Topic: Normal approximation to binomial distribution
Calculate the following probabilities using a normal approximation. P(9 ≤ X ≤ 12) where X ∼ B(21, 0.5)
Please show work as I will studying it step by step, thanks.
solution
given
X ∼ B(21, 0.5)
mean=n*p= 21*0.5=10.5
standard deviation = sqrt(n*p*(1-p))
=sqrt(21*0.5*(1-0.5)) =2.29128
SO P(9<=X<=12) = P((9-10.5)/2.29128 <(X-mean)/s <(12-.10.5)/2.29128)
=P(-0.6546 <Z< 0.6546)
=0.4844 (from standard normal table)
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