sake of discussion. An outlier might be caused by a rural family with, say, \(10\) children, living in temporary urban accommodation. This family is part of a different population.

Simple criteria exist which facilitate the detection of outliers. These criteria should be used with some caution and never automatically used simply to reject an outlier. You should always ask why such a value occurred in the first place and work to answer such a question sensibly before considering rejection. Two criteria for the detection of outliers are given below. Criterion \(1\) may be applied to data sets that are known to be normal in shape. Criterion \(2\) uses the five-number summary discussed above and may be applied to any data sets.

Criterion \(1\)

Knowing that some \(99.7%\) of a normal population lies within \(3\) standard deviations of the mean, we could treat any value further than say \(3.3\) standard deviations from the mean as on outlier. This choice essentially implies that a value has less than \(1\) in a \(1000\) of chance of occurring naturally outside the range defined by \(3.3\) standard deviations from the mean. Using standardized scores with as the potential outlier we can state the criterion

                 Accept \(x_0\) if \(|\dfrac{x_0-\bar{x}}{s_{n-1}}| \leq 3.3\)              Investigate if \(|\dfrac{x_0-\bar{x}}{s_{n-1}}|> 3.3\).

Note that \(\bar{x}\) and \(s_{n-1}\) are sample estimates of the mean and standard deviation of the population.

Criterion \(2\)

Using a five-number summary of a data set one can easily set up a criterion which may be used to classify outliers as either ‘moderate’ or ‘extreme’. The following diagram illustrates the situation where IQR is the Inter-Quartile Range.

[Image: diagram showing the situation of our distribution, we use the five-number summary of the data set in terms of the Inter-]

While all values classified as outliers should be investigated, this is particularly true of those classified as extreme outliers.

Skewness, gaps and multiple peaks

When exploring a data set, four properties worth looking for are outliers, skewness, gaps and multiple peaks. Outliers have been dealt with in some detail above so the comments below briefly address skewness, gaps and multiple peaks.

Skewness

If a skewed distribution is represented purely by two numbers, say the mean and standard deviation, then the representation will be inadequate. Remember that the term ‘skewness’ refers to the location of the ‘tail’ of a distribution.

[Image: In the picture two distributions characterised by right skew and left skew respectively]

As an example, the data set below gives the current required to burn out a component under test.

\(9.5\) | \(11.9\) | \(20.0\) | \(33.4\) | \(40.1\) | \(50.0\) | \(12.7\) | \(21.0\) | \(33.6\) | \(40.6\)
\(50.0\) | \(15.5\) | \(26.4\) | \(35.4\) | \(41.1\) | \(50.0\) | \(17.7\) | \(37.9\) | \(41.3\) | \(50.0\)
\(41.9\) | \(50.4\) | \(43.0\) | \(43.3\) | \(43.6\) | \(43.7\) | \(43.8\) | \(44.7\) | \(44.9\) | \(45.0\)
\(45.1\) | \(45.2\) | \(45.3\) | \(45.5\) | \(46.1\) | \(46.5\) | \(46.6\) | \(47.1\) | \(48.0\) | \(48.2\)
\(48.5\) | \(48.4\) | \(48.6\) | \(48.7\) | \(48.8\) | \(48.9\) | \(49.4\) | \(49.5\) | \(49.6\) | \(49.8\)

The data were obtained by measuring the current in mA applied to an electronic component under conditions of destructive testing give the following values for the mean, standard deviation, median and IQR: \[ \bar{x} = 40.72, s_{n-1} = 11.49; \text{median} = 45.05; \text{and} \text{ IQR} = 8.5.\]

The values of \(\bar{x}\) and \(s_{n-1}\) indicate that a lower average current with a greater spread will result in the destruction of the component than that indicated by the median and IQR. Clearly, further investigations are necessary to resolve this situation.

Gaps and multiple peaks

Distributions with gaps and multiple peaks can be very difficult to summarise easily. The table below summarise some 1972 data concerning adult literacy.

Interval | Frequency
\([0,10]\) | \(23\)
\((10,20]\) | \(11\)
\((20,30]\) | \(8\)
\((30,40]\) | \(9\)
\((40,50]\) | \(11\)
\((50,60]\) | \(3\)
\((60,70]\) | \(10\)
\((70,80]\) | \(14\)
\((80,90]\) | \(11\)
\((90,100]\) | \(28\)

The virtual lack of data between \(50\) and \(60\) indicates a gap and suggest that we are in fact dealing with two separate features. The usual summary statistics that you might be tempted to calculate like mean, standard deviation, median and IQR are not of much use since neither set indicates the gap or skewness. Without visual representation, such as a histogram, a single-peaked distribution tends to be assumed which is not true in this case.

