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.

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:

[tex]\mu -2\sigma = 105-2*9.108=86.785[/tex]

[tex]\mu +2\sigma = 105+2*9.108=123.215[/tex]

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

Step-by-step 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:

[tex]X \sim Binom (n = 500, p=0.21[/tex]

Part a

The expected value for this distribution is given by:

[tex] E(X) = np=500*0.21= 105[/tex]

Part b

The variance is given by:

[tex] Var(X) = np(1-p) = 500*0.21*(1-0.21)= 82.95[/tex]

And the deviation would be:

[tex] Sd(X) = \sqrt{82.95}= 9.108[/tex]

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:

[tex]\mu -2\sigma = 105-2*9.108=86.785[/tex]

[tex]\mu +2\sigma = 105+2*9.108=123.215[/tex]

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