Question

A certain flight arrives on time 8484 percent of the time. Suppose 143143 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 108108 flights are on time. ​(b) at least 108108 flights are on time. ​(c) fewer than 124124 flights are on time. ​(d) between 124124 and 128128​, inclusive are on time. ​(a) ​P(108108​)equals=0.00200.0020 ​(Round to four decimal places as​ needed.) ​(b) ​P(Xgreater than or equals≥108108​)equals=0.99800.9980 ​(Round to four decimal places as​ needed.) ​(c) ​P(Xless than<124124​)equals=0.77960.7796 ​(Round to four decimal places as​ needed.) ​(d) ​P(124124less than or equals≤Xless than or equals≤128128​)equals=0.19230.1923 ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

a) P(x=108)=0.0020

b) P(x≥108)=0.9980

c) P(x<124)=0.7794

d) P(124≤x≤128)=0.1925

Step-by-step explanation:

We know the population proportion, that is p=0.84.

We take a sample of size n=143.

We will use the normal approximation to the binomial distribution to model this problem.

The mean and standard deviation of the normal approximation to the binomial distribution will be:

$\mu=np=143*0.84=120.12\\\\\sigma=\sqrt{np(1-p)}=\sqrt{143*0.84*0.16}=\sqrt{19.22}=4.38$

a) We have to calculate the probability that exactly 108 flights are on time.

As the normal distribution considers the random variable to be continous, we have to apply the continuity correction factor.

In this case, the probability of 108 flights on time can be calculated as P(107.5<x<108.5):

$P(x=108)=P(107.5$

b) Now we have to calculate that at least 108 flights are on time.

As the probability includes 108, the continuity factor will indicates that we calculate P(x>107.5). The z-value for x=107.5 has been already calculated in point a:

$P(x\geq108)=P(x>107.5)=P(z>-2.88)=0.9980$

c) We have to calculate the probability that fewer than 124 flights are on time. According to the continuity factor, we have to calculate the probability P(x<123.5), as the flight number 124 is not included in the interval.

$P(x$

d) We have to calculate the probability that between 124 and 128 flights, inclusive, are on time.

This interval corresponds to the probability P(123.5<x<128.5)

$P(123.5$