Consider a random sample of size n-1800 from a binomial probability distribution

Question

Consider a random sample of size n-1800 from a binomial probability distribution with P 0.40, and X = number of successes. a. Find the mean and standard deviation of the number of successes. b. Find the probability that the number of successes is greater than c. With probability 0.09 the number of successes is less than how d. Find the mean and standard deviation of the proportion of e. With probability 0.20, the percentage of successes is less than (4 marks) 775. (4 marks) many? (4 marks) successes. (4 marks) what value? (4 marks)

Explanation / Answer

a) mean=np=1800*0.4=720

std deviation =(np(1-p))1/2 =20.7846

b) P(X>775) =1-P(X<=774)=1-P(Z<(774-720)/20.7846)=1-P(Z<2.5981)=1-0.9953 =0.0047

c) for 9 percentile ; z score =-1.34

therefore number of success =mean+z*std deviation =720-1.34*20.7846=692.15

d) mean of proportion =0.40

std deviation =(p*(1-p)/n)1/2 =0.0115

e) for 20th percentile ; z score =-0.8416

hence proportion of success =mean+z*std deviation =0.3903 ~ 39.03%

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