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.

Answers

Answer 1

Answer:

Answer:

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

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:

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

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

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