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Characteristics of a Poisson experiment:
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in 5 minutes. Thetime interval of interest is 5 minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in 5 minutes is 3?
Let $X$ = the number of loaves of bread put on the shelf in 5 minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in 5 minutes is
$\left(\frac{5}{30}\right)\cdot 12=2$ loaves of bread
The probability question asks you to find $\mathrm{P(x\; =\; 3)}$ .
A certain bank expects to receive 6 bad checks per day, on average. What is the probability of the bank getting fewer than 5 bad checks on any given day? Of interestis the number of checks the bank receives in 1 day, so the time interval of interest is 1 day. Let $X$ = the number of bad checks the bank receives in one day. If the bank expects to receive 6 bad checks per day then the average is 6 checks per day.The probability question asks for $P(x< 5)$ .
You notice that a news reporter says "uh", on average, 2 times per broadcast. What is the probability that the news reporter says "uh" more than 2 times per broadcast.
This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.
What is the average number of times the news reporter says "uh" during one broadcast?
2
Let $X$ = ____________. What values does $X$ take on?
Let
$X$ =
the number of times the news reporter says "uh" during one broadcast .
$x$ = 0, 1, 2, 3, ...
The probability question is $\text{P(\_\_\_\_\_\_)}$ .
$\text{P(x>2)}$
$X$ ~ $\text{P(\mu )}$
Read this as " $X$ is a random variable with a Poisson distribution." The parameter is $\mu $ (or $\lambda $ ). $\mu $ (or $\lambda $ ) = the mean for the interval of interest.
Leah's answering machine receives about 6 telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than 1 call in the next 15 minutes?
Let $X$ = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $\frac{1}{4}$ hour.)
$x$ = 0, 1, 2, 3, ...
If Leah receives, on the average, 6 telephone calls in 2 hours, and there are eight 15 minutes intervals in 2 hours, then Leah receives
$\frac{1}{8}\cdot 6=0.75$
calls in 15 minutes, on the average. So, $\mu $ = 0.75 for this problem.
$X$ ~ $\text{P(0.75)}$
Find $P(x> 1)$ . $P(x> 1)=0.1734$ (calculator or computer)
TI-83+ and TI-84: For a general discussion, see this example (Binomial) . The syntax is similar. The Poisson parameter list is ( $\mu $ for the interval of interest, number). For this problem:
Press 1- and then press 2nd DISTR. Arrow down to C:poissoncdf. Press ENTER. Enter .75,1). The result is $P(x> 1)=0.1734$ . NOTE: The TI calculators use $\lambda $ (lambda) for the mean.
The probability that Leah receives more than 1 telephone call in the next fifteen minutes is about 0.1734.
The graph of $X$ ~ $\text{P(0.75)}$ is:
The y-axis contains the probability of $x$ where $X$ = the number of calls in 15 minutes.
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