# 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)!)*px*(1-p)(n-x). In this particular case the formula is 1000!/(x!*(1000-x)!)*.01x*.99(1000-x).

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

 Problem 137 Numberof Deaths(x) Probabilitytotal=x Probabilitytotal<=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