E1C04 09/14/2010
14:7:41 Page 129
SOLUTION To estimate the probability that a single measurement will have a value within
some interval, we need to solve the integral
1
ffiffiffiffiffiffi
2p
p
Z z 1 ¼1
0
e
Àb
2 =2
db
over the interval defined by z 1 . Table 4.3 lists the solutions for this integral. Using Table 4.3 for z 1 ¼
1, we find P(z 1 ) ¼ 0.3413. However, since z 1 ¼ (x 1 À x
0 )/s, the probability that any measurement of
x will produce a value within the interval 0 x x
0
þ s is 34.13%. Since the normal distribution is
symmetric about x
0 , the probability that x will fall within the interval defined between Àz 1 d and
þz 1 d for z 1 ¼ 1 is (2)(0.3413) ¼ 0.6826 or 68.26%. Accordingly, if we made a single measurement
of x, the probability that the value found would lie within the interval x
0
À s x x
0
þ s would be
68.26%.
COMMENT Similarly, for z 1 ¼ 1.96, the probability would be 95.0%.
Example 4.3
The statistics of a well-defined varying voltage signal are given by x
0
¼ 8.5 V and s
2
¼ 2.25 V
2 . If a
single measurement of the voltage signal is made, determine the probability that the measured value
indicated will be between 10.0 and 11.5 V.
KNOWN x
0
¼ 8.5 V
s
2
¼ 2.25 V
2
ASSUMPTIONS Signal has a normal distribution about x
0 .
FIND P (10.0 x 11.5)
SOLUTION To find the probability that x will fall into the interval 10.0 x 11.5 requires
finding the area under p(x) bounded by this interval. The standard deviation of the variable is
s ¼
ffiffiffiffiffi
s 2
p ¼ 1:5 V, so our interval falls under the portion of the p(x) curve bounded by z 1 ¼ (10.0 À
8.5)/1.5 ¼ 1 and z 1 ¼ (11.5 À 8.5)/1.5 ¼ 2. From Table 4.3, the probability that a value will fall between
8.5 x 10.0 is P(8.5 x 10.0) ¼ P(z 1 ¼ 1) ¼ 0.3413. For the interval defined by 8.5 x 11.5,
P(8.5 x 11.5) ¼ P(z 1 ¼ 2) ¼ 0.4772. The area we need is just the overlap of these two intervals, so
P 10:0 x 11:5
ð
Þ ¼ P 8:5 x 11:5
ð
Þ À P 8:5 x 10:0
ð
Þ
¼ 0:4772 À 0:3413 ¼ 0:1359
So there is a 13.59% probability that the measurement will yield a value between 10.0 and 11.5 V.
COMMENT In general, the probability that a measured value will lie within an interval defined
by any two values of z 1 , such as z a and z b , is found by integrating p(x) between z a and z b . For a
normal density function, this probability is identical to the operation, P(z b ) À P(z a ).
4.4 STATISTICS OF FINITE-SIZED DATA SETS
We now try to predict the behavior of measured variable x based on a finite-sized sampling of x. We
do this by comparing the statistics from that sampling to an assumed probability density function for
the population. For example, if we recall the box of bearings discussed in Section 4.1, some two
4.4 Statistics of Finite-Sized Data Sets 129
14:7:41 Page 129
SOLUTION To estimate the probability that a single measurement will have a value within
some interval, we need to solve the integral
1
ffiffiffiffiffiffi
2p
p
Z z 1 ¼1
0
e
Àb
2 =2
db
over the interval defined by z 1 . Table 4.3 lists the solutions for this integral. Using Table 4.3 for z 1 ¼
1, we find P(z 1 ) ¼ 0.3413. However, since z 1 ¼ (x 1 À x
0 )/s, the probability that any measurement of
x will produce a value within the interval 0 x x
0
þ s is 34.13%. Since the normal distribution is
symmetric about x
0 , the probability that x will fall within the interval defined between Àz 1 d and
þz 1 d for z 1 ¼ 1 is (2)(0.3413) ¼ 0.6826 or 68.26%. Accordingly, if we made a single measurement
of x, the probability that the value found would lie within the interval x
0
À s x x
0
þ s would be
68.26%.
COMMENT Similarly, for z 1 ¼ 1.96, the probability would be 95.0%.
Example 4.3
The statistics of a well-defined varying voltage signal are given by x
0
¼ 8.5 V and s
2
¼ 2.25 V
2 . If a
single measurement of the voltage signal is made, determine the probability that the measured value
indicated will be between 10.0 and 11.5 V.
KNOWN x
0
¼ 8.5 V
s
2
¼ 2.25 V
2
ASSUMPTIONS Signal has a normal distribution about x
0 .
FIND P (10.0 x 11.5)
SOLUTION To find the probability that x will fall into the interval 10.0 x 11.5 requires
finding the area under p(x) bounded by this interval. The standard deviation of the variable is
s ¼
ffiffiffiffiffi
s 2
p ¼ 1:5 V, so our interval falls under the portion of the p(x) curve bounded by z 1 ¼ (10.0 À
8.5)/1.5 ¼ 1 and z 1 ¼ (11.5 À 8.5)/1.5 ¼ 2. From Table 4.3, the probability that a value will fall between
8.5 x 10.0 is P(8.5 x 10.0) ¼ P(z 1 ¼ 1) ¼ 0.3413. For the interval defined by 8.5 x 11.5,
P(8.5 x 11.5) ¼ P(z 1 ¼ 2) ¼ 0.4772. The area we need is just the overlap of these two intervals, so
P 10:0 x 11:5
ð
Þ ¼ P 8:5 x 11:5
ð
Þ À P 8:5 x 10:0
ð
Þ
¼ 0:4772 À 0:3413 ¼ 0:1359
So there is a 13.59% probability that the measurement will yield a value between 10.0 and 11.5 V.
COMMENT In general, the probability that a measured value will lie within an interval defined
by any two values of z 1 , such as z a and z b , is found by integrating p(x) between z a and z b . For a
normal density function, this probability is identical to the operation, P(z b ) À P(z a ).
4.4 STATISTICS OF FINITE-SIZED DATA SETS
We now try to predict the behavior of measured variable x based on a finite-sized sampling of x. We
do this by comparing the statistics from that sampling to an assumed probability density function for
the population. For example, if we recall the box of bearings discussed in Section 4.1, some two
4.4 Statistics of Finite-Sized Data Sets 129