Step 1

Given that the probability of a baby being born a boy is 0.512. We need to find the probability of exactly 7 boys in 11 births.

Let p=probability of a baby being born a boy=0.512. Therefore, q=probability of a baby being born not a boy=1-0.512=0.488.

Here n=11, and it follows binomial distribution. We know that the mean and variance of binomial distribution is

\(\mu=np=11 \times 0.512=5.632\)

\(\sigma^2=npq=11 \times 0.512 \times 0.488=2.748416\)

\(\Rightarrow \sigma=\sqrt{2.748416}\)

\(\Rightarrow \sigma=1.6578\)

Recall Normal Approximation to Binomial:

The normal distribution can be used as an approximation to the binomial distribution:

If \(X \sim B(n,p)\) and if n is large or p is close to \(\frac{1}{2}\), then X is approximately \(N(np,npq)=N(\mu,\sigma)\)

Therefore, the distribution is approximately N(5.632 , 2.748416)

Step 2

Normal Distribution:

A random variable X is said to have a normal distribution with parameters \(\mu\) and \(\sigma^2\) if its p.d.f. is given by the probability law:

\(f(x;\mu, \sigma)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} ; \ \ -\infty <x<\infty , -\infty < \mu< \infty , \sigma >0\)

Thus, the probability of exactly 7 boys in 11 births is given by

P(x=7)

\(=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(7-\mu)^2}{2\sigma^2}}\)

\(=\frac{1}{1.6578 \sqrt{2\pi}}e^{-\frac{(7-5.632)^2}{2 \times 2.748416}}\)

\(=\frac{1}{1.6578 \sqrt{2\pi}}e^{-\frac{(1.368)^2}{5.496832}}\)

\(=\frac{1}{1.6578 \sqrt{2\pi}}e^{-0.34}\)

\(=\frac{1}{1.6578 \times 2.51}e^{-0.34}\)

\(=\frac{1}{4.16}e^{-0.34}\)

\(=0.24 \times 0.712\)

\(=0.17088\)

Therefore, the probability of exactly 7 boys in 11 births is 0.17088