Chebyshev’s Inequality Calculator

Use below Chebyshev’s inqeuality calculator to calculate required probability from the given standard deviation value (k) or P(X>B) or P(A<X<B) or outside A and B.

Chebyshev's Inequality Calculator
Find k
P(X > B)
P(A< X < B) and
Outside A and B and
Results
Required Probability :

Chebyshev Inequality Formula

Let $X$ be a random variable with mean $\mu$ and finite variance $\sigma^2$. Then for any real constant $k>0$, $$ \begin{equation*} P[|X-\mu| \geq k\sigma] \leq \frac{1}{k^2}\quad \text{ and }\quad P[|X-\mu|< k\sigma] \geq 1-\frac{1}{k^2} \end{equation*} $$

OR

If $\mu$ and $\sigma$ are the mean and the standard deviation of a random variable $X,$ then for any positive constant $k$, the probability is at least $1-\dfrac{1}{k^2}$ that $X$ will take on a value within $k$ standard deviations of the mean.

The probability that the random variable $X$ is within $k$ standard deviation of the mean is given by $$ \begin{aligned} P[|X-\mu| \geq k\sigma] \leq \frac{1}{k^2}\quad \text{ and }\quad P[|X-\mu|< k\sigma] \geq 1-\frac{1}{k^2} \end{aligned} $$ where,

  • $\mu$ is the mean of $X$,
  • $\sigma$ is the standard deviation of $X$,
  • $k$ is the real constant greater than 0.

Chebyshev’s Inequality Theorem

If $g(x)$ is a non-negative function and $f(x)$ be p.m.f. or p.d.f. of a random variable $X$, having finite expectation and if $k$ is any positive real constant, then $$ \begin{equation*} P[g(x)\geq k] \leq \frac{E[g(x)]}{k}\quad \text{ and }\quad P[g(x) < k] \geq 1-\frac{E[g(x)]}{k} \end{equation*} $$

Chebyshev’s Inequality Theorem Proof

Discrete Case

Let $X$ be a discrete random variable with p.m.f. $f(x)$. Let $g(x)$ be a non-negative function of $X$. Then $E[g(x)] = \sum g(x)f(x)$.

Let us introduce a new random variable $Y$ such that $$ \begin{equation*} Y=\left\{ \begin{array}{ll} 0, & \hbox{$g(x) < k$;} \\ k, & \hbox{$g(x) \geq k$.} \end{array} \right. \end{equation*} $$ Hence, $$ \begin{eqnarray*} E(Y) & = & 0\cdot P[Y=0] + k \times P[Y=k]\\ &=& k\times P[Y=k]\\ & = & k \times P[g(x)\geq k]. \end{eqnarray*} $$ But $g(x) \geq Y$. Therefore $E[g(x)] \geq E[Y]$. $$ \begin{eqnarray*} \therefore E[g(x)] &\geq & k\times P[g(x) \geq k]\\ \Rightarrow P[g(x)\geq k] &\leq & \dfrac{E[g(x)]}{k}. \end{eqnarray*} $$ By taking complement of the above probability, we get $$ \begin{eqnarray*} P[g(x) < k] & = & 1- P[g(x) \geq k]\\ \text{i.e., } P[g(x) < k] &\geq & 1 -\frac{E[g(x)]}{k}. \end{eqnarray*} $$

Continuous Case

Let $X$ be a continuous random variable with p.d.f. $f(x)$ and $g(x)$ be a non-negative function of $X$. Let $S$ be the set of all $x$ for which $g(x)\geq k$, i.e., $S = { x | g(x) \geq k}$, so $\overline{S} = { x | g(x) < k}$.

$$ \begin{eqnarray*} E[g(x)] &=& \int_{-\infty}^{\infty} g(x) f(x) \; dx \\ &=& \int_S g(x) f(x) \;dx +\int_{\overline{S}}g(x) f(x) \; dx\\ &\geq & \int_S g(x) f(x) \;dx \;\;\text{ (since both the integrals are positive)}\\ & \geq & k \int_S f(x) \;dx \;\; \text{ (since $g(x) \geq k$)}\\ & \geq & k \times P[g(x)\geq k]. \end{eqnarray*} $$ Therefore, $$ \begin{equation*} P[g(x)\geq k]\leq \frac{E[g(x)]}{k}. \end{equation*} $$ Taking complement, we get $$ \begin{equation*} P[g(x)< k]\geq 1-\frac{E[g(x)]}{k}. \end{equation*} $$

Another Form of Chebyshev’s Inequality

Let $X$ be a random variable with mean $\mu$ and finite variance $\sigma^2$. Then for any real constant $k>0$, $$ \begin{equation*} P[|X-\mu| \geq k] \leq \frac{\sigma^2}{k^2} \end{equation*} $$ and $$ \begin{equation*} P[|X-\mu|< k] \geq 1-\frac{\sigma^2}{k^2} \end{equation*} $$

First inequality gives upper bound for the probability whereas the second inequality gives lower bound for the probability.

