Need help in R to do part c and d but in R programming language:

Solutions so far:

\fDaily counts = fyily 21 ..- yng n= 365. a there are M number of frogs in the pond. Pityis) probability of seeing a frogaday - . To Assumptions : the number of frogs seen each day is independent of the previous day number The probability of seeing a frog each each day is the same for 365 days: There are only two possible outcomes ( feeing a frog or not seeing a frog !) b) Likelihood L( O,M) = M OF (1 - 0 ) M - TC - TT ( M) ET (1 - 0)M - ZY " Taking the Log of Likelihood 6 2 ( AM) =/ log ) + ti logo + nM - dog (-0) = ZYi - ( nM - Z Y C ) 1 - 0 Equating to zero ( MM - Zti ) 1- 0 (nM - Epi ) 9 = Etc ( 1 - 3) nMe = Zy MLE ON = ETC nM nMTo prove its a maximum 226 - - Zyi nM - ZY" 9 2 1- 92 = - ZY - ( 1 - 0 ) 2 Which is negative hence On is a Makmum Likelihood estimator. C From the data. n = 20 I t: = Sum counts =52' OM = Iti nM 52 20 M 2 2 - 6 M Average number of frogs 52= M (To get the MU am I Z ti/ nxmy where Ey wall sum counts and n= 365 / M = ?) ( since , ten toled the link and if didn't open as I was finishing the question.)\f