Question

A certain flight arrives on time 86 percent of the time. Suppose 154 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 136 flights are on time. ​(b) at least 136 flights are on time. ​(c) fewer than 137 flights are on time. ​(d) between 137 and 146​, inclusive are on time. ​(a) ​P(136​)equals=nothing ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

Step-by-step explanation:

Let X be no of flights that arrives on time 86 percent of the time. Suppose 154 flights are randomly selected

Mean = np and variance = npq

Hence X is N(132.44,4.31)

Without continuity correction if we do P(X=136) =0

with cont correction

a)P(x=136) = $P(135.5<x<136.5)\\=P(0.57<Z<0.88)\\=0.3106-0.2157\\=0.0949$

b)P(atleast136)=$P(X\geq 135.5)\\=0.5-0.2157\\=0.2843$

c)$P(X\leq 137)\\=P(X\leq 136.5)\\=0.8106$

d)$P(137<x<146)=P(136.5<x<146.5)=P(0.88<Z<3.26)\\=1-0.3106\\\\=0.6894$