Nw 35. On-Time Flights According to flightstats.com, American Airlines flights f
ID: 3357506 • Letter: N
Question
Nw 35. On-Time Flights According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 15 flights are randomly selected, and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Find and interpret the probability that exactly 10 flights are (c) Find and interpret the probability that fewer than 10 flights (d) Find and interpret the probability that at least 10 flights are (e) Find and interpret the probability that between 8 and 10 on time. are on time. on time flights, inclusive, are on time.Explanation / Answer
35.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 15 * 0.8
= 12
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 15 * 0.8 * 0.2
= 2.4
III.
standard deviation = sqrt( variance ) = sqrt(2.4)
=1.549193
a.
American airlines flights from from dallas to chicago are on time 80% of the time,suppose 15 flights are randomly selected,recorded on time
here n=15 indepedent trials,probability of success = 0.8 so that we will take binomial distribution
b.
the probability that exactly 10 flights are on time
P( X = 10 ) = ( 15 10 ) * ( 0.8^10) * ( 1 - 0.8 )^5
= 0.1032
c.
the probability that fewer than 10 flights on time
P( X < 10) = P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 9 ) * 0.8^9 * ( 1- 0.8 ) ^6 + ( 15 8 ) * 0.8^8 * ( 1- 0.8 ) ^7 + ( 15 7 ) * 0.8^7 * ( 1- 0.8 ) ^8 + ( 15 6 ) * 0.8^6 * ( 1- 0.8 ) ^9 + ( 15 5 ) * 0.8^5 * ( 1- 0.8 ) ^10 + ( 15 4 ) * 0.8^4 * ( 1- 0.8 ) ^11 + ( 15 3 ) * 0.8^3 * ( 1- 0.8 ) ^12 + ( 15 2 ) * 0.8^2 * ( 1- 0.8 ) ^13 + ( 15 1 ) * 0.8^1 * ( 1- 0.8 ) ^14 + ( 15 0 ) * 0.8^0 * ( 1- 0.8 ) ^15
= 0.0611
d.
the probability that atleast 10 flights are on time
P( X < 10) = P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 15 9 ) * 0.8^9 * ( 1- 0.8 ) ^6 + ( 15 8 ) * 0.8^8 * ( 1- 0.8 ) ^7 + ( 15 7 ) * 0.8^7 * ( 1- 0.8 ) ^8 + ( 15 6 ) * 0.8^6 * ( 1- 0.8 ) ^9 + ( 15 5 ) * 0.8^5 * ( 1- 0.8 ) ^10 + ( 15 4 ) * 0.8^4 * ( 1- 0.8 ) ^11 + ( 15 3 ) * 0.8^3 * ( 1- 0.8 ) ^12 + ( 15 2 ) * 0.8^2 * ( 1- 0.8 ) ^13 + ( 15 1 ) * 0.8^1 * ( 1- 0.8 ) ^14 + ( 15 0 ) * 0.8^0 * ( 1- 0.8 ) ^15
= 0.0611
P( X > = 10 ) = 1 - P( X < 10) = 0.9389
e.
the probability that between 8 and 10 flights are on time
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 15 * 0.8 = 12
standard deviation ( npq )= 15*0.8*0.2 = 1.5492
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8) = (8-12)/1.5492
= -4/1.5492 = -2.582
= P ( Z <-2.582) From Standard Normal Table
= 0.00491
P(X < 10) = (10-12)/1.5492
= -2/1.5492 = -1.291
= P ( Z <-1.291) From Standard Normal Table
= 0.09835
P(8 < X < 10) = 0.09835-0.00491 = 0.0934
interpretations
Binomial Experiment-
A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties:
1.The experiment consists of n repeated trials.
2.Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
3.The probability of success, denoted by P, is the same on every trial.
4.The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
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