Suppose the probability distribution px of random variable X
Suppose the probability distribution p(x) of random variable X is:
x
7
5
9
11
P(x)
.10
.30
.40
.20
a. Calculate the mean and standard deviation of X.
 b. Let f(x) = (5x + 7). Calculate the probability distribution of f(x).
 c. Calculate the mean and standard deviation of f(x). Use the \"short method.\"
| x | 7 | 5 | 9 | 11 | 
| P(x) | .10 | .30 | .40 | .20 | 
Solution
Consider the table:
x   P(x)   x P(x)   x^2 P(x)
 7   0.1   0.7   4.9
 5   0.3   1.5   7.5
 9   0.4   3.6   32.4
 11   0.2   2.2   24.2
Sum [xP(x)] = 8
 Sum[x^2 P(x)] = 69
As
Mean = Sum(x P(x))
Summing column 3,
Mean = 8 [ANSWER]
The standard deviation meanwhile is
standard deviation =sqrt{ Sum (x^2 P(x)) - [Sum (x P(x)]^2 }
= sqrt(69 - 8^2)
= 2.236067977 [ANSWER, standard deviation]
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b.
Substituting the values of f(x) = 5x + 7 to the x column,
f(x)   P[f(x)]
 42   0.1
 32   0.3
 52   0.4
 62   0.2
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c.
Mean (5x + 7) = 5 Mean (x) + 7 = 5(8) + 7 = 47 [ANSWER]
s (5x + 7) = s(5x) = 5 s(x) = 5*2.236067977 = 11.18033989 [ANSWER]


