In 2005, 0.76% of all airline flights were on-time. if we choose a simple random sample of 2000 flights, find the probability (to four decimal places, using normal chart, no continuity correction)

(a) at least 79% of the sample's flights were on time

(b) at most 1580 of the sample's flights were on time

(c) the sample proportion of on-time flights (p-hat) differs from the truth by more than three percent

Step-by-step explanation:

Here X no of flight that reach on time has two outcomes only and each trial is independent of the other.

Since sample size is large we approximate to normal with mean =np and variance =npq

i.e. X is N(152,19.10)

is used for converting to std normal variate

Continuity correction of 0.5 on either side of interval would be applied for x

a)

=0.00

b)=1.00

c) p is N(0.76, sqrtpq/n)=N(0.76,0.0095)

P(p diff>0.03) =

=0.00

answer:

.5 or 5%

Step-by-step explanation:

Use the simple interest rate formula I/PT = R, I is your interest which is 18 and P is the original price which is 400 and then T is your time/period which is 9 if you put those numbers in you get 1/200 then you multiply that by a hundred and you get .5 to check your work use this formula I = Prn and r is your rate which is .5 and you get 1,800 you divide that by 100 and you get 18. Hope this helps and is not too late :)