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Worked solution

Scale one observation or combine independent variables

Independent random variables satisfy X∼N(100,49)X\sim N(100,49) and Y∼N(110,576)Y\sim N(110,576). Find (i) P(3X<250)P(3X<250), (ii) P(X+Y>200)P(X+Y>200), (iii) P(Y>X)P(Y>X), (iv) P(2X+Y<300)P(2X+Y<300).

CJC · Lecture · Example 7

(i) Method 1: transform the distribution

3X∼N(3(100),32(49))=N(300,441)3X\sim N(3(100),3^2(49))=N(300,441). Thus P(3X<250)=0.00863P(3X<250)=0.00863 to three significant figures. The second normal parameter is variance; use standard deviation 441=21\sqrt{441}=21 for a normalcdf calculation.

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(i) Method 2: transform the event

Alternatively use the original distribution: P(3X<250)=P(X<250/3)=0.00863P(3X<250)=P(X<250/3)=0.00863 to three significant figures. Both source methods are retained.

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(ii) Add independent variables

X+Y∼N(100+110,49+576)=N(210,625)X+Y\sim N(100+110,49+576)=N(210,625). Consequently P(X+Y>200)=0.655P(X+Y>200)=0.655 to three significant figures.

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(iii) Subtract means but add independent variances

Y−X∼N(110−100,576+49)=N(10,625)Y-X\sim N(110-100,576+49)=N(10,625). Rewrite P(Y>X)=P(Y−X>0)=0.655P(Y>X)=P(Y-X>0)=0.655 to three significant figures. The variance of a difference is positive and adds here because X and Y are independent.

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(iv) Square the coefficient in the variance

2X+Y∼N(2(100)+110,22(49)+576)=N(310,772)2X+Y\sim N(2(100)+110,2^2(49)+576)=N(310,772). Hence P(2X+Y<300)=0.359P(2X+Y<300)=0.359 to three significant figures.

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Common mistakes

Do not subtract variances when subtracting independent variables, or treat 3X as three independent observations.

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