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Statistics and Probability
Upper level probability

vazelinahS

Answered question

2021-05-19

The probability of a baby being born a boy is 0.512. Consider the problem of finding the probability of exactly 7 boys in 11 births. Solve that problem using normal approximation to the binomial.

Answer & Explanation

Jeffrey Jordon

Expert2021-11-11Added 2605 answers

Step 1

Given that the probability of a baby being born a boy is 0.512. The likelihood of exactly 7 boys in every 11 births must be determined.

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

$μ = n p = 11 \times 0.512 = 5.632$

$σ^{2} = n p q = 11 \times 0.512 \times 0.488 = 2.748416$

$\Rightarrow σ = \sqrt{2.748416}$

$\Rightarrow σ = 1.6578$

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 $N (n p, n p q) = N (μ, σ)$

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 $μ$ and $σ^{2}$ if its p.d.f. is given by the probability law:

$f (x; μ, σ) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}}; - \infty < x < \infty, - \infty < μ < \infty, σ > 0$

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

P(x=7)

$= \frac{1}{σ \sqrt{2 π}} e^{- \frac{(7 - μ)^{2}}{2 σ^{2}}}$

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

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

$= \frac{1}{1.6578 \sqrt{2 π}} e^{- 0.34}$

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

$= \frac{1}{4.16} e^{- 0.34}$

$= 0.24 \times 0.712$

$= 0.17088$

Consequently, the likelihood of precisely 7 boys in 11 births is 0.17088

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