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Worked solution

Find a probability within one standard deviation of a binomial mean

A box contains two red balls among seven balls. A ball is drawn randomly, its colour noted, then returned and mixed with the others. Repeat 50 times.

(i) Find the expected red count μ and standard deviation σ.

(ii) Find P(μ−σ<X<μ+σ)P(\mu-\sigma<X<\mu+\sigma).

CJC · Lecture · Example 9

(i) Use the replacement model and both moments

Let X count red draws among 50. Replacement and mixing support the independent-redraw model X∼B(50,2/7)X\sim B(50,2/7). The Teacher completes E(X)=np=50(2/7)=100/7E(X)=np=50(2/7)=100/7 and σ2=np(1−p)=50(2/7)(5/7)=500/49\sigma^2=np(1-p)=50(2/7)(5/7)=500/49. Hence σ=500/49=105/7\sigma=\sqrt{500/49}=10\sqrt5/7.

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(ii) Intersect the open interval with the integer support

P(μ−σ<X<μ+σ)=P(100/7−105/7<X<100/7+105/7)P(\mu-\sigma<X<\mu+\sigma)=P(100/7-10\sqrt5/7<X<100/7+10\sqrt5/7). The source approximates the bounds as 11.091 and 17.480; the exact bounds select the same integers. Thus the event is 12≤X≤1712\le X\le17, and its probability is P(X≤17)−P(X≤11)=0.650P(X\le17)-P(X\le11)=0.650 to 3 significant figures.

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Common mistakes

The mean and standard deviation need not be integers; the included count values must be.

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