Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (igreen):

Solved

OpenStudy (igreen):

What are the expected number of flights on time?

OpenStudy (acxbox22):

use proportions

OpenStudy (mathstudent55):

Since 84% of the flights in the random sample are on time, we expect the same percentage to be on time in general. 84% means 84 out of 100.

OpenStudy (igreen):

Whoa..major lag

OpenStudy (mathstudent55):

\(\dfrac{84}{100} = \dfrac{x}{203} \)

OpenStudy (astrophysics):

Your fav igreen! XD

OpenStudy (igreen):

I came up with .84 * 203 or 170.52

OpenStudy (astrophysics):

That sounds right

OpenStudy (mathstudent55):

You can think of the proportion as follows: 84 flights out of 100 are on time just like x flights out of 203 are on time.

OpenStudy (astrophysics):

170.52

OpenStudy (igreen):

I was trying to say that before anyone else commented..for some reason I can only make one comment and it glitches and I have to reload..

OpenStudy (mathstudent55):

Correct. I'd round it off to 171 because I don't know what 0.52 of a flight is.

OpenStudy (igreen):

Okay, now it asks: "What is the standard deviation?"

OpenStudy (igreen):

Any idea on how to find that? @mathstudent55

OpenStudy (igreen):

What about you? @Astrophysics

OpenStudy (astrophysics):

Standard deviation is the average difference between any point of data and the mean, been a long time since I did statistics but I'm sure we can figure it out. \[\sigma = \sqrt{\frac{ 1 }{ N }\sum_{i=1}^{N}(x_i-\mu)^2}\]

OpenStudy (astrophysics):

But for this the easiest way I see is we can just use \[\sigma = \sqrt{npq}\] where \[\mu = \sqrt{np}\] that should be easy enough :P

OpenStudy (igreen):

Ahhh

OpenStudy (igreen):

I got 5.22

OpenStudy (astrophysics):

That sounds good!

OpenStudy (igreen):

\(\sf \sqrt{260 \times 0.84 \times 0.16}\)

OpenStudy (igreen):

Thanks!

OpenStudy (astrophysics):

Np

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!