That is, Z is a standard in .NET

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1. That is, Z is a standard
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Creating a new random variable by this transformation is referred to as standardizing. The random variable Z represents the distance of X from its mean in terms of standard deviations. It is the key step to calculate a probability for an arbitrary normal random variable. EXAMPLE 4-13 Suppose the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 milliamperes and a variance of 4 (milliamperes)2. What is the probability that a measurement will exceed 13 milliamperes Let X denote the current in milliamperes. The requested probability can be represented as P(X 13). Let Z (X 10) 2. The relationship between the several values of X and the transformed values of Z are shown in Fig. 4-15. We note that X 13 corresponds to Z 1.5. Therefore, from Appendix Table II, P1X 132 P1Z 1.52 1 P1Z 1.52 1 0.93319 0.06681 13. That is,
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Rather than using Fig. 4-15, the probability can be found from the inequality X P1X 132 Pa 1X 102 2 113 102 2 b P1Z 1.52
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Distribution of Z =
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0 1.5 Distribution of X
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Standardizing a normal random variable.
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In the preceding example, the value 13 is transformed to 1.5 by standardizing, and 1.5 is often referred to as the z-value associated with a probability. The following summarizes the calculation of probabilities derived from normal random variables.
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Suppose X is a normal random variable with mean P 1X x2 Pa X x
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and variance b P1Z z2
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1x 2 where Z is a standard normal random variable, and z is the z-value obtained by standardizing X. The probability is obtained by entering Appendix Table II with z 1x 2 .
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Continuing the previous example, what is the probability that a current measurement is between 9 and 11 milliamperes From Fig. 4-15, or by proceeding algebraically, we have P19 X 112 P119 102 2 1X P1 0.5 Z 0.52 0.69146 0.30854 102 2 111 P1Z 0.52 0.38292 102 22 P1Z 0.52
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Determine the value for which the probability that a current measurement is below 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 probability expression can be written as P1X x2 P11X P1Z 0.98 102 2 1x 1x 102 22 102 22
Appendix Table II is used to nd the z-value such that P(Z bility from Table II results in P1Z 2.052 0.97982
0.98. The nearest proba-
x 10 = 2.05 2
Figure 4-16 Determining the value of x to meet a speci ed probability.
Therefore, (x 10) 2 for x. The result is
2.05, and the standardizing transformation is used in reverse to solve
x EXAMPLE 4-15
14.1 milliamperes
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 P1N 0.92 Pa N 0.45 0.9 b 0.45 P1Z 22 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 nd x such that P1 x N x2 0.99. A graph is shown in Fig. 4-17. Now, P1 x N x2 P1 x 0.45 P1 x 0.45 N 0.45 x 0.452 Z x 0.452 0.99
From Appendix Table II P 1 2.58 Z 2.582 0.99
Distribution of N
Figure 4-17 Determining the value of x to meet a speci ed probability.
Standardized distribution of
N 0.45
Therefore, x 0.45 and x 2.5810.452 1.16 2.58
Suppose a digital 1 is represented as a shift in the mean of the noise distribution to 1.8 volts. What is the probability that a digital 1 is not detected Let the random variable S denote the voltage when a digital 1 is transmitted. Then, P1S 0.92 Pa S 1.8 0.45 0.9 1.8 b 0.45 P1Z 22 0.02275
This probability can be interpreted as the probability of a missed signal. EXAMPLE 4-16 The diameter of a shaft in an optical storage drive is normally distributed with mean 0.2508 inch and standard deviation 0.0005 inch. The speci cations on the shaft are 0.2500 0.0015 inch. What proportion of shafts conforms to speci cations Let X denote the shaft diameter in inches. The requested probability is shown in Fig. 4-18 and P10.2485 X 0.25152 Pa 0.2485 0.2508 0.2515 0.2508 Z b 0.0005 0.0005 P1 4.6 Z 1.42 P1Z 1.42 P1Z 4.62 0.91924 0.0000 0.91924
Most of the nonconforming shafts are too large, because the process mean is located very near to the upper speci cation limit. If the process is centered so that the process mean is equal to the target value of 0.2500, P10.2485 X 0.25152 Pa 0.2485 0.2500 0.0005 Z 0.2515 0.2500 b 0.0005
P1 3 Z 32 P1Z 32 P1Z 32 0.99865 0.00135 0.9973 By recentering the process, the yield is increased to approximately 99.73%.
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