# Probability given a mean and standard deviation of a random variable

The number of transactions handled by a bank teller in a day is a random variable with a mean of 80 and standard deviation of 5.

What can be said that the teller will handle at least 400 transactions in a day?

What can be said about the probability that the teller will handle between 70 and 90 transactions in a day?

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Read about the Markov inequality which says that the probability that a nonnegative random variable $X$ takes on values $\alpha$ or larger is bounded above by $\mu/\alpha$ where $\mu$ is the mean of $X$. Similarly, read about the Chebyshev inequality which bounds the probability that $X$ differs from its mean $\mu$ by $\alpha$ or more in terms of the standard deviation. – Dilip Sarwate Oct 7 '12 at 1:20

the probability that the teller will handle at least 400 transactions per day is close to 0, Im 95% confident that the teller will handle between 70 to 90 transactions in a day.

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Why are you fractionally confident that the teller will handle between 70 and 90 transactions a day? How can you be sure that the probability of handling at least 400 transactions is close to 0? Please explain. – Dilip Sarwate Oct 7 '12 at 1:22
@DilipSarwate sorry, i take that back. I think i had assumed the distribution to be normal... – user133466 Oct 7 '12 at 1:34
@DilipSarwate had the distribution been normal, i'd have been right? – user133466 Oct 7 '12 at 1:40

The probability that the teller will handle at least 400 transactions in a day is the following. $$P(|X| \ge 400) \le \frac{80}{400} = \frac{1}{5}$$

The probability that the teller will handle between 70 and 90 transactions in a day is the following. $$P(70 < X < 90) \ge 1-\frac{5^2}{100} = \frac{3}{4}$$

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Where did I go wrong? – idealistikz Oct 9 '12 at 3:09
You are asked what can be said about $P\{70 \leq Y \leq 90\} = P\{|Y-80|\leq 10\}$ or perhaps $P\{70 < Y < 90\} = P\{|Y-80| < 10\}$ depending on how you want to interpret the phrase "between $70$ and $90$". You have correctly bounded $P\{|Y-80|\geq 10\}$. Now turn the inequality around to get a lower bound on the probability requested in the problem – Dilip Sarwate Oct 9 '12 at 12:17
@DilipSarwate, can I assume the distribution is symmetric since the mean is 80? – idealistikz Oct 10 '12 at 2:49
No, if you assume a symmetric distribution, then since the teller cannot handle a negative number of transactions, the maximum number of transactions is restricted to be $160$ or less. You might want to give some thought to the fact that an $8$ hour work day has $480$ minutes (not considering lunch break, coffee break, etc) and so a teller who handles $400$ transactions in a day is processing them at an average rate of one a minute: highly unlikely! – Dilip Sarwate Oct 10 '12 at 2:56
$90+60 = 2*80$ Try again. My calculator says that $90+60=2*75$. – Dilip Sarwate Oct 10 '12 at 3:13