Question 2. Using both latest_poverty and population, create a four-column table called recent_poverty_total with one row for each country in latest_poverty. The four columns should have the following labels and contents: 1. geo contains the 3-letter country code, 1. poverty_percent contains the most recent poverty percent, 1. population_total contains the population of the country in 2010, 1. poverty_total contains the number of people in poverty rounded to the nearest integer, based on the 2010 population and most recent poverty rate. In [61]: latest_poverty . join ('geo' , population) .group ( 'geo' , first) Out [61] : geo | time first | poverty_percent first | time_2 first | population_total first ago | 2009 | 43.37 | 1800 | 1567028 alb | 2012 | 0.46 1600 | 200000 arg | 2011 | 1.41 1800 | 534000 arm | 2012 | 1.75 1800 413326 aus I 2003 | 1.36 1600 450000 aut I 2004 | 0.34 1600 2500000 aze 2008 | 0. 31 1800 879960 bdi | 2006 | 81.32 1800 899097 bel | 2000 1 0.5 1600 1600000 ben | 2012 | 51 . 61 1800 | 636559 . .. (135 rows omitted) In [74]: poverty_and_pop = latest_poverty . join ( 'geo' , population) . where ('time_2', are . equal_to (2 recent_poverty_total = poverty_and_pop. with_column ( 'poverty_total', np. round((poverty recent_poverty_total