- #1
TeenieBopper
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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
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.
μ=.5
σ=.01
k= number of standard deviations from the mean= 2
Tchebysheff's theorem:
P(|Y-μ|<kσ)>= 1- [itex]\frac{1}{k^{2}}[/itex]
Plugging everything in, I get P(|Y-μ|<(2)(.01))>= 1-[itex]\frac{1}{2^{2}}[/itex]
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
Let Y be a random variable such that
p(-1)= [itex]\frac{1}{18}[/itex]
p(0)= [itex]\frac{16}{18}[/itex]
p(1)= [itex]\frac{1}{18}[/itex]
a) show that E(Y)=0 and V(Y)=[itex]\frac{1}{9}[/itex]
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.
Tchebysheff's Theorem
P(|Y-μ|<kσ)>=1-[itex]\frac{1}{k^{2}}[/itex] or P{|Y-μ|>=kσ)<=[itex]\frac{1}{k^{2}}[/itex]
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)=[itex]\frac{16}{18}[/itex], I found k= [itex]\frac{3\sqrt{2}}{4}[/itex]. But I don't know if this number is even useful.
Thanks in advance for any help I get.
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- [itex]\frac{1}{k^{2}}[/itex]
The Attempt at a Solution
Plugging everything in, I get P(|Y-μ|<(2)(.01))>= 1-[itex]\frac{1}{2^{2}}[/itex]
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)= [itex]\frac{1}{18}[/itex]
p(0)= [itex]\frac{16}{18}[/itex]
p(1)= [itex]\frac{1}{18}[/itex]
a) show that E(Y)=0 and V(Y)=[itex]\frac{1}{9}[/itex]
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-[itex]\frac{1}{k^{2}}[/itex] or P{|Y-μ|>=kσ)<=[itex]\frac{1}{k^{2}}[/itex]
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)=[itex]\frac{16}{18}[/itex], I found k= [itex]\frac{3\sqrt{2}}{4}[/itex]. But I don't know if this number is even useful.
Thanks in advance for any help I get.