Distribution of shoe production

Question

I know that the number of shoes produced in a factory during one week is a random variable with mean $300$.

a) What can I say about the probability that this week’s production will be at least $800$?

b) If in addition I know that the variance of this week’s production is $150$, what can I say about the probability that this week’s production will be between $200$ and $400$?

For a) I understand that I need to compute $P(X >= 800)$ but without any other informations how am I supposed to compute a number ?

Answer 1

Without further information, you can't compute these probabilities exactly. You can, however, get some inequalities. For $a$, for example, let $p$ be the probability that production is at least $800$. Since production can't be negative we see that the mean production, $\mu$ is at least $p\times 800$. Thus we have $$800p≤300\implies p≤\frac 38$$.

For $b$ you have more information, so you can do a bit better. We can invoke Chebyshev's Inequality. We have $\sigma =\sqrt {150}=12.24744871$ and we are interested in $k=\frac {100}{\sigma}=8.164965809$ from which we deduce that $$P(|X-300|≥100)≤\frac 1{8.164965809^2}=.015$$

Thus the probability you want is bounded below by $.985$

Answer 2

Clearly $X$ is nonnegative, so Markov's inequality yields $P(X \ge 800) \le \frac{E[X]}{800} = \frac{3}{8}$.

Now can we say anything more about this? We actually can't in the first case. If $0 \le p \le \frac{3}{8}$, we can find a nonnegative integer-valued random variable $X$ with mean $300$ that satisfies $P(X \ge 800) = p$. Simply let $p_2 = 1 - \frac{8}{3} p$ and $p_3 = 1 - p - p_2$ (convince yourself that $p_2$ and $p_3$ are both in the interval $[0, 1]$. Now let $P(X = 800) = p$, $P(X = 300) = p_2$ and $P(X = 0) = p_3$. Then $$E[X] = 800p + 300p_2 = 800p + 300 - 800p = 300,$$ which is what we wanted.