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```ANATOMY OF A DISCRETE RANDOM VARIABLE:
Calculations for the Mean, m, and Standard Deviation, s
The mean of a discrete random variable, denoted by m. is actually the
mean of its probability distribution. The mean of a discrete random
variable is also called its expected value, and is denoted E(x). The
mean or expected value of a discrete random variable is the value that
we expect to observe per repetition on average if we perform an
experiment a large number of times. We use the following formula to
calculate the mean.
m   xP(x)
Example:
Let x denote the number of breakdowns
for an ice cream machine during a given
week. The probability distribution of x
is shown to the right.
s   x 2 P( x)  m 2
x
0
1
2
3
P(x)
.15
.20
.35
.30
Using the data from the problem to the left, the following table shows all the
calculations needed for the computation of the standard deviation of x.
Note that
 P( x)  1
To calculate the mean (or expected value) of a discrete random variable x,
we multiply each value of x by the corresponding probability. We then
take the sum of the resulting products. This sum gives the mean (or
expected value) of the discrete random variable x as is shown below.
x
P(x)
xP(x)
0
.15
0(.15) = .00
1
.20
1(.20) = .20
2
.35
2(.35) = .70
3
.30
3(.30) = .90
 xP( x)  1.80
The mean (or expected value) is
The standard deviation of a discrete random variable, denoted by s
measures the spread of its probability distribution. A higher value for the
standard deviation of a discrete random variable x indicates that x can
assume values over a larger range about the mean. On the other hand, a
smaller value for the standard deviation indicates that most of the values that
x can assume are clustered closely about the mean. We use the following
formula to calculate the standard deviation.
m   xP( X )  1.80  E( x)
IN WORDS: We may expect this machine to break down on average
1.80 times during a given week. This DOES NOT mean this machine will
break down exactly 1.80 times during a given week. (Actually, a machine
cannot break down 1.8 times). This simply means that if we observe for
many weeks, this machine will break down a different number of times
during different weeks. If we take the average of all these weeks, we can
expect this average to be 1.80 times.
xP(x)
P(x)
x P( x)
x
_____________________________________________
x
2
2
0
.15
.00
0
.00
1
.20
.20
1
.20
2
.35
.70
4
1.40
_____________________________________________
3
.30
.90
9
2.70
 xP( x)  1.80
 x2 P(x)  4.30
We perform the following steps to compute the standard deviation:
Step 1) Compute the mean. The sum of the products xP(x), recorded in
the third column, gives the mean of x (m = S xP(x) = 1.80).
Step 2) Square each value of x and record it in the fourth column. Then
multiply the values of x2 by the corresponding values of P(x). The
resulting values of x2P(x) are recorded in the fifth column. The sum of the
fifth column is the S x2P(x) = 4.30
Step 3) Substitute the values of m and S x2P(x) in the formula for the
standard deviation of x.
s   x P( x)  m 2  4.30  (1.80) 2  1.06  1.03
(rounded)
```
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