Question

A certain flight arrives on time 88 percent of the time. Suppose 120 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 114 flights are on time. ​(b) at least 114 flights are on time. ​(c) fewer than 97 flights are on time. ​(d) between 97 and 105​, inclusive are on time.

​(a)

​P(114​)equals=nothing

​(Round to four decimal places as​ needed.)

​(b)

​P(Xgreater than or equals≥114​)equals=nothing

​(Round to four decimal places as​ needed.)

​(c)

​P(Xless than<97​)equals=nothing

​(Round to four decimal places as​ needed.)

​(d)

​P(97less than or equals≤Xless than or equals≤1105​)equals=nothing

​

Answer 1

Solution:
Given in the question
P(Arrives on time) = 0.88
Number of sample = 120
By using the normal approximation to the binomial distribution
Mean() = n*p = 0.88*120 = 105.6
Standard deviation() = sqrt(n*p*(1-p)) = sqrt(120*0.88*0.12) = 3.56
Solution(a)
We need to calculate probability of exactly 114 flight on time, according to normal distribution, P(X=114) = 0, because normal distribution provide us cumulative probability not for exact raw score.
Solution(b)
We need to calculate at least 114 flights are on time
P(X>=114) = ?
First, we will calculate Z-score which can be calculated as
Z= (X-)/ = (114-105.6)/3.56 = 2.36
From Z table we found a p-value
P(X>=114) = 0.0091
So there is a 0.91% probability that at least 114 flights are on time.
Solution(c)
We need to calculate fewer than 97 flights are on time
P(X<97) =?
First, we will calculate Z-score which can be calculated as
Z= (X-)/ = (97-105.6)/3.56 = -2.42
From Z table we found a p-value
P(X<97) = 0.0078
So there is a 0.78% probability that fewer than 97 flights are on time.
Solution(d)
We need to calculate between 97 and 105 flights are on time
P(97<=X<=105) =P(X<=105) - P(X<=97)
First, we will calculate Z-score which can be calculated as
Z= (X-)/ = (97-105.6)/3.56 = -2.42
Z= (X-)/ = (105-105.6)/3.56 = -0.17
From Z table we found a p-value
P(97<=X<=105) =P(X<=105) - P(X<=97) = 0.4325 - 0.0078 = 0.4247
So there is a 42.47% probability that at least 114 flights are on time.