Stratified Sampling: Total and mean estimate of population

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Question: In a sample survey designed to estimate the total number of cattle, the universe of 2072 farms was stratified into 5 strata on the basis of the total acreage of farms. In the hth stratum (h = 1,2,...,5), a simple random sample (with replacement) of farms (of size $n_h$) was taen in proportion to the total number of farms in the stratum ($N_h$), the total sample size being $ n = \Sigma n_h = 500$; for each stratum, the total number of cattle in the sample farms ($\large \Sigma_{i=1}^{n_h}y_{hi})$ are given in the table below. Estimate the total and average number of cattle per farm, along with their standard error.

\begin{array}{rcl} \mbox{Stratum (acres)} &\mbox{# farms in stratum (Nh)} & \mbox{# farms sampled} & \mbox{number of cows in sample}\\ \ 0-15 & 635 & 153 & 619\\ 16-30 & 570 & 138 & 1423 \\ 31-50 & 475 & 115 & 1578 \\ 51-75 & 303 & 73 & 1691 \\ 76-100 & 89 & 21 & 603 \\ \mbox{All Strata} & 2072 \mbox{(N)} & 500 & 6094 \\ \end{array}

Attempt:

Average cattle per stratum:

0-15 acres: $\large \frac{619}{153} = 4.046$

16-30 acres: $\large \frac{1423}{138} = 10.312$

31-50 acres: $\large \frac{1578}{115} = 13.722$

51-75 acres: $\large \frac{1691}{73} = 23.164$

76-100 acres: $\large \frac{603}{21} = 28.714$

Then the average number of cows is: $\frac{1}{2072}[(4.046)(635)+(10.312)(570)+(13.722)(475)+(23.164)(303)+(28.714)(89)]$

$\bar{y}=11.843 \approx 12$

Total number of cattle is $2072 * 11.843 = 24538.696 \approx 24539$.