Question

Use Chebyshev's Inequality to get a lower bound for the number of times a fair coin must be tossed in order for the probability to be at least 0.90 that the ratio of the observed number of heads to the total number of tosses be between 0.4 and 0.6. 

Let X be a random variable with μ=10 and σ=4. A sample of size 100 is taken from this population. Find the probability that the sum of these 100 observations is less than 900 .

Answer 1

1.

Let sample size =n

Then

P(0.4 0.9

So

P(p - 0.5 <0.1) > 0.9

Now

Comparing with chebyshev inequality

P(p - EP)

Where

op) = 1 0.5 * 0.5

E(p)=0.5

So on comparison

2 = 160

Now

k*0(p) = 1 0.5*0.5 * Vn = 0.1 = n= = = 5.55

So n~6

2.

4 = 10, J = 4, n = 4

Let

$T= \sum_{i=1}^{n}X_i$

We have to find P(T<900)

100 E(T) - ΣΕ(Χ.) = 100 και 10 = 1000

Var(T) = { var(Xi) = 100 * 4? = 1600

Hence

OT = 40

Now

900 — 1000, 900 — E(Т), 2) от = PT < 900) - Рт – E(Т) От P(x < -2.5) P(Z <

=0.006