Suppos that X, the amount of liquid apples contained in a container of commerical apple juice, is a random variable having mean 4 grams. A) what can be said about the probability that a given container contains more than 6 grams of liquid apple? B) if var(x) =4 grams^2 what can be said about the probability that a given container will contain between 3 and 5 grams of liquid apple? Solution Well, the questions can be answered easily once you know the following inequalities: 1. P(X>=a) <= E[X]/a (Markove Inequality: http://en.wikipedia.org/wiki/Markov%27s_inequality) 2. P( |X-E[X]| >= k*StdDev(X) ) <= 1/k*k (Chebyshev\'s Inequality: http://en.wikipedia.org/wiki/Chebyshev%27s_inequality) Hence, the correct solutions would be: A. Let X denote the grams of liquid apple. Hence, E[X] = mean = 4 Thus by Markov\'s inequality, P(X>=6) <= E[X]/6, i.e., P(container has more than 6 grams of liquid apple) <= 2/3 B. StdDev(X)^2 = Var(X) Thus, StdDev(X) = Var(X)^0.5 = 2 Now, we are interested in 3= k*StdDev(X)) <= 1/k*k. Thus, P(|X-E[X]| < k*StdDev(X)) = 1 - P(|X-E[X]| >= k*StdDev(X)) >= 1 - 1/k*k Hence, Probability(Weight of liquid apples is between 3 and 5 grams) >= 1 - 4 >= -3. But we already know that Probability of any event is non-negative. Hence, Probability(Weight of liquid apples is between 3 and 5 grams) >= 0. Thus, with the given peice of information, one cannot say anything with respect to the 2nd probability..

1. Suppos that X, the amount of liquid apples contained in a container of commerical apple juice, is
a random variable having mean 4 grams.
A) what can be said about the probability that a given container contains more than 6 grams of
liquid apple?
B) if var(x) =4 grams^2 what can be said about the probability that a given container will contain
between 3 and 5 grams of liquid apple?
Solution
Well, the questions can be answered easily once you know the following inequalities:
1. P(X>=a) <= E[X]/a (Markove Inequality:
http://en.wikipedia.org/wiki/Markov%27s_inequality)
2. P( |X-E[X]| >= k*StdDev(X) ) <= 1/k*k (Chebyshev's Inequality:
http://en.wikipedia.org/wiki/Chebyshev%27s_inequality)
Hence, the correct solutions would be:
A. Let X denote the grams of liquid apple. Hence, E[X] = mean = 4
Thus by Markov's inequality, P(X>=6) <= E[X]/6,
i.e., P(container has more than 6 grams of liquid apple) <= 2/3
B. StdDev(X)^2 = Var(X)
Thus, StdDev(X) = Var(X)^0.5 = 2
Now, we are interested in 3= k*StdDev(X)) <= 1/k*k.
Thus, P(|X-E[X]| < k*StdDev(X)) = 1 - P(|X-E[X]| >= k*StdDev(X))
>= 1 - 1/k*k
Hence, Probability(Weight of liquid apples is between 3 and 5 grams) >= 1 - 4
>= -3.

2. But we already know that Probability of any event is non-negative.
Hence, Probability(Weight of liquid apples is between 3 and 5 grams) >= 0.
Thus, with the given peice of information, one cannot say anything with respect to the 2nd
probability.