Question

Question: Suppose that the number of cans of soda pop filled in a day at a bottling plant is a random variable with an expected value of 10,000 and a variance of 1000.

a) Use Markov's inequality to obtain an upper bound on the probability that the plant will fill more than 11,000 cans on a particular day.

b) Use Chebyshev's inequality to obtain a lower bound on the probability that the plant will fill between 9000 and 11,000 cans on a particular day.

Short answer

Answer

a. So, the probability that the plant will fill more than 11,000 cans is 0.9091.

b. So, the probability that the plant will fill between 9,000 and 11,000 cans is 0.999

Step-by-step solution

In the problem given  

The plant will fill between 9,000 and 11,000 cans on a particular.

The definition and the formula for the given problem

Markov's inequality\(P(X(s) \ge a) \le E(X)/a\) says that for a positive random variable X and any positive real number a, the probability that X is greater than or equal to a is less than or equal to the expected value of X divided by a.

Chebyshev's inequality: it states that minimum of just \(75\% \) of values must lie within two standard deviations of the mean and \(89\% \) within three standard deviations.

Determining the sum in expanded form

(a) Apply Markov's inequality with a=11000:

Markov's inequality:

\(P(X(s) \ge a) \le \frac{{E(X)}}{a}\)

\(\begin{array}{c}E(X) = 10000\\V(X) = 1000\end{array}\)

\(\begin{array}{c}P(X(s) \ge 11,000) \le \frac{{E(X)}}{{11,000}}\\ = \frac{{10,000}}{{11,000}}\\ = \frac{{10}}{{11}}\\ \approx 0.9091\end{array}\)

Using Markov's inequality to obtain an upper bound on the probability.  

(b) Apply Chebyshev's inequality with r=1000 :

Chebyshev's inequality:

\(P(|X(s) - E(X)| \ge r) \le \frac{{V(X)}}{{{r^2}}}\)

\(\begin{aligned}{*{20}{c}}{P(9000 < X < 11,000)}{ = P( - 1000 < X - 10,000 < 1000)}\\{}&{ = P( - 1000 < X - E(X) < 1000)}\\{}&{ = P(|X(s) - E(X)| < 1000)}\\{}&{ = 1 - P(|X(s) - E(X)| \ge 1000)\quad {\rm{ Complement rule }}}\\{}&{ \ge 1 - \frac{{V(X)}}{{{{1000}^2}}}}\\{}&{ = 1 - \frac{{1000}}{{{{1000}^2}}}}\\{}&{ = 1 - \frac{1}{{1000}}}\\{}&{ = \frac{{999}}{{1000}}}\\{}&{ = 0.999}\end{aligned}\)

Result

1. \(\frac{{10}}{{11}} \approx 0.9091\)
2. \(\frac{{999}}{{1000}} \approx 0.999\)