Probability Distributions - Oxford University Press

Figure 5.5
➜ Excel Solution
Mileage data Cells B4:B9 Values
Frequency, f Cells C4:C9 Values
Relative frequency Cell D4:D9 Formula: =C4/$C$11
Copy formula from D4:D9
Total f Cell C11 Formula: =SUM(C4:C9)
Total relative frequency Cell D11 Formula: =SUM(D4:D9)
Figure 5.5 illustrates the calculation
process.
We observe from Figure 5.5
that the relative frequency
for 440–460 miles travelled is
0.283333. This implies that we
have a chance or probability
of 34/120 that the miles travelled
lies within this class.
Note Thus, relative frequencies provide estimates of the probability for that class,
or value, to occur. If we were to plot the histogram of relative frequencies we would
in fact be plotting out the probabilities for each event e.g. P(400–420 miles) = 0.10,
P(420–440 miles) = 0.225.
Mileage travelled by 120 salesmen
The distribution of probabilities
given in Table 5.3 and graphi­
0.300000
cally represented in Figure 5.6
0.250000
are different ways of illustrating
0.200000
the probability distribution.
0.150000
Whereas for the frequency
0.100000
distribution the area under
0.050000
the histogram is proportional
0.000000
400–420 420–440 450–460 460–480 480–500 500–520 to the total frequency, for the
Miles travelled, X
probability distribution the
Figure 5.6
area is proportional to total
probability (= 1.0).
Given a particular probability distribution we can determine the probability for any
event associated with it. Thus, P(400–460 miles) = P(400 ≤ x ≤ 460) = Area under the distribution
from 400 to 460 = P(400–420) + P(420–440) + P(440–460) = 0.10 + 0.225 + 0.283 =
0.608. Thus, we have a probability estimate of 61% for the mileage travelled to lie between
400 and 460 miles.
Relative frequency or Probability
Probability Distributions
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