Question 1199772
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Let X be a binomial random variable with n = 75 and p = 0.6
a) What is μ?
b) What is σ2?
c) What is the probability of P(x≥52), without continuity correction and with continuity correction.
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(a)  μ is the mean       μ = n*p = 0.6*0.75 = 45.



(b)  σ2 is the variance  σ2 = n*p*(1-p)= 75*0.6*(1-0.6) = 18.



(c)  P(x>=52) without continuity correction = normalcdf(52, 9999, μ, σ) = normalcdf(52, 9999, 45, 4.24264).

     Use your calculator.  Here 4.24264 = {{{sqrt(18)}}}.



     P(x>=52) with continuity correction = normalcdf(51.5, 9999, μ, σ) = normalcdf(51.5, 9999, 45, 4.24264).

     Use your calculator, again.
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Answered in full.