# 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