Final comments on data representations

1. You should not rely on summary statistics such as the mean and standard deviation or median and IQR alone to represent a data set. Remember that if a distribution has outliers, gaps, skewness or multiple peaks, then the shape is probably more important than location and spread.
2. The shape of a distribution is better shown visually than numerically. Remember that a histogram and box-and-whisker plots emphasises the detail contained in the tails of a distribution.

Exercise

Consider the data set given below:

[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]

1. Write the five-number summary of our data set in the table below (it is convenient to place the items in the data set in rank order to find the five-number summary). Write the results with two decimal points.

Lowest Value |
Lower Quartile |
Median |
Upper Quartile |
Highest Value |
Inter-Quartile Range |

2. Does the box-and-whisker diagram tell you that the data set you are working with is symmetrical?
Feedback: (a) - The lowest value is [value]. - The lower quartile has rank position \(8\) and its value is [value]. - The median is between rank positions \(15\) and \(16\) and its value is \(\dfrac{[value]+[value]}{2} = [value]\). - The upper quartile has rank position \(23\) and its value is [value]. - The maximum is [value]. - The inter-quartile range is [value]. (b) The plot indicates that the distribution is not symmetrical, for example, you would expect the median value to appear midway between the hinges for a symmetrical distribution.

Exercise

Find the five-number summary and the mid-spread, high-spread and low-spread for the distribution given below:

[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]

Lowest Value |
Lower Hinge |
Low-Spread |
Median |
Mid-Spread |
Upper Hinge |
High-Spread |
Highest Value |
Feedback: - The lowest value is \([value]\) - The Lower Hinge is the 25th percentile, so it is given by \(([value]+[value])/2\) - The Low-Spread is the difference between the lowest value and the median: \(([value]+[value])/2-[value]\) - The Median is the middle number of the distribution:\(([value]+[value])/2\), in this case, since we have thirty elements the median is the average of the 15th and 16th elements of the distribution. - The Mid-Spread spread is given by the mean absolute deviation, which is the average distance to the mean:\(([value]+[value]/2-([value]+[value])/2\) - The Upper Hinge is the 75th percentile: \(([value]+[value])/2\) - The High-Spread is the difference between the highest value and the median\([value]-([value]+[value])/2\) - The highest value is \([value]\)

Exercise

Consider the following rounded data:

[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]

1. Use the summary to check the data for normality.

Lowest Value |
Lower Hinge |
Low-Spread |
Hinge to Extreme |
Median |
Median to Lower Hinge |
Median to Upper Hinge |
Upper Hinge |
High-Spread |
Hinge to Extreme |
Highest Value |

2. Is the distribution of the data symmetric?
Feedback: a) Data for normality - The lowest value is \([value]\) - The Lower Hinge is the 25th percentile: \(([value]+[value])/2\) - The Low-Spread is the difference between the lowest value and the median: \(([value]+[value])/2-[value]\) - The Hinge to Extreme is the difference Lower Hinge-Lowest Value:\(([value]+[value])/2-[value]\) - The Median is the middle number of the distribution: \(([value]+[value])/2\) - The Median to Lower Hinge is the difference Median-Lower Hinge: \(([value]+[value])/2-([value]+[value])/2\) - The Median to Upper Hinge is the difference Upper Hinge-Median: \(([value]+[value])/2-([value]+[value])/2\) - The Upper Hinge is the 75th percentile: \(([value]+[value])/2\) - The High-Spread is the difference between the highest value and the median: \([value]-([value]+[value])/2\) - The Hinge to Extreme is the difference Highest value-Upper Hinge: \([value]]-([value]+[value])/2\) - The highest value is \([value]\) b) - While there are no hard-and-fast rules for comparing figures such as those obtained here, many authors suggest that the figures should be within \(10%\) of each other before normality can be assumed. This is clearly not the case here. We conclude that the distribution of data being investigated is not symmetrical. In fact, the figures above suggest that the distribution is skewed to the left, a fact supported by the steam-and-leaf diagram of the same data to be found above. Remember that the term 'skewness' refers to the location of the 'tail' of a distribution. [Image: Right skew on the left and left skew on the right are represented]

Exercise

Manufacturing processes generally result in a certain amount of wasted material. For reasons of cost, companies need to keep such wastage to a minimum. The following (sorted) data were gathered over a 30-day period by a manufacturing company. The members given represent the percentage wastage of the amount of material used in the manufacturing process.

