Question

suppose that 32% of all births in a certain region take place by caesarian section each year.
a. in a random sample of 800 births, how many, on average, will take place by caesanan section?
b. what is the standard deviation of the number of caesarian section births in a sample of 800 births?
c. use your answers to parts a and b to form an interval that is likely to contain the number of caesarian section births in a sample of 800 births.

a. on average, out of 800 births will take place by caesarian section.
(round to the nearest whole number as needed.)

Explanation:

Step1: Calculate the average number of Caesarian section births

The formula for the mean ($\mu$) of a binomial distribution is $\mu = np$, where $n$ is the number of trials and $p$ is the probability of success.
Here, $n = 800$ and $p=0.32$.
So, $\mu=800\times0.32 = 256$.

Step2: Calculate the standard deviation of the number of Caesarian section births

The formula for the standard deviation ($\sigma$) of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$.
Substitute $n = 800$ and $p = 0.32$ into the formula:
$\sigma=\sqrt{800\times0.32\times(1 - 0.32)}=\sqrt{800\times0.32\times0.68}=\sqrt{174.08}\approx13.19$.

Step3: Form the interval

For a normal - approximation of a binomial distribution (since $np = 256\geq10$ and $n(1 - p)=800\times(1 - 0.32)=544\geq10$), we use the interval $\mu\pm2\sigma$.
Substitute $\mu = 256$ and $\sigma\approx13.19$ into the interval formula:
Lower bound: $256-2\times13.19=256 - 26.38 = 229.62$
Upper bound: $256+2\times13.19=256 + 26.38 = 282.38$

Answer:

a. On average, $256$ out of 800 births will take place by Caesarian section.
b. The standard deviation is approximately $13.19$.
c. The interval is $(230,282)$ (after rounding the bounds from part b to the nearest whole number).