Question

Suppose that 2121​% of all births in a certain region take place by Caesarian section each year. a. In a random sample of 500500 ​births, how​ many, on​ average, will take place by Caesarian​ section? b. What is the standard deviation of the number of Caesarian section births in a sample of 500500 ​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 500500 births.

Answer 1

a)
`E(X) = np=500*0.21= 105`

b)
`Sd(X) = √(82.95)= 9.108`

c) Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

`\mu -2\sigma = 105-2*9.108=86.785`

`\mu +2\sigma = 105+2*9.108=123.215`

So we expect about 86 and 123 most of the numbers of Caesarian section births

Explanation:

For this case we can define the random variable X as the number of births in the Caesarian​ section and from the data given we know that the distribution of X is:

`X \sim Binom (n = 500, p=0.21`

Part a

The expected value for this distribution is given by:

`E(X) = np=500*0.21= 105`

Part b

The variance is given by:

`Var(X) = np(1-p) = 500*0.21*(1-0.21)= 82.95`

And the deviation would be:

`Sd(X) = √(82.95)= 9.108`

Part c

Assuming a the normality assumption we will have within 2 deviations from the mean most of the data from the distribution and the interval for this case would be:

`\mu -2\sigma = 105-2*9.108=86.785`

`\mu +2\sigma = 105+2*9.108=123.215`

So we expect about 86 and 123 most of the numbers of Caesarian section births