Daily Losses (\(\%\))

[value]
[value]
[value]

1. Find the mean and compute the standard deviation of the percentage losses of the material over this period. Give each answer to exactly 3 decimal places.
\(\bar{x} =\)
\(s_{n-1} =\)
2. Assuming that the losses are roughly normally distributed, apply an appropriate criterion to decide whether any of the losses are smaller of larger than might be expected by chance. Select all values that are outliers, or none if all values are as expected.
Feedback: Using standardized scores with \(x_0\) as the potential outlier we need to calculate the quantity \(|\dfrac{x-\bar{x}}{s_{n-1}}|\) and then accept \(x_0\) as a member of the distribution if \(|\dfrac{x-\bar{x}}{s_{n-1}}|<3.3\). We calculate: \(x\) | \(|\dfrac{x-\bar{x}}{s_{n-1}}|\) [value] | [value] [value] | [value] \(\dots\) | \(\dots\) [value] | [value] The calculation shows that all values of \(|\dfrac{x-\bar{x}}{s_{n-1}}|<3.3\) and so we conclude that the daily losses are within the range indicated by chance variation.

Exercise

The following data give the lifetimes in hours of \(50\) electric lamps.

[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]
[value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value] | [value]

1. Which is the mean lifetime of these data?   Write the result in the box (Round up each answer and enter 2 decimal places. If the answer has less than 2 decimal places, then enter zero(s) at the end)

2. Does the mean lifetime give a good indication of the expected lifetime of a lamp?
Feedback: 1. The mean lifetime is given by the usual formula. So the mean lifetime is [value]. 2. Yes.The mean lifetime gives a reasonable indication of what can be expected since the distribution is fairly symmetrical.

Exercise

During the winter of \(1893/94\), Lord Rayleigh conducted an investigation into the density of nitrogen gas taken from various sources. He had previously found discrepancies between the density of nitrogen obtained by chemical decomposition and nitrogen obtained by removing oxygen from the air. Lord Rayleigh’s investigations led to the discovery of argon. The raw data obtained during his investigations are given below.

Date | Source | Weight | Date | Source | Weight
\(29/11/93\) | NO | \(2.30143\) | \(26/12/93\) | N\(_2\)O | \(2.29889\)
\(05/12/93\) | NO | \(2.29816\) | \(28/12/93\) | N\(_2\)O | \(2.29940\)
\(06/12/93\) | NO | \(2.30182\) | \(09/01/94\) | NH\(_4\)NO\(_2\) | \(2.29849\)
\(08/12/93\) | NO | \(2.29890\) | \(13/01/94\) | NH\(_4\)NO\(_2\) | \(2.29889\)
\(12/12/93\) | Air | \(2.31017\) | \(29/01/94\) | Air | \(2.31024\)
\(14/12/93\) | Air | \(2.30986\) | \(30/01/94\) | Air | \(2.31030\)
\(19/12/93\) | Air | \(2.31010\) | \(01/02/94\) | Air | \(2.31028\)
\(22/12/93\) | Air | \(2.31001\) |  |  |

1. Organise the data into a frequency table using the classes \(2.29-2.30\), \(2.30-2.31\), \(2.31-2.32\). Fill in the blank in the following statement.
The highest class is obtained entirely from
2. Classify the data according to the two sources 'Air' and 'Other'. Order each data set and hence find the median, the hinges and the mid-spreads for each data set. Keep 5 decimal places.

 | Air | Other
Median |  |
Upper Quartile |  |
Lower Quartile |  |
IQR |  |
Feedback: Comment: What do the box-and-whisker plots tell you about the nitrogen obtained from the two sources? Box-and-whisker plot tells us that some other element is present in Air which is responsible for the additional weight. This additional element subsequently proved to be the inert gas argon.

Exercise

1. Is the variance measured in the same units as the mean?
2. Is the mean measured in the same units as the median?
3. Is the standard deviation measured in the same units as the mode?
4. Is the mode measured in the same units as the inter-quartile range?
5. Is the standard deviation measured in the same units as the minimum value?
6. Is the inter-quartile range measured in the same units as the quartiles?
Feedback: Look at the formulas of the various quantities to see in which unit they are expressed.

The steam-and-leaf diagram

You have already seen a basic stem-and-leaf diagram and you know that it shows the shape of distributions well. Here you will learn how to handle larger amounts of data to form stem-and-leaf diagrams. As you will be able to verify, one set of data can give rise to more than one stem-and-leaf diagram and highlight different aspects of the data. Look at the data set below:

\(11\)  \(9\)  \(6\)  \(27\)  \(17\)  \(2\)  \(19\)  \(12\)  \(8\)  \(17\)  \(3\)  \(10\)  \(23\)  \(6\)  \(18\)

\(13\)  \(11\)  \(22\)  \(13\)  \(19\)  \(4\)  \(12\)  \(23\)  \(34\)  \(19\)  \(15\)  \(7\)  \(40\)  \(16\)  \(20\).

