I. In the US, 30% of births are Cesarean deliveries. (a) If 10 mothers are selec

Question

I. In the US, 30% of births are Cesarean deliveries. (a) If 10 mothers are selected at random, what is the probability that more than 4 of them will have Cesareans? (b) If 20 mothers are selected at random, what is the probability that 7 or less of them will have Cesareans? (d) Suppose 100 mothers are selected at random, use the normal approximation to estimate the probability that more than 40 of them will have Cesareans. (e) EXPLAIN why the answer to (d) is higher or lower than the answer to (a) (0 Suppose 200 mothers are selected at random, use the normal approximation to estimate the probability that 70 or lessof them will have Cesareans (8) EXPLAIN why the answer to (0 is higher or lower than the answer to (b) MacBook Air 30 p 1

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
a.
X ~ B(10, 0.30)
P( X < = 4) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 10 4 ) * 0.3^4 * ( 1- 0.3 ) ^6 + ( 10 3 ) * 0.3^3 * ( 1- 0.3 ) ^7 + ( 10 2 ) * 0.3^2 * ( 1- 0.3 ) ^8 + ( 10 1 ) * 0.3^1 * ( 1- 0.3 ) ^9 + ( 10 0 ) * 0.3^0 * ( 1- 0.3 ) ^10
= 0.85
P( X > 4) = 1 - P ( X <=4) = 1 -0.85 = 0.15
b.
P( X < = 7) = P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 20 7 ) * 0.3^7 * ( 1- 0.3 ) ^13 + ( 20 6 ) * 0.3^6 * ( 1- 0.3 ) ^14 + ( 20 5 ) * 0.3^5 * ( 1- 0.3 ) ^15 + ( 20 4 ) * 0.3^4 * ( 1- 0.3 ) ^16 + ( 20 3 ) * 0.3^3 * ( 1- 0.3 ) ^17 + ( 20 2 ) * 0.3^2 * ( 1- 0.3 ) ^18 + ( 20 1 ) * 0.3^1 * ( 1- 0.3 ) ^19 + ( 20 0 ) * 0.3^0 * ( 1- 0.3 ) ^20
= 0.772
d.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 100 * 0.3 = 30
standard deviation ( npq )= 100*0.3*0.7 = 4.5826
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X < 40) = (40-30)/4.5826
= 10/4.5826= 2.1822
= P ( Z <2.1822) From Standard NOrmal Table
= 0.9855
P(X > = 40) = (1 - P(X < 40))
= 1 - 0.9855 = 0.0145

e.
increasing the sample size can decreases the chances of higher probability occurance

f.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 200 * 0.3 = 60
standard deviation ( npq )= 200*0.3*0.7 = 6.4807
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X > 0.7) = (0.7-60)/6.4807
= -59.3/6.4807 = -9.15024612
= P ( Z >-9.15024612) From Standard Normal Table
= 1
P(X < = 0.7) = (1 - P(X > 0.7))
= 1 - 0 = 1

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