Question

A certain flight arrives on time 81 percent of the time. Suppose 133 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 118 flights are on time.

​(b) at least 118 flights are on time.

​(c) fewer than 117 flights are on time.

​(d) between 117 and 120 inclusive are on time.

Answer 1

here mean of distribution=μ=np=		107.73
and standard deviation σ=sqrt(np(1-p))=		4.52
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

probability =P(117.5<X<118.5)=P((117.5-107.73)/4.524)<Z<(118.5-107.73)/4.524)=P(2.16<Z<2.38)=0.9913-0.9846=0.0067

b)

probability =P(X>117.5)=P(Z>(117.5-107.73)/4.524)=P(Z>2.16)=1-P(Z<2.16)=1-0.9846=0.0154

c)

probability =P(X<116.5)=(Z<(116.5-107.73)/4.524)=P(Z<1.94)=0.9738

d)

probability =P(116.5<X<120.5)=P((116.5-107.73)/4.524)<Z<(120.5-107.73)/4.524)=P(1.94<Z<2.82)=0.9976-0.9738=0.0238