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.

a)

b)

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:

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:

Part a

The expected value for this distribution is given by:

Part b

The variance is given by:

And the deviation would be:

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:

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

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.

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