082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
964 Chapter 29 Statistics and probability distributionsN(x)N(x)N(x) (a) (b)(a)mx(b)mxm 1 m 2 xFigure29.10Two typical normal curves.Figure29.11Two normalcurveswith the samestandarddeviationbutdifferentmeans.Letxbeacontinuous random variable with a normal distribution,N(x). ThenN(x) = 1σ √ 2π e−(x−µ)2 /2σ 2−∞<x<∞where µ =expected(mean)valueofx, σ =standarddeviationofx.Figure29.10showstwo typical normal curves. All normal distributions are bell shaped and symmetricalabout µ.In Figure 29.10(a) the values ofxare grouped very closely to the mean. Such a distributionhas a low standard deviation. Conversely, in Figure 29.10(b) the values of thevariable are spread widely about the mean and so the distribution has a high standarddeviation.Figure29.11showstwonormaldistributions.Theyhavethesamestandarddeviationbutdifferentmeans.ThemeanofthedistributioninFigure29.11(a)is µ 1whilethemeanof that in Figure 29.11(b) is µ 2. Note that the domain ofN(x) is (−∞,∞); that is, thedomain is all real numbers. As for all distribution curves the total area under the curveis1.29.14.1 ThestandardnormalA normal distribution is determined uniquely by specifying the mean and standard deviation.The probability thatxliesinthe interval [a,b] isP(axb)=∫ baN(x)dxThemathematicalformofthenormaldistributionmakesanalyticintegrationimpossible,so all probabilities must be computed numerically. As these numerical values wouldchange every time the value of µ or σ was altered some standardization is required. Tothis end we introduce the standard normal. The standard normal has a mean of 0 anda standard deviation of 1.Considertheprobabilitythattherandomvariable,x,hasavaluelessthanz.Forconveniencewecall thisA(z).A(z) =P(x <z) =∫ z−∞N(x)dxFigure29.12illustratesA(z).ValuesofA(z)havebeencomputednumericallyandtabulated.TheyaregiveninTable29.7.Usingthetableandthesymmetricalpropertyofthedistribution,probabilities can becalculated.
29.14 The normal distribution 965N(x)A(z)0zxFigure29.12A(z) =P(x <z) = ∫ z−∞ N(x)dx.Example29.20 Thecontinuousrandomvariablexhasastandardnormaldistribution.Calculatetheprobabilitythat(a) x < 1.2 (b) x > 1.2 (c) x > −1.2 (d) x < −1.2Solution (a) From Table 29.7P(x < 1.2) = 0.8849This isdepicted inFigure 29.13.(b) P(x > 1.2) = 1 −0.8849 = 0.1151This isshown inFigure 29.14.(c) By symmetryP(x > −1.2) isidentical toP(x < 1.2) (see Figure 29.15). So,P(x > −1.2) = 0.8849(d) Using part(c)we findP(x < −1.2) = 1 −P(x > −1.2) = 0.1151(see Figure 29.16).N(x)N(x)Figure29.13P(x < 1.2) = 0.8849.0 1.2 xFigure29.14P(x > 1.2) = 0.1151.0 1.2 xN(x)N(x)–1.2 0xFigure29.15P(x > −1.2) = 0.8849.–1.2 0xFigure29.16P(x < −1.2) = 0.1151.
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29.14 The normal distribution 965
N(x)
A(z)
0
z
x
Figure29.12
A(z) =P(x <z) = ∫ z
−∞ N(x)dx.
Example29.20 Thecontinuousrandomvariablexhasastandardnormaldistribution.Calculatetheprobability
that
(a) x < 1.2 (b) x > 1.2 (c) x > −1.2 (d) x < −1.2
Solution (a) From Table 29.7
P(x < 1.2) = 0.8849
This isdepicted inFigure 29.13.
(b) P(x > 1.2) = 1 −0.8849 = 0.1151
This isshown inFigure 29.14.
(c) By symmetryP(x > −1.2) isidentical toP(x < 1.2) (see Figure 29.15). So,
P(x > −1.2) = 0.8849
(d) Using part(c)we find
P(x < −1.2) = 1 −P(x > −1.2) = 0.1151
(see Figure 29.16).
N(x)
N(x)
Figure29.13
P(x < 1.2) = 0.8849.
0 1.2 x
Figure29.14
P(x > 1.2) = 0.1151.
0 1.2 x
N(x)
N(x)
–1.2 0
x
Figure29.15
P(x > −1.2) = 0.8849.
–1.2 0
x
Figure29.16
P(x < −1.2) = 0.1151.
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