Paper 2 Section B Practice: Normal Distribution — Answers

Questions: Normal Distribution Drills.

Question 1

(a)

$$P(498<X<514)=P\left(-\frac{7}{6}<Z<\frac{3}{2}\right)\approx \mathrm{0.811.}$$

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(b) Since $P(X<a)=0.02$,

$$a=505+6{\Phi }^{-1}(0.02)\approx \mathrm{492.7.}$$

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(c) For independent packet masses $X_1,X_2$,

$$X_1+X_2\sim N(1010,72).$$

Therefore

$$P(X_1+X_2<1000)=P\left(Z<\frac{-10}{\sqrt{72}}\right)\approx \mathrm{0.119.}$$

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Question 2

(a) Independence allows the variances to add:

$$A-B\sim N(8,5^2+4^2)=N(8,41).$$

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(b)

$$P(A-B>15)=P\left(Z>\frac{7}{\sqrt{41}}\right)\approx \mathrm{0.137.}$$

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(c)

$$E(W)=2(40)+3(32)=176,$$

and, using independence,

$$\mathrm{Var}(W)=2^2(5^2)+3^2(4^2)=244.$$

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(d) Since $W\sim N(176,244)$ and the upper-tail probability is $0.10$,

$$w=176+\sqrt{244}{\Phi }^{-1}(0.90)\approx \mathrm{196.0.}$$

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