Solving Chebyshev's Inequality with 900 Coin Flips - 35/36 Probability

[Gémeaux]

Hi.

I'm having a bit of trouble with Chebyshev's inequality and I was wondering if someone could point me in the right direction as to how to answer this question:

Use chebyshev's theorem to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.40 and 0.60.

By the looks of it, I have to use E(Y)=p, variance(Y)=p(1-p)/n.

I'm not quite sure how to approach this as my lecturer didn't cover this in much depth.

Thank you.

 

[HallsofIvy]

Do you at least know what Chebyshev's theorem says?

 

[HallsofIvy]

Do you at least know what Chebyshev's theorem says?

My understanding of it is "For any population or sample, at least (1 - (1 / k)2) of the observations in the data set fall within k standard deviations of the mean, where k greater than or equal to 1."

Okay, for 900 flips of an unbiased coin what will the expected number of heads be? (No, it is not p, it is E(Y)=pn. And p= ? n= ?). What will the standard deviation be? (yes, the variance is variance(Y)=p(1-p)/n. And so the standard deviation is?

Now that you have figured out what the mean and standard deviation for this problem are, the proportion of head in 900 flips between .4 and .6 means that the actual number of heads is what? How far on either side of the mean are those? How many standard deviations (k) is that? Finally, what is 1/k²?