Minimum Number of Coin Flips for Probability Assertion

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Homework Help Overview

The problem involves determining the minimum number of times a balanced coin must be flipped to assert with a specific probability that the difference between the proportion of tails and 0.50 is at least 0.04. The discussion centers around the application of probability inequalities and the properties of random variables related to coin flips.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of a probability theorem and its implications for determining the number of flips needed. Initial confusion about the definitions of the random variable and parameters is noted, with attempts to clarify the correct expressions for mean and standard deviation. Questions arise regarding the interpretation of the inequality and how to solve for the number of flips.

Discussion Status

Some participants have made progress in understanding the problem, with one clarifying the correct expressions for mean and standard deviation. However, there remains uncertainty about the implications of the derived equations and whether the calculations are correct. A potential mistake in the expression for standard deviation is also raised, indicating ongoing exploration of the topic.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the amount of direct assistance they can provide. There is a focus on ensuring the correct interpretation of the random variable involved in the problem.

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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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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.
 
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
 
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
\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:
oops i think i made a mistake.

would it be σ = 1/(2sqrt(n))

instead of

σ = 1/sqrt(2n) ?
 

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