# Chebyshev's theorem problem

1. Sep 27, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
How many times do we have to flip a balanced coin to be able to assert with a probability of at most .01 that the difference between the proportion of tails and .50 will be at least .04?

2. Relevant equations
P( |X-μ| ≥ kσ ) ≤ 1/k^2

3. The attempt at a solution

I am very confused about how to use this theorem. So far I have only managed to figure out bits and pieces.
I know that
P ≤ .01 = 1/k^2 so
k^2 = 1/.01 = 100
k = 10

also μ = np and since the coin is balanced, p = 1/2 so
μ = n/2
also σ^2 = np(1-p) = n/2(1/2) = n/4
so σ = sqrt(n)/2

plugging this all into the inequality I get

P( |X-n/2| ≥ (10)(sqrt(n)/2) ) ≤ .01

P( |X-n/2| ≥ (5)sqrt(n) ) ≤ .01

But I am still confused about what this means or how I can solve for n (which is what I think I need to be solving for.) please help :(

2. Sep 27, 2015

### davidmoore63@y

The random variable X is the proportion of flips that are tails, not the number of tails. Thus mu= 0.5 not n/2 and sigma = sqrt (p(1-p)/n). Try again using these.

3. Sep 27, 2015

### toothpaste666

Now things are starting to make sense, thank you so much.
so
μ = 1/2
and σ = 1/sqrt(2n)

The inequality becomes
P( |X-1/2| ≥ 10/sqrt(2n) ) ≤ .01
so

10/sqrt(2n) = .04

10/.04 = sqrt(2n)

250 = sqrt(2n)

62500 = 2n

n = 31250

4. Sep 27, 2015

### Ray Vickson

It is just as easy to continue using X = number of tails, but to write the desired condition correctly. You want to know the probability of the event
$$\left\{ \left| \frac{X}{n} - \frac{1}{2} \right| > .04 \right\}$$
Using $\mu = n/2, \sigma = (1/2) \sqrt{n}$ we can re-write the event above is the same as
$$\{ |X - \mu| > 0.04\, n \} = \{ |X - \mu| > 0.08 \sqrt{n} \; (1/2) \sqrt{n} \} = \{ |X - \mu| > 0.08 \sqrt{n} \: \sigma \}$$

Last edited: Sep 27, 2015
5. Sep 27, 2015

### toothpaste666

oops i think i made a mistake.

would it be σ = 1/(2sqrt(n))