Question

A certain flight arrives on time 86 percent of the time. Suppose 187 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 148 flights are on time. ​(b) at least 148 flights are on time. ​(c) fewer than 173 flights are on time. ​(d) between 173 and 174​, inclusive are on time. ​(a) ​P(148​)equals 4.745 ​(Round to four decimal places as​ needed.)

Answer 1

Solution:

n = 187

p = 86

binomial probabilitydistribution

Formula:

P_{(k out of n )}= n!*p^k * q^n-k / k! *(n - k)!

( a )

P( x = 148 ) = 187!*0.86¹⁴⁸ * 0.14^187-148 / 148! *(187 - 148)!

= 0.0009

( b )

P( x $\geq$ 148 ) = 187!*0.86¹⁴⁸ * 0.14^187-148 / 148! *(187 - 148)! .....+187!*0.86187 * 0.14^187-187 / 187! *(187 - 187)!

= 0.9990

( c )

P( x < 173 ) = 187!*0.86¹⁷³ * 0.14^187-173 / 173! *(187 - 173)!+ ....187!*0.86⁰ * 0.14^187-0 / 0! *(187 - 0)!

=  0.9879

( d )

P( 173<x<174) = 187!*0.86¹⁷³ * 0.14^187-173 / 173! *(187 - 173)! + 187!*0.86¹⁷⁴ * 0.14^187-174 / 174! *(187 - 174)!

= 0.0060+0.0033

= 0.0093