Example 1 Chebyshev’s Inequality Calculator

The ages of members of gym have a mean of 45 years and a standard deviation of 11 years. What can you conclude about the percentage of gym members aged between 28.5 and 61.5?

Solution

Given that $\mu=45$ and $\sigma = 11$.

The probability that the gym members age is between $28.5$ and $61.5$ is $$ \begin{aligned} P(28.5 < X < 61.5) &= P(28.5-45<X-\mu< 61.5-45)\\ &= P(-16.5<(X-\mu)< 16.5)\\ &=P\big(|X-\mu|< 16.5\big) \end{aligned} $$

Comparing this with the Chebyshev’s inequality theorem, we get

$$ \begin{aligned} & k\sigma = 16.5\\ \Rightarrow & k =\frac{16.5}{\sigma}\\ \Rightarrow & k =\frac{16.5}{11}\\ \Rightarrow & k =1.5 \end{aligned} $$

Therefore, by Chebyshev’s inequality calculator,

$$ \begin{aligned} P(28.5 < X < 61.5) &=P\big(|X-\mu|< 16.5\big)\\ &\geq 1-\frac{1}{k^2}\\ &\geq 1-\frac{1}{1.5^2}\\ &\geq 1-0.4444\\ &\geq 0.5556 \end{aligned} $$

Thus, the percentage of gym members aged between $28.5$ and $61.5$ is at least $55.56$.

Example 2 Chebyshev Inequality Theorem Calculator

The daily production of electric motors at a certain factory averaged 120, with a standard deviation of 10.

a. What can be said about the fraction of days on which the production level falls between 100 and 140?

b. Find the shortest interval certain to contain at least 90% of the daily production levels.

Solution

Given that $\mu=120$ and $\sigma = 10$.

a. The probability that the production level falls between $100$ and $140$ is

$$ \begin{aligned} P(100 < X < 140) &= P(100-120<X-\mu< 140-120)\\ &= P(-20<(X-\mu)< 20)\\ &=P\big(|X-\mu|< 20\big) \end{aligned} $$

Comparing this with the Chebyshev’s inequality theorem, we get

$$ \begin{aligned} & k\sigma = 20\\ \Rightarrow & k =\frac{20}{\sigma}\\ \Rightarrow & k =\frac{20}{10}\\ \Rightarrow & k =2 \end{aligned} $$

Therefore, by Chebyshev’s inequality calculator,

$$ \begin{aligned} P(100 < X < 140) &=P\big(|X-\mu|< 20\big)\\ &\geq 1-\frac{1}{k^2}\\ &\geq 1-\frac{1}{2^2}\\ &\geq 1-0.25\\ &\geq 0.75\\ \end{aligned} $$

Thus, the percentage of days on which the production level falls between $100$ and $140$ is at least $75$.

b. We want to find the value of $k$ such that shortest interval certain to contain at least 90% of the daily production levels.

Using Chebyshev’s inequality formula,

$$ \begin{aligned} P(|X-120|<10k)\geq 1-\frac{1}{k^2}=0.9 \end{aligned} $$

$$ \begin{aligned} & 1-\frac{1}{k^2}=0.9 \\ \Rightarrow & \frac{1}{k^2} = 0.1\\ \Rightarrow & k^2 = 10\\ \Rightarrow & k = \sqrt{10}\\ \Rightarrow & k = 3.16\\ \end{aligned} $$

Using the Chebyshev’s inequality formula

$$ \begin{aligned} & P(|X-120|<10\times 3.16)\geq 0.9\\ \Rightarrow & P(|X-120|<31.6)\geq 0.9\\ \Rightarrow & P(-31.6<X-120<31.6)\geq 0.9\\ \Rightarrow & P(-31.6+120<X<31.6+120)\geq 0.9\\ \Rightarrow & P(88.4<X<151.6)\geq 0.9\\ \end{aligned} $$

Thus, the shortest interval $(88.4,151.6)$ will contain at least 90% of the daily production levels.

Conclusion

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