Problem 137 Solution

Posted June 24, 1999

Binomial Distribution Solution

This is a basic binomial distibution problem. Given n people and a probability of death of p the probability of x deaths is n!/(x!*(n-x)!)*p^x*(1-p)^(n-x). In this particular case the formula is 1000!/(x!*(1000-x)!)*.01^x*.99^(1000-x).

The table below shows the probability of specific numbers of death using this formula:

Problem 137
Number of Deaths (x)	Probability total=x	Probability total<=x
0	0.00004317	0.00004317
1	0.00043607	0.00047924
2	0.00220019	0.00267943
3	0.00739322	0.01007265
4	0.01861375	0.02868640
5	0.03745311	0.06613951
6	0.06273711	0.12887663
7	0.08998657	0.21886319
8	0.11282407	0.33168726
9	0.12561333	0.45730059
10	0.12574021	0.58304080
11	0.11430928	0.69735009
12	0.09516152	0.79251160
13	0.07305328	0.86556489
14	0.05202279	0.91758768
15	0.03454173	0.95212941
16	0.02147955	0.97360896
17	0.01255845	0.98616742
18	0.00692759	0.99309501
19	0.00361663	0.99671164
20	0.00179188	0.99850352
21	0.00084466	0.99934817
22	0.00037967	0.99972784
23	0.00016307	0.99989092
24	0.00006705	0.99995797
25	0.00002644	0.99998441
26	0.00001002	0.99999443
27	0.00000365	0.99999808
28	0.00000128	0.99999936
29	0.00000043	0.99999979
30	0.00000014	0.99999994
31	0.00000004	0.99999998
32	0.00000001	0.99999999
33	0.00000000	1.00000000
34	0.00000000	1.00000000

The table shows that if 14 graves were dug then the probability of not running out would be 0.91758768 (the smallest number greater or equal to 90%). So they must dig 14 graves to have a 90% of not running out. To have a 95% chance they must dig 15 and to have a 99% chance they must dig 18.

Normal Distribution Solution

We can expect that the number of deaths to approximate a normal distribution curve. The mean is 1000*.01=10. The standard deviation is (1000*(.01)*(.99))^1/2 =~ 3.1464 . Let g be the number of graves such that the probability of running out is less than 90%. Let d be the number of deaths.

Pr(d <= g+.5) = 0.90 (The 0.5 is added because the number of deaths must be an integer)
Pr(d-9.5 <= g-10) = 0.90
Pr((d-9.5)/3.1464 <= (g-10)/3.1464) = 0.90
Pr(Z <= (g-9.5)/3.1464) = 0.90
(g-9.5)/3.1464 = 1.28
g =~ 13.527

The number of graves must be an integer, thus we round up to 14.

Michael Shackleford, ASA