Question

While he was a prisoner of war during World War II, John Kerrich tossed a coin 10,000 times. He got 5067 heads. If the coin is perfectly balanced, the probability of a head is 0.5. Is there reason to think that Kerrich's coin was not balanced? To answer this question, use a Normal distribution to estimate the probability that tossing a balanced coin 10,000 times would give a count of heads at least this far from 5000 (that is, at least 5067 heads or at most 4933 heads.)

Solution

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Given:

n=10000n=10000

p=0.5p=0.5

The mean of a binomial variable is the product of the sample size nn and the probability pp:

μX=np=10000(0.5)=5000\mu_X=np=10000(0.5)=5000

The standard deviation of a binomial variable is the square root of the product of the sample size nn, the probability pp and the probability 1p1-p:

σX=np(1p)=10000(0.5)(10.5)=50\sigma_X=\sqrt{np(1-p)}=\sqrt{10000(0.5)(1-0.5)}=50

The z-score is the value decreased by the mean divided by the standard deviation:

z=xμσ=49335000501.34z=\dfrac{x-\mu}{\sigma}=\dfrac{4933-5000}{50}\approx -1.34

z=xμσ=50675000501.34z=\dfrac{x-\mu}{\sigma}=\dfrac{5067-5000}{50}\approx 1.34

Determine the corresponding probability using table A:

P(X4933 or X5067)=P(Z1.34 or X1.34)=2×P(Z<1.34)P(X\leq 4933\text{ or } X\geq 5067)=P(Z\leq -1.34\text{ or } X\geq 1.34)=2\times P(Z<-1.34)

=2×0.0901=0.1802=18.02%=2\times 0.0901=0.1802=18.02\%

There is no reason to think that Kerrich's coin was not balanced, because the probability is more than 5% and thus the event is likely to occur.

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