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Worked solution

Find a central height interval and a large-sample tail limit

Girl heights are normal with mean 155 cm and standard deviation 10 cm. Find

(i) the symmetric interval about the mean containing 40% of heights, marks

(ii) the probability exactly two of four randomly chosen girls exceed 170 cm, marks

(iii) the probability that the total height of n randomly chosen girls exceeds 150 n cm, for large n. marks

[2009/NYJC/II/6(modified)]

CJC · Review problems · Question 7

Answer

(i)approximately (149.756,160.244)cm, rounded (150,160)cm; (ii)0.0233; (iii)probability tends to 1.

(i) Split the outside probability equally

Let X∼N(155,102)X\sim N(155,10^2) and a be the half-width. P(155−a<X<155+a)=0.4P(155-a<X<155+a)=0.4 implies P(X<155−a)=0.3P(X<155-a)=0.3. The lower quantile is 155−a≈149.7559949, so a≈5.24 cm to 3 significant figures. The interval is approximately (149.756,160.244)cm. The source reports rounded endpoints (150,160)cm to 3 significant figures; these are rounded boundaries, not an exactly 40% interval.

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(ii) Count independent heights above 170

Let Y count those above 170 among four independent observations. Y∼B(4,P(X>170))=B(4,0.0668072287)Y\sim B(4,P(X>170))=B(4,0.0668072287). Thus P(Y=2)=0.0233P(Y=2)=0.0233 to 3 significant figures.

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(iii) Standardise the independent sum

T=X1+⋯+Xn∼N(155n,100n)T=X_1+\cdots+X_n\sim N(155n,100n). Then P(T>150n)=P((T−155n)/100n>(150n−155n)/100n)=P(Z>−0.5n)P(T>150n)=P((T-155n)/\sqrt{100n}>(150n-155n)/\sqrt{100n})=P(Z>-0.5\sqrt n).

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(iii) Interpret “large n” as a limit

As n grows, −0.5n→−∞-0.5\sqrt n\to-\infty, so the probability tends to 1. The source writes P(Z>−∞)=1P(Z>-\infty)=1; for any finite n the probability is below 1, with 1 the large-n approximation.

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Height normal curve centred at 155 cm. The central 40% lies between 155−a and 155+a with equal half-width a. Each outside tail has probability 0.30.
Height normal curve centred at 155 cm. The central 40% lies between 155−a and 155+a with equal half-width a. Each outside tail has probability 0.30.
Height normal curve centred at 155 cm. The central 40% lies between 155−a and 155+a with equal half-width a. Each outside tail has probability 0.30.

Common mistakes

The sum standard deviation is√(100 n), and the large-n probability approaches 1 rather than equalling 1 at finite n.

Solution notes

All asks/marks, complete source symmetry/quantile/width/rounded boundaries, binomial model, full standardisation and square roots/limit retained. Central 40% figure and both half-width a labels reconstructed. Rounding and finite-n equality qualified explicitly.

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