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Worked solution

Distinguish a binomial tail from a CLT mean tail

In the stated probability model, a flu patient has a 98% chance of full recovery. Find the probability that (i) more than 2 in a sample of 20 do not recover fully marks; (ii) in fifty samples of 20 patients, the mean number not recovering fully is greater than 0.5. marks

[2009/NYJC/II/9(a)]

CJC · Review problems · Question 4

Answer

(i) 0.007070.00707; (ii) printed CLT 0.1290.129, exact binomial 0.1099331102…0.1099331102\ldots.

(i) Exact binomial tail

Under the source’s independent-outcome model, X∼B(20,0.02)X\sim B(20,0.02). Hence P(X>2)=1−P(X≤2)=0.0070686934…≈0.00707P(X>2)=1-P(X\le2)=0.0070686934\ldots\approx0.00707, to 3 s.f.

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(ii) Printed CLT approximation

E(X)=0.4E(X)=0.4 and Var⁡(X)=20(0.02)(0.98)=0.392\operatorname{Var}(X)=20(0.02)(0.98)=0.392. Across 50 independent samples, Xˉ∼˙N(0.4,0.392/50)=N(0.4,0.00784)\bar X\mathrel{\dot\sim}N(0.4,0.392/50)=N(0.4,0.00784). Thus P(Xˉ>0.5)≈0.129P(\bar X>0.5)\approx0.129, as printed.

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(ii) Authored exact total comparison

The total over the fifty independent samples is T∼B(1000,0.02)T\sim B(1000,0.02). The strict mean event is T>25T>25, or T≥26T\ge26. Its exact probability is 0.1099331102… . This is distinct from the printed unadjusted normal approximation.

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Common mistakes

A CLT calculation for a discrete low-count distribution is approximate; do not label it an exact binomial result.

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