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 |

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81
percent of the time. Suppose
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,
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81
percent of the time. Suppose
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(a) exactly
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(b) at least
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flights are on time.
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(a) exactly 117 flights are on time.
(b) at least 117
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(d) between 96 and 108, inclusive are on time.

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135 flights are randomly selected. Use the normal approximation to
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inclusive are on time.

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