Wolfram Alpha:

```Box and Whisker Plots
---------------------

Q1                Q2                Q3
1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17
3.9 4.1 4.2 4.3 4.3 4.4 4.4 4.4 4.4 4.5 4.5 4.6 4.7 4.8 4.9 5.0 5.1

The median is Q2 = 4.4.

The first half has eight values, so the median is the average of
the middle two:

Q1 = (4.3 + 4.3)/2 = 4.3

Likewise, the median of the second half is:

Q3 = (4.7 + 4.8)/2 = 4.75

The interquartile range is Q3 - Q1 = 4.75 - 4.3 = 0.45

Thw min and max are 3.9 and 5.1 respectively.

Therefore, the BW plot looks like:

------------
|---|    |       |----|
------------
-------+---+----+-------+----+-----------
3.9 4.3  4.4     4.75 5.1

Outliers
--------

One general rule of thumb to determine if a data point is an
outlier is to define LIMITS as follows:

UL = Q3 + (1.5 * IQR)

LL = Q1 - (1.5 * IQR)

Any data points above (>) and below (<) these limits can then be
considered to be outliers.

Consider:

------------
|---|    |       |----|   o
------------
-------+---+----+-------+----+---+--------
44  78   80      85   96  99

UL = 85 + (1.5 * 7) = 95.5

LL = 78 - (1.5 * 7) = 67.5

Therefore, 99 would be correctly classified as an outlier.

Now, supposing we were given a 5 number summary as follows:

Min Q1 M  Q3 Max
--- -- -- -- ---
44  78 80 85 96

The question is "can we construct a box plot using only
this information?".

UL = 85 + (1.5 * 7) = 95.5

LL = 78 - (1.5 * 7) = 67.5

The answer is 'no' because 44 falls below 67.5 and must
be an outlier meaning that we cannot determine the min
value for the whisker.  A similar argument applies for
96.

What about 66 78 79 86 95?

UL = 86 + (1.5 * 8) = 98

LL = 78 - (1.5 * 8) = 66

Now, the answer is 'yes' because 66 is coincident with
the lower limit meaning that it not an outlier and can
be regarded as the min.  Similarly, 95 is the less than
the upper limit and, therefore, is a valid max.

```