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An accounting professor claims that no more than one quarter of undergraduate business
students will major in accounting. hence, p = 1/4 = 0.25
This is binomial sampling, ut n is large so we will use the normal approximation with
mean μ= np = 1200*0.25 = 300 and standard deviation σ = √np(1-p) =√(1200*0.25*0.75)
= 15
Using continuity correction
P(X ≥ 336) = P[X > 335.5] =P[(X-300)/15 > (335.5 – 300)/15]
= P[Z > 2.3667]
= 1 – P[Z < 2.3667]
= 1 – 0.9910
= 0.009
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