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You take a fair die to a party and announce that you will roll it 25 times.

You will record each outcome and at the end average the 25 outcomes together to get their arithmetical mean.

You offer a bet: The player puts down 1:if mean exceeds 4, you will give him $21 back, but otherwise, he loses his dollar.Is this a good bet for the player(Using Central limit theorem CTL) ?

I tried to solve check whether it is true:

I have some hints for you and I suggest you work from there. I might have made errors as well. If you need more help or find an error, leave a comment.

First we write the probability mass function for a having the result $N$ on a single die:

$$prob(N_i)=1/6,, textwith N=1,2,3,4,5,6$$

This is a Discrete Uniform Distribution. All results are equally likely since the die is fair. Now consider the result of summing the faces of 25 rolls of the die and taking the average. The average $S$ is given by the sum of $N_i$ with $i=1,...25$ over 25 where the $N_i$ are i.i.d. distributed as above.

$$S= fracN_1 + N_2 + ... +N_2525$$

So we ask what is $prob(S>4)$.

For that you first need the probability distribution of $S$. Here is where the hint with the Central Limit Theorem comes in handy. It tells us that we can approximate the density of $prob(S)$ can be approximated as Gaussian with

$$prob(S) sim mathcalN(mu,sigma^2/25)$$

where $mu$ is the expected value of the uniform distribution from above and $sigma^2$ is its variance. You can calculate them using the information on this Wikipedia page.

EDIT
If you do not want to use the Gaussian approximation, have a look at the Bates Distribution as suggested by @wolfies.

Do you think you can take it from here?

You are effectively taking the probability for a single dice roll being larger than 4 (which is indeed $frac13$). But that is a different thing than the mean of several dice rolls being larger than 4 (you can express this also as the sum of $x$ dice rolls being larger than $4x$).

Two dice rolls example

See for instance the possible outcomes of two dice rolls, where only $frac1036 < frac13$ have a mean above 4 (or total above 8)

$$beginarrayc
& 1 & 2 & 3 & 4 & 5 & 6 \
hline
1 & 2 & 3 & 4 & 5 & 6 & 7 \
2 & 3 & 4 & 5 & 6 & 7 & 8 \
3 & 4 & 5 & 6 & 7 & 8 & colorred9 \
4 & 5 & 6 & 7 & 8 & colorred9 & colorred10 \
5 & 6 & 7 & 8 & colorred9 & colorred10 & colorred11\
6 & 7 & 8 & colorred9 & colorred10 & colorred11 & colorred12
endarray$$

More dice rolls

The image below shows how this continues for more dice rolls, by plotting the probability of the sum of dice rolls $X$.

25 dice rolls

• 
  Explicit: You can calculate this explicitly by computing a table like above for the two dice rolls, but then for many instead. This site there has already been a question about this (How to easily determine the results distribution for multiple dice?). Such calculation will give you a probability of: $$P(barx>4) = frac18231483546232988166^25 approx 0.0641 > frac121 $$
  


• 
  

  Approximation with normal distribution: In the image above you might note the bell shape curve of the normal distribution. The normal distribution is actually a quite good approximation for the mean of a dice roll (in fact the normal distribution was first described in relation to the approximation of coin flips, where deMoivre used a precursor of the normal distribution to approximated the binomial distribution, and you might see the dice roll as a multivariate generalization of the coin flip).

  
  
  

  This is 'what the question wants you to do': Use the normal distribution as an approximation for the mean of dice rolls. (and then use the resulting expression for the normal distribution to compute $P(barx>4)$). The question mentions CLT (the central limit theorem), and when you look that up you may find expressions for "approximating the mean of a sample based on the variance of the distribution" (if one would be pedantic then one could say that this approximation is not exactly the same as the 'central limit theorem', but many people mention/use this term when they employ this type of approximation).