4.5 The Poisson Distribution

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 five minutes.  The time interval of interest is five minutes.  What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?

Solution:

Let [latex]X[/latex] be the number of loaves of bread put on the shelf in five minutes.  If the average number of loaves put on the shelf in 30 minutes (half an hour) is 12, then the average number of loaves put on the shelf in five minutes is [latex]\displaystyle{\frac{5}{30} \times 12=2}[/latex] loaves of bread.

The probability question asks you to find [latex]\displaystyle{P(x=3)}[/latex].

Leah receives about six telephone calls every two hours.

1. What is the probability that Leah receives exactly 4 calls in the next two hours?
2. What is the probability that Leah receives at most 9 calls in the next two hours?
3. What is the probability that Leach receives at most 2 calls in the next hour?

Solution:

1. The average number of calls in any two hour period is 6, so [latex]\lambda=6[/latex].
   

   Function	Poisson.dist	Answer
   Field 1	4	0.1339
   Field 2	6
   Field 3	false

   The probability that Leah receives 4 calls in the next two hours is 13.39%.

2. The average number of calls in any two hour period is 6, so [latex]\lambda=6[/latex].
   

   Function	Poisson.dist	Answer
   Field 1	9	0.9161
   Field 2	6
   Field 3	true

   The probability that Leah receives at most 6 calls in the next two hours is 91.61%.

3. The average number of calls in any two hour period is 6.  So the average number of calls in one hour is [latex]\displaystyle{\frac{6}{2}=3}[/latex].
   

   Function	Poisson.dist	Answer
   Field 1	2	0.4232
   Field 2	3
   Field 3	true

   The probability that Leah receives at most 6 calls in the next two hours is 42.32%.

According to Baydin, an email management company, an email user gets, on average, 147 emails over a six hour period.

1. What is the probability that an email user receives fewer than 160 emails over an six hour period?
2. What is the probability that an email user receives more than 40 emails over a two hour period?
3. What is the probability that an email user receives at least 600 emails over a 24 hour period?
4. What is the probability that an email user receives between 150 and 200 emails over a six hour period?

Solution:

1. The average over a six hour period is 147.  We want to find [latex]\displaystyle{P(x \lt 160)}[/latex].  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  Because [latex]x[/latex] must be an integer (it is the number of emails), [latex]x \lt 160[/latex] is the same as [latex]x \leq 159[/latex].  So [latex]\displaystyle{P(x \lt 160)=P(x \leq 159)}[/latex] and [latex]\displaystyle{P(x \leq 159)}[/latex] is a probability we can calculate with the Poisson.dist functon.
   

   Function	Poisson.dist	Answer
   Field 1	159	0.8486
   Field 2	147
   Field 3	true

   The probability a user receives fewer than 160 emails over a six hour period is 84.86%.

2. The average over a two hour period is [latex]\displaystyle{\frac{147}{3}=49}[/latex].  We want to find [latex]\displaystyle{P(x \gt 40)}[/latex].  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  The complement of [latex]\gt[/latex] is [latex]\leq[/latex], so [latex]\displaystyle{P(x \gt 40)=1-P(x \leq 40)}[/latex] and [latex]\displaystyle{P(x \leq 40)}[/latex] is a probability we can calculate with the Poisson.dist function.
   

   Function	1-Poisson.dist	Answer
   Field 1	40	0.8902
   Field 2	49
   Field 3	true

   The probability a user receives more than 40 emails over a two hour period is 89.02%.

3. The average over a 24 hour period is [latex]147 \times 4=588[/latex].  We want to find [latex]\displaystyle{P(x \geq 600)}[/latex].  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  The complement of [latex]\geq[/latex] is [latex]\lt[/latex], so [latex]\displaystyle{P(x \geq 600)=1-P(x\lt600)}[/latex].  Because [latex]x[/latex] must be an integer (it is the number of emails), [latex]x\lt600[/latex] is the same as [latex]x \leq 599[/latex].  So [latex]\displaystyle{P(x \geq 600)=1-P(x \lt 600)=1-P(x \leq 599)}[/latex] and [latex]\displaystyle{P(x \leq 599)}[/latex] is a probability we can calculate with the Poisson.dist function.
   

   Function	1-Poisson.dist	Answer
   Field 1	599	0.3158
   Field 2	588
   Field 3	true

   The probability a user receives at least 600 emails over a 24-hour period is 31.58%.

4. We want to find [latex]\displaystyle{P(150 \leq x \leq 200)}[/latex].  We cannot find this probability directly in Excel because the Poisson.dist function can only calculate [latex]=[/latex] or [latex]\leq[/latex] probabilities.  But, [latex]\displaystyle{P(150 \leq x \leq 200)=P(x \leq 200)-P(x \leq 149)}[/latex].  So we can calculate [latex]\displaystyle{P(150 \leq x \leq 200)}[/latex] as the difference of two Poisson.dist functions.
   

   Function	Poisson.dist	-Poisson.dist	Answer
   Field 1	200	149	0.4132
   Field 2	147	147
   Field 3	true	true

   The probability a user receives between 150 and 200 emails over a six hour period is 41.32%.