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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 = .
P(x>=52) with continuity correction = normalcdf(51.5, 9999, μ, σ) = normalcdf(51.5, 9999, 45, 4.24264).
Use your calculator, again.
Answered in full.