Page 146 - Applied statistics and probability for engineers

124 Chapter 4/Continuous Random Variables and Probability Distributions

Example 4-14 Normally Distributed Current Continuing Example 4-13, what is the probability that a current mea-
surement is between 9 and 11 milliamperes? From Fig. 4-15, or by proceeding algebraically, we have
(
10 / 2 < 11 − )
P 9 10 2 / <(
<
P 9 < X 11) = ( ( − ) X − ) ( 10 / 2)
0 5)
= ( 0 5 < Z < . P Z < 0 5) − (Z < − .
0 5) = (
P − .
.
0
P
0 69146 − .
= . 0 30854 = .
0 38292
Determine the value for which the probability that a current measurement is less than this value is 0.98. The requested
(
value is shown graphically in Fig. 4-16. We need the value of x such that P X < x) = 0 98. . By standardizing, this prob-
ability expression can be written as
(
/
P X < x) = ( ( 10 / < ( x − ) ) 2
10
P X − ) 2
P Z (
= ( , x − ) ) =10 / 2 0 98
.
(
Appendix Table III is used to i nd the z-value such that P Z < z) = 0 98. . The nearest probability from Table III results in
(
P Z < 2 06) = 0 97982
.
.
(
.
10
/
Therefore, x − ) 2 = 2 05, and the standardizing transformation is used in reverse to solve for x. The result is
=
x = ( 2 2 05 ) + 10 14 1. mA
.
Example 4-15 Signal Detection Assume that in the detection of a digital signal, the background noise follows
a normal distribution with a mean of 0 volt and standard deviation of 0.45 volt. The system assumes
a digital 1 has been transmitted when the voltage exceeds 0.9. What is the probability of detecting a digital 1 when
none was sent?
Let the random variable N denote the voltage of noise. The requested probability is
0 9 . ⎞
N
(
−
.
.
P Z > 2) =
P N > 0 9) = P ⎛ ⎜ ⎝ 0 45 > 0 45⎠ ⎟ = ( 1 0 97725 = 0 02275
.
.
.
This probability can be described as the probability of a false detection.
Determine symmetric bounds about 0 that include 99% of all noise readings. The question requires us to i nd x such
that P − ( x < N < x) = .99. A graph is shown in Fig. 4-17. Now,
0
P − ( x < N < x) = ( x .45 < N / 0 .45 < x / 0 45
. )
P − / 0
= P − ( x 0 45/ . , Z , x 0 45/ . ) = 0 9. 99
From Appendix Table III,
(
P − .2 58 < Z < 2 . ) = .99
5
8
0
Therefore,
.
x / 0 45 = 2 58
.
and
x = . (0 . ) = .16
4
5
2 58
1
Suppose that when a digital 1 signal is transmitted, the mean of the noise distribution shifts to 1.8 volts. What is the prob-
ability that a digital 1 is not detected? Let the random variable S denote the voltage when a digital 1 is transmitted. Then,
. −
(
1 8
.
P S < 0 9 . ) = P ⎛ ⎜ ⎝ S − . < 0 9 1 8 . ⎞ ⎟ = ( − 0 02275
P Z < 2) =
.
0 45 ⎠
.
0 45