# Estimate of population size after 20 weeks.

## Question:

Estimate of population size after 20 weeks.

Week | N1 | N2 | N3 | Lamb 1 | Lamb 2 | Lamb 3 |
---|---|---|---|---|---|---|

8 | 392 | 449 | 476 | |||

9 | 616 | 679 | 726 | 1.568924414 | 1.513031954 | 1.553016087 |

10 | 852 | 902 | 990 | 1.384103165 | 1.328708274 | 1.364209078 |

11 | 1066 | 1072 | 1201 | 1.250768017 | 1.188363381 | 1.213335863 |

12 | 1211 | 1179 | 1339 | 1.136776624 | 1.0999052555 | 1.114815032 |

13 | 1300 | 1239 | 1418 | 1.072808078 | 1.050613686 | 1.058680776 |

14 | 1348 | 1270 | 1459 | 1.03748048 | 1.025017023 | 1.0291547 |

15 | 1374 | 1285 | 1480 | 1.018949754 | 1.012210773 | 1.1.014275675 |

16 | 1387 | 1293 | 1490 | 1.009491876 | 1.005923038 | 1.006939563 |

17 | 1400 | 1295 | 1492 | 1.009436426 | 1.001679153 | 1.001151106 |

18 | 1398 | 1304 | 1501 | 0.998571429 | 1.006949807 | 1.006032172 |

19 | 1390 | 1312 | 1499 | 0.994277539 | 1.006134969 | 0.998667555 |

Average Lamb | 1.1346898 | 1.112594301 | 1.1236616 | |||

Overall Lamb | 1.11116565 | 1.09353344 | 1.10199982 | |||

r | 0.12635931 | 0.106694497 | 0.116592639 | |||

r | 0.105409599 | 0.089414142 | 0.097126548 |

## Data fit

This is a problem that a rather ambiguous approach. Therefore, the best way to obtain information from the data provided is to fit the data. This is a statistical process by which a specific mathematical behavior is assigned to a group of experimental points.

## Answer and Explanation:

In our case, the experimental points are the number of lambs in each herd as a function of time. If there were much more information, a specific behavior could be assigned to the number of lambs, however in this case, this information is not available. Therefore a stretched asymptotic exponential equation of the type was chosen:

{eq}y=a(1-e^{-bx})^c {/eq}

For herd 1 we have:

{eq}N1(t)=1418\left ( 1-e^{-(0.52)t} \right )^{84}\\ N1(20\text{ weeks})=1418\left ( 1-e^{-(0.52)(20)} \right )^{84}\\ \therefore N1(20\text{ weeks})=1414\text{ lambs} {/eq}

For herd 2 we have:

{eq}N2(t)=1315\left ( 1-e^{-(0.56)t} \right )^{94}\\ N2(20\text{ weeks})=1315\left ( 1-e^{-(0.56)(20)} \right )^{94}\\ \therefore N2(20\text{ weeks})=1313\text{ lambs} {/eq}

For herd 3 we have:

{eq}N3(t)=1516\left ( 1-e^{-(0.54)t} \right )^{94}\\ N3(20\text{ weeks})=1516\left ( 1-e^{-(0.54)(20)} \right )^{94}\\ \therefore N3(20\text{ weeks})=1513\text{ lambs} {/eq}