Supplementary Exercise 4.48 of IPS7e
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X = class of randomly chosen son of a father in class 1

(a) The probability for class 5, 0.01, corresponds to a percentage of 1%.

(b) The probabilities sum to 1, and are all >=0, so the probability
distribution is valid.

(c) X<=3 expresses at most social class 3.
    P(X<=3) = P(X=1) + P(X=2) + P(X=3)
            = 0.48 + 0.38 + 0.08 = 0.94.

(d) X<3 expresses lower social class than 3.
    P(X<3) = P(X=1) + P(X=2)
           = 0.48 + 0.38 = 0.86.

(e) X>=4 expresses one of the two highest classes.
    P(X>=4) = P(X=4) + P(X=5)
            = 0.05 + 0.01 = 0.06.

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Addition: graphical representation of the distribution

The best depiction of the distribution is a bar graph (or bar chart). 
In PSLS and IPS, the somewhat misleading term "probability histogram" is used
for graphs of discrete probability distributions. The term is misleading
because these are not (and should not look like) histograms. There are no
values in between the discrete outcomes, and if the graph is drawn with
bars (instead of vertical lines) the bars should NOT join. 

If the values 1,2,3,4,5 are typed into column c1 (labeled "x") and the 
probabilities into column c2 (labeled "p(x)"), the following Minitab 
command (which we also used for bar charts of observed data) produces the graph:

MTB > Chart ( 'p(x)' ) * 'x';
SUBC>   Summarized;
SUBC>   Bar.