Please answer all parts. This Question: 12 pts 10f 10 (°complete) Given a random

Question

Please answer all parts. This Question: 12 pts 10f 10 (°complete) Given a random sample size of n 900 from a binomial probability distribution with P 0.20, complete parts a. through e. below. E Click the icon to view the standard normal distribution table. a. Find the probability that the number of successes is greater than 205. P(X > 205)- (Round to four decimal places as needed) b. Find the probability that the number of successes is fewer than 160 PK 160)-(Round to four decimal places as needed) c. Find the probability that the number of successes is between 170 and 204 P(170

Explanation / Answer

n = 900

p = .20

a. P(X>205) = 900C206*(.2^206)*(.8^695)....(900C900)*(.2^0)*.8^0 = 0.01796

b. P(X<160) = summation of all P(X=0,1,2..159) = 900C0 *.2^0 *.8^900 +...+900C159 *.2^159 *.8^741 = 0.0422

c. P(170<X<204) = P(X<=203) - P(X<=169) = 0.9734 - 0.1913 = .7821

d. P(X<c) = .10

So, Lets see we, will use normal distribution aprioxmation to convert this distribution:

P(Z<(c-np)/sqrt(npq) ) = .10

P(Z< (c-.2*900)/sqrt(.2*900*.8)) = .10

We have a Z of -1.28 at P(X<c) = .10

So, (c-180)/sqrt(144) = -1.28

c = -1.28*sqrt(144)+180 = 164.64 or 165

So, answer is 165

e. Similarly, P(X>c) = .07

We have Z value of 1.475

So, 1.475 = (c-180)/sqrt(144)

c = 1.475*sqrt(144) +180 = 197.7 or 198

Answer is 198

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