What Is the Lower Bound for a Product's Lifetime Using Tsebyshev's Inequality?

[TheSodesa]

Homework Statement

The average lifetime of a product $T=7.5$ (years). The variance of the lifetime $\sigma^{2} = 0.41$.

Using Tsebyshev's inequality, determine the lower bound for the probability, that the product lasts at least 5 years.

Homework Equations

Tsebyshev's inequality:
\begin{equation}
P(|X-\mu | \geq t) \leq \frac{\sigma^{2}}{t^2}
\iff
P(|X-\mu | < t) \geq 1 - \frac{\sigma^{2}}{t^2}
\end{equation}
where $\mu$ is the expected value, $\sigma^2$ is the variance and $t$ is the deviation from the expected value.

The Attempt at a Solution

We want
\begin{align*}
P(T \geq 5)
&\geq P(5 \leq T \leq 10)\\
&= P(|T-7.5| \leq 2.5) &|t=2.5\\
&\geq 1 - \frac{\sigma^2}{t^2} &| Tsebyshev\\
&= 1-\frac{0.41^2}{2.5^2}\\
&\approx 0.973
\end{align*}
This is apparently not the correct answer, and I'm not sure what I'm doing wrong. We obviously want the deviation from the mean to be less than 2.5; otherwise $|T-\mu|$ would produce a value that is below 5.

 

[mfb]

Edit: Why did you square the 0.41?

Some manual tuning: The worst case is some fraction p failing "just before" 5 years and all other machines failing after X years, X a bit larger than 7.5

The condition for the mean is then 5p+(1-p)*X = 7.5.
The condition for the variance is p*2.5^2 + (1-p)*(X-7.5)^2 = 0.41

Solving gives p=0.062 and X=7.66.

 

[micromass]

Why do you think the answer is wrong?

You can use it for both. One gives a lower bound, the other gives an upper bound.

 

[micromass]

Homework Statement

The average lifetime of a product $T=7.5$ (years). The variance of the lifetime $\sigma^{2} = 0.41$.

Using Tsebyshev's inequality, determine the lower bound for the probability, that the product lasts at least 5 years.

Homework Equations

Tsebyshev's inequality:
\begin{equation}
P(|X-\mu | \geq t) \leq \frac{\sigma^{2}}{t^2}
\iff
P(|X-\mu | < t) \geq 1 - \frac{\sigma^{2}}{t^2}
\end{equation}
where $\mu$ is the expected value, $\sigma^2$ is the variance and $t$ is the deviation from the expected value.

The Attempt at a Solution

We want
\begin{align*}
P(T \geq 5)
&\geq P(5 \leq T \leq 10)\\
&= P(|T-7.5| \leq 2.5) &|t=2.5\\
&\geq 1 - \frac{\sigma^2}{t^2} &| Tsebyshev\\
&= 1-\frac{0.41^2}{2.5^2}\\
&\approx 0.973
\end{align*}
This is apparently not the correct answer, and I'm not sure what I'm doing wrong. We obviously want the deviation from the mean to be less than 2.5; otherwise $|T-\mu|$ would produce a value that is below 5.

You are given $\sigma^2 = 0.41$. But later on, you write $0.41^2$ at the end of your computation. There is no need for that square.

 

[TheSodesa]

Chebyshev's inequality makes a statement about larger deviations. You can use it for P(|T-7.5|>2.5) but not for P(|T-7.5|<2.5).

So what I could do is calculate $P(T<5)$ and go from there?

\begin{align*}
P(T<5)
&\leq P(T<5 \text{ and } T>10)\\
&= P(|T-\mu| >= 2.5)\\
&\leq \frac{\sigma^2}{2.5^2}\\
&=\frac{0.41^2}{2.5^2}\\
&= 0.026896
\end{align*}
But then I'm stuck with the shaded area shaded blue in this picture:

and I need the one in the middle. At least that's my understanding.

 

[TheSodesa]

You are given $\sigma^2 = 0.41$. But later on, you write $0.41^2$ at the end of your computation. There is no need for that square.

Oh. My. Gauss!

Thank you. How typical of me...