Question

According to an? airline, flights on a certain route are on time 90?% of the time. Suppose 99

flights are randomly selected and the number of on time flights is recorded. Use technology to find the probabilities. Use the Tech Help button for further assistance.

**?(a)** Determine whether this is a binomial
experiment.

**?(b)** Find and interpret the probability that
exactly 77 flights are on time.

**?(c)** Find and interpret the probability that at
least 77 flights are on time.

**?(d)** Find and interpret the probability that
fewer than 77 flights are on time.

**?(e)** Find and interpret the probability that
between 66 and 88 flights, inclusive, are on time.

Answer #1

**Solution:**

n = 99

p = 0.9

**( a )**

**It is a binomial probability distribution**

**Formula:**

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

**( b )**

P( x = 77 ) = 99!*0.9^{77} * 0.1^{99-77} / 77!
*(99 - 77)! = **0.0002**

**( c ) **

P( x
77 ) = 99!*0.9^{77} * 0.1^{99-77} / 77! *(99 -
77)!+99!*0.9^{78} * 0.1^{99-78} / 78! *(99 - 78)!
+.....

99!*0.9^{99} * 0.1^{99-99} / 99!
*(99 - 99)!

= **0.9999**

**( d )**

P( x <77 ) = 99!*0.9^{76} * 0.1^{99-76} / 76!
*(99 - 76)!+99!*0.9^{75} * 0.1^{99-75} / 75! *(99 -
75)! +.....

99!*0.9^{0} * 0.1^{99-0} / 0! *(99 -
0)!

= **0.0001**

**( e )**

P( 66
x
88 ) = 99!*0.9^{66} * 0.1^{99-66} / 66! *(99 -
66)!+99!*0.9^{67} * 0.1^{99-67} / 67! *(99 - 67)!
+..... 99!*0.9^{88} * 0.1^{99-88} / 88! *(99 -
88)!

= **0.4037**

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