Tchebysheff's Theorem questions (statistics)

[TeenieBopper]

I have two different problems involving Tchebysheff's Theorem. Hopefully there isn't a rule about asking two different questions in one post.

Number 1

Homework Statement

The US mint produces dimes with an average diameter of .5 inch and a standard deviation of .01. Using Tchebysheff's theorem, find a lower bound for the number of coins in a lot of 400 coins that are expected to have a diameter between .48 and .52.

Homework Equations

μ=.5
σ=.01
k= number of standard deviations from the mean= 2
Tchebysheff's theorem:
P(|Y-μ|<kσ)>= 1- \frac{1}{k^{2}}

The Attempt at a Solution

Plugging everything in, I get P(|Y-μ|<(2)(.01))>= 1-\frac{1}{2^{2}}
Simplifying, it becomes P(|Y-μ|<.02)>= .75

The lower bound of the probability of a coin being under two standard deviations from the mean is .75. What's throwing me is the lot of 400 coins. I don't know how to take the result of Tchebysheff's theorem and apply it to the sample. Or am I just overthinking it and I just need to multiply 400*.75?

Question 2

Homework Statement

Let Y be a random variable such that
p(-1)= \frac{1}{18}
p(0)= \frac{16}{18}
p(1)= \frac{1}{18}

a) show that E(Y)=0 and V(Y)=\frac{1}{9}

b) Use the probability distribution of Y to calculate P(|Y-μ|>=3σ). Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when k = 3.

Homework Equations

Tchebysheff's Theorem
P(|Y-μ|<kσ)>=1-\frac{1}{k^{2}} or P{|Y-μ|>=kσ)<=\frac{1}{k^{2}}

The Attempt at a Solution

I thought I could just plug numbers into find E(Y) and V(Y). However, in order to get E(Y)= 0, I'd need k=0, which gives me an undefined fraction. Instead, using p(0)=\frac{16}{18}, I found k= \frac{3\sqrt{2}}{4}. But I don't know if this number is even useful.

Thanks in advance for any help I get.

 

[canis89]

The first problem asks you to determine the lower bound of the expected number of coins with diameters between .48 and .52, out of 400 coins. Note that such number is random.

You can think of this number as the number of trials of getting coins with such specification from 400 trials.

Under the assumption that each coin is produced independently from one another (or, each trial is independent from one another), this number is a binomial random variable with parameters:

n:=number of independent trials=400
p:=success probability=probability of getting a coin with diameter between .48 and .52. (the one you obtain by using Tchebysheff's Theorem is the lower bound of p, say, p')

The expected value of this variable is n*p>=n*p'. Since the problem asks you to determine the lower bound of the expected value of such number, then you have n*p'=400*0.75 as the answer.

For the second problem, do not get confused by 2b). Find the expectation and variance using usual formula for 2a).

For 2b), plug these values into Tchebysheff's inequality. Then, compare the lower bound of the probability with the actual value of the probability that you can get using the cumulated probability mass function (or by any logical way if it's too complicated; it's easy).