Question

Find the probability that of the next 200 children born, (a) less than 40% will be boys, (b) between 43% and 57% will be girls, (c) more than 54% will be boys. Assume equal probabilities for births of boys and girls

Answer 1

If $X$ is a binomial random variable with mean $μ = np$ and variance $\sigma^2=np(1-p),$ then the limiting form of the distribution of

$$Z=\dfrac{X-np}{\sqrt{np(1-p)}},$$

as $n\to \infin,$ is the standard normal distribution $n(z; 0, 1).$

In practice, the approximation is adequate provided that both $np\geq10$ and $np(1-p)\geq 10.$

a) $n=200, p=0.5, \mu=np=200(0.5)=100,$

$\sigma^2=np(1-p)=200(0.5)(1-0.5)=50$

$$P(X<80)\approx P(X<80-0.5)$$

$$\approx P(Z<-2.8991378)$$

$$\approx0.001871$$

b) $n=200, p=0.5, \mu=np=200(0.5)=100,$

$\sigma^2=np(1-p)=200(0.5)(1-0.5)=50$

$$P(86<X<114)=P(X<114)-P(X\leq 86)$$

$$\approx P(X<114-0.5)-P(X<86+0.5)$$

$$=P(Z<\dfrac{113.5-100}{\sqrt{50}})-P(Z<\dfrac{86.5-100}{\sqrt{50}})$$

$$\approx P(Z<1.90918831)-P(Z<-1.90918831)$$

$$\approx0.9718811-0.0281189\approx0.943762$$

c) $n=200, p=0.5, \mu=np=200(0.5)=100,$

$\sigma^2=np(1-p)=200(0.5)(1-0.5)=50$

$$P(X>108)=1-P(X\leq 108)$$

$$\approx 1-P(X<108+0.5)$$

$$=1-P(Z<\dfrac{108.5-100}{\sqrt{50}})$$

$$\approx1-P(Z<1.20208153)$$

$$\approx 0.114666$$