Statistics help

Question

American Airlines' flights from Chicago to Dallas are on time 90 % of the time. Suppose 14 flights are randomly selected, and the number on-time flights is recorded.

1.The probability that at least 2 flights are on time is =

2.The probability that at most 11 flights are on time is =

3.The probability that exactly 11 flights are on time is =

Answer 1

Hi Emily,

This is a binomial case, meaning you have "success" (on-time) and "failure" (not). Formula for that is:

P(X=x)= C(n,x) p^x*q^n-x

x= Number of "successes" you are interested in

n=Total Sample Size

p=probability of success

q=probability of failure=1-p

C(n, x)= n!/(x!(n-x)!

!=Factorial=4*3*2*1--you can input factorial (!) into TI-80s series calculators

Question 1

This is a three-step calculation. I will do step one as an example and then direct you for steps 2 and 3. You want the probability that at least two flights are on time. That means we need the individual probabilities of 2 flights up to all 14 flights being on time. We could compute individually from 2 to 14, but that is A LOT of work. Instead, we can apply the Complement Rule, aka the "One Minus Trick." The probability of at least two flights being on time is equal to one minus the probability that less than two will be on time. Mathematically:

P(X>=2) = 1-P(X<2)

Now we must compute only two probabilities instead of twelve: P(X=1) and P(X=0).

For P(X=1):

x=1

n=14

p=0.9

q=1-0.9=0.1

P(X=1)= C(10,1)*0.9¹*0.1⁹

C(10,1)= 10!/(9!*1!) = 10

P(X=1) = 10*0.9¹*0.1⁹

P(X=1)= 0.000000009

Now, compute P(X=0) on your own. Use the same equation, but:

x=0

n=10

p=0.9

q=0.1

C(n,x)= C(10,0)

Once you have that probability, add it to the above. Then, subtract that total from 1 to get your "final answer" for the probability of at least two flights arriving on time.

Question #2

This is basically the same as above. You want 11 flights at most, so we need to get the probabilities for 12, 13, and 14, and apply our One Minus Trick. I will do P(X=12) as an example and leave P(X=13) and P(X=14) to you.

P(X=x)=C(n,x)*p^x*q^n-x

x=12

n=14

p=0.9

q=0.1

P(X=12) = C(14,12)*0.9¹²*0.1²

C(14,12) = 14!/(12!*2!)

C(14,12) = 91

P(X=12)= 91*0.9¹²*0.1²

P(X=12)= 0.257

Now, compute P(X=13) and P(X=14) on your own. All variables are as above except X, which you must change to 13 and 14. Then, add the three probabilities together:

P(X=12) + P(X=13) + P(X=14)

Then, subtract the result from 1 and you will have P(X<=11).

Question #3

We do not need to do any supplemental subtraction here. This is just:

x=11

n=14

p=0.9

q=0.1

P(X=11)= C(14,11)*0.9¹¹*0.1³

P(X=11)= 364*0.9¹¹*0.1³

P(X=11) = 0.114

I hope this helps.