How to solve 1.96\sqrt((0.65(1-0.65))/(1016))~~0.029
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65 = 0.35$ Show more…
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Suppose x has distribution with μ = 69 and σ = 15. Random samples of size 71 are drawn: Calculate the following probability: Round your answer to 4 decimal places P(65 < x̄ < 66)
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Theory: Suppose that X ~ Binomial(n, p). If n is big and p, q >= 10/n then X and P^ = X/n are approximately normal: X ≈ Normal( np , sqrt(np(1-p)) ) P^ ≈ Normal( p , sqrt(p(1-p)/n) ) Application: A sample of size 65 is drawn from a population, finding 29 occurrences. What is the sample proportion? p^ = 29/65 What is the standard error of the sample proportion? sigma_p^ = sqrt(29/65*(1-(29/65))/65) Give the 92% confidence interval for the population proportion p. ( 29/65-(2.05)*0.06165669 , 29/65+(2.05)*0.06165669 ) If the total population is 90000, then what is the 92% confidence interval for the total number of occurrences? ( , )
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