Minimum Number of Coin Flips for Probability Assertion

  • Thread starter toothpaste666
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In summary: Yes, you are correct. The correct equation is sigma = 1/(2sqrt(n)). This was a typo on my part and I apologize for the confusion.
  • #1
toothpaste666
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Homework Statement


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?

Homework Equations


P( |X-μ| ≥ kσ ) ≤ 1/k^2

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 :(
 
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  • #2
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
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
toothpaste666 said:
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

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
[tex] \left\{ \left| \frac{X}{n} - \frac{1}{2} \right| > .04 \right\} [/tex]
Using ##\mu = n/2, \sigma = (1/2) \sqrt{n}## we can re-write the event above is the same as
[tex] \{ |X - \mu| > 0.04\, n \} = \{ |X - \mu| > 0.08 \sqrt{n} \; (1/2) \sqrt{n} \} = \{ |X - \mu| > 0.08 \sqrt{n} \: \sigma \} [/tex]
 
Last edited:
  • #5
oops i think i made a mistake.

would it be σ = 1/(2sqrt(n))

instead of

σ = 1/sqrt(2n) ?
 

1. What is Chebyshev's theorem problem?

Chebyshev's theorem is a mathematical concept that describes the proportion of data that falls within a certain number of standard deviations from the mean in a normal distribution. It states that at least (1 - 1/k^2) of the data will fall within k standard deviations from the mean, where k is any positive number greater than 1.

2. How is Chebyshev's theorem used?

Chebyshev's theorem is used to determine the proportion of data that falls within a certain range in a normal distribution, without knowing the exact shape of the distribution. It is helpful in understanding the spread of data and identifying outliers.

3. What is the difference between Chebyshev's theorem and the Empirical Rule?

The Empirical Rule is a more specific version of Chebyshev's theorem, and it only applies to normal distributions. It states that approximately 68% of the data falls within 1 standard deviation from the mean, 95% falls within 2 standard deviations, and 99.7% falls within 3 standard deviations.

4. Can Chebyshev's theorem be used for any type of data?

Yes, Chebyshev's theorem can be used for any type of data, regardless of the shape of the distribution. However, it is most accurate when used with a large sample size and a symmetrical distribution.

5. How is Chebyshev's theorem related to the Central Limit Theorem?

Chebyshev's theorem is a special case of the Central Limit Theorem, which states that as the sample size increases, the sampling distribution of the mean will approach a normal distribution, regardless of the shape of the population distribution. This means that Chebyshev's theorem can be applied to any population distribution, as long as the sample size is large enough.

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