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This Distributome Activity illustrates an application of probability theory to study Colorblindness.

This Distributome Activity? illustrates an application of probability theory to study Colorblindness.

Colorblindness results from an abnormality on the X chromosome. The condition is thus rarer in women since a woman would need to have the abnormality on both of her X chromosomes in order to be colorblind (whether a woman has the abnormality on one X chromosome is essentially independent of having it on the other).

Suppose that p is the probability that a randomly selected man is colorblind.

1. 100 men are selected at random. What is the distribution of XM = the number of these men that are colorblind? <solution>XM~Binomial(100,p)</solution>
2. 100 women are selected at random. What is the distribution of XF = the number of these women that are colorblind?

Hint: the chance that an individual woman is colorblind is p2.

Solution: XF~Binomial(100,p2)

1. To estimate the probability that a randomly selected woman is colorblind, you might use the proportion of colorblind women in a sample of n women. What is the variance of this estimator? <solution> XF~Binomial(n,p2). Thus Var(XF/n)=p2(1--?p2)/n.

2. Alternatively, to estimate the probability that a randomly selected woman is colorblind, you might use the square of the proportion of colorblind men in a sample of n men. Explain why this estimate makes sense. What is the variance of this estimator?

Hint: The moment generating function can be used to find the fourth moment about the origin.

Hint: We want to estimate p2 and XM/n estimates p so it makes sense to use (XM/n)2 as the estimator (in fact it will be the maximum likelihood estimate). We have Var[( XM n )2 ] = n!4[E(XM 4 ) ! (E(XM 2 ))2 ].Take q=1--?p. Then the fourth moment about the origin of a binomial is E(X4)= np(q--?6pq2+7npq--?11np2q+6n2p2q+n3p3) and the second moment is E(X2)=np(q+np). Thus Var[( XM n )2 ] = n!3(pq + 6(n !1)p2q2 + 4n(n !1)p3q).

1. For large samples, is it better to use a sample of men or a sample of women to estimate the probability that a randomly selected women is colorblind? Explain.

Hint: Show that a normal approximation is valid for both and then compare the variances.

Solution: For large n the ratio of the variances for the estimate in part c to the estimate in part d is Var(XF / n) Var((XM / n)2 ) ! p2 (1" p2 ) 4p3q = 1+ p 4p . When this ratio is greater than 1, the estimator based on the sample of men will be better. Since this happens for any p< 1/3, which is clearly the case for colorblindness, it is better to use a sample of men to estimate the probability that a random woman is colorblind.

Note: you can also use the delta method to find the approximate variance for the estimator above.