Question

When a balanced coin is flipped 10,000 times, find the lower bound of the probability that the proportion of heads obtained will fall between 0.45 and 0.55. Hint, use Chebyshev.

Answer 1

i am okay doing it myself here you go

Explanation:

Given that

n= 10000, P= 1/2 or 0.5

Let us find mean and S.D.

mean μ= np

= 10000(0.5) = 5000

S.D. σ =√npq

= √(1000(1/2)(1/2))= 50

μ-kσ

5000-10(50)= 4500

μ+kσ

5000+10(50)= 5500

4500/10000= 0.45

and

5500/10000= 0.5

Applying Chebishev's inequality

P(|x-μ|<kσ)>_ 1-1/k^2

P(|x-5000|<500)>_ 1-1/100 = 0.99