Question

Data from

1414

cities were combined for a​ 20-year period, and the total

280280

​city-years included a total of

8080

homicides. After finding the mean number of homicides per​ city-year, find the probability that a randomly selected​ city-year has the following numbers of​ homicides, then compare the actual results to those expected by using the Poisson​ probabilities:

Homicides each​ city-year	a. 0	b. 1	c. 2	d. 3	e. 4
Actual results	211	59	9	1	0

a. ​(Round to four decimal places as​ needed.)

​P(0)equals=

​P(1)equals=

​P(2)equals=

​P(3)equals=

​P(4)equals=

Answer 1

Solution :

x	f(x)	xf(x)
0	211	0
1	59	59
2	9	18
3	1	3
4	0	0
total	280	80

mean =xf/ f =80/280 = 0.28

actual result expected result

  P(0)=e-0.28*0.280/0! =0.7557 , 0.7557*280 = 211.296

P(1)=e-0.28*0.281/1! =0.2116 , 0.2116*280 = 59.248

P(2)=e-0.28*0.282/2! =0.0296 , 0.0296*280 = 8.288

P(3)=e-0.28*0.283/3! =0.0027 , 0.0027*280 = 0.756

P(4)=e-0.28*0.284/4! =0.0001 , 0.0001*280 = 0.028

There for the results from poission distribution closly match to actual results.

Let me know in the comment section if anything is not clear. I will reply ASAP!
If you liked the answer, please give a thumbs-up. This will be quite encouraging for me.Thank-you!