Using the numbers to the left of the stem to represent \(10\)s and the numbers to the right to represent units, we obtain the stem-and-leaf diagram shown below:

[Image: stem and leaf diagram of the data set (11, 9, 6, 27, 17, 2, 19, 12, 8, 17, 3, 10, 23, 6, 18, 13, 11, 22, 13, 19, 4, 12,]

Notice that the skewed nature of the data stands out immediately. What also stands out are the following:

- the \(10\)s class has the highest number of members;
- the modal (most frequently occurring) value is \(19\);
- the \(30\)s and \(40\)s tie for the least number of members (one each).

This is not new information, we could have written these down after properly inspecting the original raw data. The advantage of the stem-and-leaf diagram is that it enables to express these properties in a clear and obvious way. As a further illustrative example, look at the idea in the table below which we will use to draw two stem-and-leaf diagrams.

\(9.5\) | \(11.9\) | \(20.0\) | \(33.4\) | \(40.1\) | \(50.0\) | \(12.7\) | \(21.0\) | \(33.6\) | \(40.6\)
\(50.0\) | \(15.5\) | \(26.4\) | \(35.4\) | \(41.1\) | \(50.0\) | \(17.7\) | \(37.9\) | \(41.3\) | \(50.0\)
\(41.9\) | \(50.4\) | \(43.0\) | \(43.3\) | \(43.6\) | \(43.7\) | \(43.8\) | \(44.7\) | \(44.9\) | \(45.0\)
\(45.1\) | \(45.2\) | \(45.3\) | \(45.5\) | \(46.1\) | \(46.5\) | \(46.6\) | \(47.1\) | \(48.0\) | \(48.2\)
\(48.5\) | \(48.4\) | \(48.6\) | \(48.7\) | \(48.8\) | \(48.9\) | \(49.4\) | \(49.5\) | \(49.6\) | \(49.8\)

Drawing steam-and-leaf diagram

We can start by looking at the data as it is displayed by a stem-and-leaf diagram. Here we will use two-digit leaves with the first digit representing units and the second digit representing tenths. The tens are represented by the numbers to the left of the stem.

[Image]

Notice that all we have really done is rank the data from the lowest value to the highest value reading from top to bottom. This particular display has over half of its members crushed into one class - the \(4\)-class.

It may be informative to split the classes and look more closely at the data. This can be done by:

1. rounding the raw data to two figures;
2. splitting each class according to the rule

               second digit \(0 - 4\) ......*       second digit \(5 - 9\) .....\(\bullet\)

The rounded raw now appears as follows

\(10\) | \(12\) | \(20\) | \(33\) | \(40\) | \(50\) | \(13\) | \(21\) | \(34\) | \(41\)
\(50\) | \(16\) | \(26\) | \(35\) | \(41\) | \(50\) | \(18\) | \(38\) | \(41\) | \(50\)
\(42\) | \(50\) | \(43\) | \(43\) | \(44\) | \(44\) | \(44\) | \(45\) | \(45\) | \(45\)
\(45\) | \(45\) | \(45\) | \(46\) | \(46\) | \(47\) | \(47\) | \(47\) | \(48\) | \(48\)
\(49\) | \(48\) | \(49\) | \(49\) | \(49\) | \(49\) | \(49\) | \(50\) | \(50\) | \(50\)

The stem and leaf diagram now becomes

[Image: Stem and leaf diagram of the data in the table before the picture]

Essentially, the classes have been split according to the usual rule for rounding decimals. This process can make certain information contained in the data a little more obvious than the previous stem and leaf diagram. For example:

- the values in the \(3\)-class are evenly distributed between both halves of the class in the sense that each half has two members;
- the \(4\)-class is split in the ratio \(2:1\) in favour of the upper half of the class;
- the values in the \(5\)-class are all in the lower half of the class.

You should have realised that:

- this is not new information - the new display has merely highlighted certain aspects of the raw data;
- some of the conclusions may have been affected by the rounding process.

Looking at the original stem and leaf diagram of the Inter-party data it is easy to produce a five-number summary of the data. The summary is:

1. The lowest value, this is \(9.50\);
2. The lower hinge, this is \(39\) (to find the lower hinge average the \(12\)th and \(13\)th values);
3. The median, this is \(45.05\) (the average of the \(25\)th and \(26\)th values);
4. The upper hinge, this is \(48.55\) (to find the upper hinge average the \(37\)th and \(38\)th values);
5. The highest values, this is \(50.04\).

The corresponding spreads are:

1. The low-spread, this is \(45.05-9.50=35.55\);
2. The mid-spread, this is \(48.55-39.00=9.55\);
3. The high-spread, this is \(50.40-45.05=5.35\).

Notice that the spreads indicate a considerable deviation from normality. For an ideal normal distribution, we would expect:

- The distances between the median and hinges to be equal
- The high-spread and low-spread to be equal
- The distance between the hinges and the extremes to be equal

as shown in the following diagram

[Image: normal distribution with the five essential statistics and the spreads]