A gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the gender-selection technique, 904 births consisted of 467 baby girls and 437 baby boys. In analyzing these results, assume that boys and girls are equally likely. a. Find the probability of getting exactly 467 girls in 904 births. b. Find the probability of getting 467 or more girls in 904 births. If boys and girls are equally likely, is 467 girls in 904 births unusually high? c. Which probability is relevant for trying to determine whether the technique is effective: the result from part (a) or the result from part (b)? d. Based on the results, does it appear that the gender-selection technique is effective?

Lets call Y the amount of baby girls that were given birht. Y is a random variable with Binomial distribution, with n = 904 and p ) 1/2, because we are assuming that it is equally likely to give births to boys than to girls.

We need to calculate P(Y = 467), which can be obtained from calculating

[tex] {904 \choose 467} * \frac{1}{2}^{467} * (1-\frac{1}{2})^{(904-467)} = {904 \choose 467} * \frac{1}{2}^{904} [/tex]

This computation cant be donde easily, even with the help of a computer, so i will use an approximation using Normal variables.

First, since we will use a continuous variable, we need to make a correction so the probability obtained inst just 0 because of being punctual.

P(Y = 467) = P(Y ∈ [466.5,467.5])

Now, Y is a sum of 904 independent bernoulli variables Y1,Y2,...., Y904 with parameter p=1/2. The expected values of this bernoulli variables is E(Yi) = p = 1/2, and the variance is V(Yi) = p*(1-p) = 1/4. Hence, Sd(Yi) = √(1/4) = 1/2.

When we calculate the expected value and the variance of a sum of independent and identically distributed random variables, we can just calculate them for one, and multiply by the total amount.

According to the central limit theorem the distribution of Y, which is a sum of a lot of independent and identically distributed random variables, has distribution practically Normal, with mean equal to the mean of Y, that is 904*1/2 = 452, and variance equal to the Variance of Y, that is 904*1/4 = 226. Hence, Sd(Y) = √(226) = 15.033

Lets call Z the approximation, Z has distribution N(452,15.033), and we conclude

P(Y = 467) = P(466.5 < Y < 467.5) = P(466.5 < Z < 467.5)

To calculate this probability, we standarize Z. The random variable

[tex] X = \frac{Z-452}{15.033} [/tex]

has distribution N(0,1), and its cummulative function, Ф, is tabulated. You can find the values of Ф in the attached file.

[tex] P(466.5 < Z < 467.5) = P( \frac{466.5-452}{15.033} < X < \frac{467.5-452}{15.033}) ) P(0.9645 < X < 1.031) = Ф(1.031) - Ф(0.9645) = 0.8485 - 0.8315 = 0.017 [/tex]

Hence, the probability of having 467 births is 0.017

b) Now we want to calculate P(Y ≥ 467). We will use the approximation Z and the standarization X again. First, we need to correct the continuity

[tex] P(Y \geq 467) = P(Y \geq 467.5) = P(Z \geq 467.5) = P(X \geq \frac{467.5-452}{15.033} ) = P(X \geq 1.031) = 1 - \phi(1.031) = 1-0.8485 = 0.1515 [/tex]

So, P(Y ≥ 467) is just 0.1515, a pretty low number. This means that it 467girls from 904 births is a pretty high nummber.

c) If you calculate a punctual probability, you are most likely to get a low number, even from the most frequent results, because there are many resuts possible. The result from part b is more effective to determine something because it uses a conjuntion of results rather than just one or a few.

d) The results does seem to be pretty effective, because we obtained 467 births of a total of 904, and we calculate that obtaining a similar result or higher is unlikely.