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Worked solution

Recover a component normal distribution from a sum of two observations

X∼N(20,σ2)X\sim N(20,\sigma^2).

(i) Given P(12<X<k)=P(X>k)=0.42065P(12<X<k)=P(X>k)=0.42065, show that σ=8 to the nearest whole number, and find k. marks

X=Y1+Y2−4X=Y_1+Y_2-4, where Y1,Y2Y_1,Y_2 are independent observations of a normal variable Y. W is independent of Y and has normal mean 25 and variance 16.

(ii) Using σ=8, find E(Y)E(Y) and Var⁡(Y)\operatorname{Var}(Y), then P(W>2Y)P(W>2Y). marks

[2009/RI/II/8]

CJC · Review problems · Question 8

Answer

(i)σ≈8(nearest whole number),k≈21.6; (ii)E(Y)=12E(Y)=12, variance 32, probability 0.533 (3 s.f.).

(i) Find the remaining lower-tail area

P(X<12)=1−0.42065−0.42065=0.1587P(X<12)=1-0.42065-0.42065=0.1587. Thus P((X−20)/σ<(12−20)/σ)=P(Z<−8/σ)=0.1587P((X-20)/\sigma<(12-20)/\sigma)=P(Z<-8/\sigma)=0.1587. The source quantile gives −8/σ=−0.99982-8/\sigma=-0.99982, hence σ≈8 to the nearest whole number.

How did this part go?

(i) Find the upper-tail cutoff

Using the source rounded model X∼N(20,82)X\sim N(20,8^2), P(X>k)=0.42065P(X>k)=0.42065 implies P(X≤k)=1−0.42065P(X\le k)=1-0.42065. Therefore k≈21.6 to 3 significant figures. Using the unrounded σ from the first quantile gives the same stated rounding.

How did this part go?

(ii) Recover Y mean and variance

Let Y∼N(a,b2)Y\sim N(a,b^2). From E(X)=E(Y1+Y2−4)E(X)=E(Y_1+Y_2-4), 20=2a−420=2a-4, so a=12. From Var⁡(X)=Var⁡(Y1+Y2−4)\operatorname{Var}(X)=\operatorname{Var}(Y_1+Y_2-4), 82=2b28^2=2b^2, so b2=32b^2=32 and b=32b=\sqrt{32}. Thus E(Y)=12E(Y)=12, Var⁡(Y)=32\operatorname{Var}(Y)=32, and Y∼N(12,32)Y\sim N(12,32).

How did this part go?

(ii) Compare W with one scaled Y

W∼N(25,16)W\sim N(25,16). Then E(W−2Y)=E(W)−2E(Y)=25−2(12)=1E(W-2Y)=E(W)-2E(Y)=25-2(12)=1. Independence gives Var⁡(W−2Y)=Var⁡(W)+22Var⁡(Y)=16+4(32)=144\operatorname{Var}(W-2Y)=\operatorname{Var}(W)+2^2\operatorname{Var}(Y)=16+4(32)=144. Therefore W−2Y∼N(1,144)W-2Y\sim N(1,144) and P(W>2Y)=P(W−2Y>0)=0.533P(W>2Y)=P(W-2Y>0)=0.533 to 3 significant figures. The source prints “Var E(Y)” in this line; the intended quantity is Var(Y), as its substitution 32 confirms.

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Compare your sketch

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Normal curve centred at 20. Region 12<X<k has area 0.42065 and the upper tail X>k also has area 0.42065; k lies just above 20. The remaining lower tail is 0.1587.
Normal curve centred at 20. Region 12<X<k has area 0.42065 and the upper tail X>k also has area 0.42065; k lies just above 20. The remaining lower tail is 0.1587.
Normal curve centred at 20. Region 12<X<k has area 0.42065 and the upper tail X>k also has area 0.42065; k lies just above 20. The remaining lower tail is 0.1587.

Common mistakes

Y₁+Y₂ has twice Var(Y);2 Y has four times Var(Y). The source’s “Var E(Y)” is a typographical slip.

Solution notes

Both asks/marks, all remaining-area/quantile stages, rounding, complete component-recovery equations including√32, W difference moments and event retained. Source VarE(Y) typo qualified. Both equal-probability regions and labelled cutoffs reconstructed accessibly.

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