Good evening. I have tried to solve this problem with many approaches but I have not succeeded yet.
A man wants to estimate the mean number of water in meters cubed in a village. Sample used is $45$ tanks and std deviation is $2$. I have to use central limit theorem to find the probability that the error will be less than 1m cubed and less than $0.5$ m cubed.
I started out by solving to find the mean by saying standard deviation/sq root of $n = \frac{2}{6.71} = 0.298$.
I am kind of stuck as to what to do next do I have to find the $z$ value and if so what would be the values needed? Please help.
I think that you are assuming that the amount of water in a tank is normally distributed with unknown mean $\mu$ and unknown standard deviation $\sigma$, is that right? I don't think that that follows directly from the central limit theorem without some additional assumptions, but it is the usual form for this sort of homework problem. If I am wrong about that, please disregard the remainder of this answer.
The man is going to estimate $\mu$ by taking the sample mean $\bar{x}$ of the $45$ values he has measured. We want to know that probability that the error $\vert \bar{x}-\mu\vert$ is less than $1$.
To do this, we have to consider the distribution of the statistic $\bar{x}$. As I think you know, $\bar{x}$ is distributed normally with mean $\mu$ and standard deviation $\sigma/\sqrt{45}$.
If we knew $\sigma$, it would be easy from here, since the statistic $$Z=\frac{\bar{x}-\mu}{\sigma/\sqrt{45}}$$ would have the standard normal deviation.
However, it's at this point that we run into a problem: we do not know $\sigma$! What we do know is $s$, the sample standard deviation. Fortunately for us, it is for precisely this situation that the $t$-distribution was invented. So, instead of the $Z$ statistic above, we use the following $T$-statistic: $$T=\frac{\bar{x}-\mu}{s/\sqrt{45}}.$$ This statistic follows the $t$-distribution with 44 degrees of freedom.
(The statistic $s$ is an estimator for the parameter $\sigma$, but like any estimator, it is imperfect. This imperfection is a source of noise. If we were to pretend that $s=\sigma$, then we would underestimate the variability in our distribution for $\bar{x}$, and thus underestimate the probability that $\vert \bar{x}-\mu\vert > 1$.)
Since we want to know the probability that $\vert \bar{x}-\mu\vert<1$ we want the probabilities of the events that $\frac{-1}{s/\sqrt{45}}<T<\frac{1}{s/\sqrt{45}}$. Since $s=2$, and applying the fact that a $t$ distribution is symmetric about $0$, this becomes: $$1-2\cdot P\left[ T<\frac{-1}{0.298}\right].$$
Going to the first $t$-distribution calculator I found on Google (http://stattrek.com/online-calculator/t-distribution.aspx), I found that this probability is $1-2\times 0.0008=0.9984$.
I'll leave you to work out for yourself the probability that the error is less than $0.5$.
Since you mention "finding the $z$-value", I guess that there is a chance that your class expects you to use the $Z$ statistic rather than the $T$-statistic, by assuming that $\sigma=2$.
In this case, we know that $\bar{x}$ is distributed normally with mean $\mu$ and standard deviation $2/\sqrt{45}=0.298$ and the error term $\bar{x}-\mu$ is distributed normally with mean $0$ and standard deviation $0.298$. You want to find the probability that the absolute value of the error term is less than $1$. This is the same as the probability that $$\vert Z\vert<\frac{1}{2/\sqrt{45}}=3.354 $$ where $Z$ has the standard normal distribution. Using a $z$-caluclator, or tables, we see that $P[Z<-3.354]=0.0004$, and so the probability that $\vert Z\vert < 3.354$ is $0.9992$.
Note that this is higher than the probability calculated using the $t$-distribution, since by assuming that $\sigma=2$, we are making stronger assumptions about the tightness of the distribution of the estimate $\bar{x}$.