Chebyshev’s theorem

Question

Suppose the mean noon-time temperature for September days in San Diego is 24∘ and the standard deviation is 4.9. (Temperature in this problem is measured in degrees celsius)

On September 26, 1963, the all-time record of noon-time temperature in San Diego of 44∘ was hit. Assume the temperature distribution is symmetric around the mean, what is the Chebyshev bound for the probability of breaking (or tieing) this record?

I am having a hard time understanding this. Could someone explain to me how to do this?

Answer 1

I think you want the bound on probability provided by Chebyshev, which states that

$$ Pr(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2} $$ which, in English, says that the probability that the deviation of the random variable $X$ from its expectation by more than $k$ times the standard deviation is less than 1 over this same constant $k$, but this time squared.

In your case, $\mu=24$ and $\sigma=4.9$. You want to know:

$$ Pr(X\geq 44)=Pr(X-24\geq 20) $$

So then, $$ k\sigma=20\Leftrightarrow k=\frac{20}{4.9} $$

Note, however, that in the first equation we have

$$ Pr(|X-\mu|\geq k\sigma)=2Pr(X-\mu\geq k\sigma) $$ since you have a symmetric distribution and $X$ can deviate in either direction. Combining this:

$$ Pr(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2} \Longrightarrow 2Pr\left(X-24\geq 20\right)\leq \frac{1}{\left(\frac{20}{4.9}\right)^2}\\ \Longrightarrow Pr\left(X-24\geq 20\right)\leq \frac{1}{2\left(\frac{20}{4.9}\right)^2}\approx 0.03 $$

Answer 2

$P(X\geq 44) = P(X-24 \geq 20) \leq \frac{E(X-24)^2}{20^2} = \frac{4.9^2}{400} \leq \frac{25}{400} = 16^{-1}$