Question

What is the probability that out of 250 babies born, 110 or fewer will be boys?

Answer 1

Answer:

A. 3.3%

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

250 babies, boys and girls equally as likely:

This means that $n = 250, p = 0.5$.

Mean and standard deviation:

$\mu = E(X) = np = 250*0.5 = 125$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{250*0.5*0.5}$

Probability that out of 250 babies born, 110 or fewer will be boys?

Using continuity correction, this is P(X < 110 + 0.5) = P(X < 110.5), which is the p-value of Z when X = 110.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{110.5 - 125}{\sqrt{250*0.5*0.5}}$

$Z = -1.83$

$Z = -1.83$ has a p-value of 0.033

0.033*100% = 3.3%, so option A.