Unbiased estimatior for $bar{x} $ from a Random Sample with unequal selection probability

A small population with twice as many children as adults. Suppose there are 4 each of children of ages 5, 10, and 15; and that there are 3 each of adults of ages 25 and 45. That means the average weight in the population of $36$ is $[4(20+35+50) + 3(90+85)]/18 = 52.5$kg.

What sample size? Also, suppose we are take a random sample of $n = 3$ from this population. (A clue that we should use $n = 3$ is having been given a sample of size three in the problem.)

The population of 36 weights is kg as follows:

kg=c(rep(c(20,35,50), each=4), rep(c(90,85), each=3))
mean(kg)
[1] 52.5

Simulation results. If we take many samples of size 3 from this population, we can get a good approximation of the sampling distribution.

set.seed(2020)
m = 10^5;  n=3;  s.3 = wt = numeric(m)
for(i in 1:m) {
  x = sort(sample(kg, 3))
  s.3[i] = sum(x == c(35,50,85))
  wt[i] = mean(x) }
mean(s.3==3)    # prob sample has 35,50,85
[1] 0.058995    # aprx 1/17
1/17
[1] 0.05882353  # exact 1/17
mean(wt)
[1] 52.50809    # aprx 52.5
2*sd(wt)/1000   
[1] 0.02902334

Probability of specified sample. With a million samples, one can expect 2 or 3 places of accuracy. One can show by simple combinatorics that the probability of getting one each of the weights $35, 50, 85$ (in some order) is $1/7,$ which is consistent with the simulation.

Unbiased estimator. Also, the mean weight in the population is $52.5.$ The simulation approximates $E(bar X_3)= 52.508 pm 0.029,$ with a 95% margin of simulation error.

If sampling had been with replacement, it is obvious that the mean of the sample of $n=3$ would be an unbiased estimate of the population weight $52.2.$ It is not hard to show that the same is true for sampling without replacement, and I will leave that to you.

Confidence intervals. I don't know what you have studied about confidence intervals. The sample mean of the specified sample of three observations is $bar X_3 = 56.67;$ it should be the center of a CI for the true mean weight of the population. Using it's standard error you should be able to get some style of CI.

Three observations are hardly enough for a good bootstrap CI, but if you know about bootstrapping this part of the problem may be a prompt to do whatever kind of bootstrap you may have studied. A naive percentile 75% nonparametric bootstrap CI can be found as follows (repeatedly re-sampling with replacement from the sample of three). This CI is $(40.0, 73.3),$ which does cover the known population mean.

set.seed(721)
re.avg = replicate(10^4, mean(sample(c(35,50,85), 3, rep=T)))
quantile(re.avg, c(.125, .875))
   12.5%    87.5% 
40.00000 73.33333