My work
I don't know how to use central limit theorem. What I think normally, The bounds are 30 , 120 , 300 and their mean is 150 and standard deviation is 112.24 So $$Z = \frac{160-150}{112.24}=0.089$$
So, $$P(Z>0.089) =1-0.535=0.465$$
Can anyone tell me did I solve corectly ?
You can comment and let me correct in comment by giving hint also
You did almost a good job! I will explain it, so other people with the same problem can follow what you are doing.
You dit it right for the expectation. The expectation of 1 day is $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 5 + \frac{1}{3} \cdot 10 =5$. You have 30 days, so the expectation over 30 days is $30 \cdot 5 = 150$ (since expectation values are addable).
You made a small mistake using the variance. Var$(X) = \mathbb{E}[X^2]- \mathbb{E}[X]^2$. We want to find it for only one day now, you'll find $\text{Var}(X) = (\frac{1}{3}\cdot 1^2 + \frac{1}{3}\cdot 4^2 + \frac{1}{3}\cdot 10^2) - 5^2 = 14$. The property op the variance is the following: $$\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) \ \text{when X and Y are independent}$$ So what you need to do is only take $30 \cdot \text{Var}(X) = 30 \cdot 14 = 420$. For the standard deviation, the following holds: $\sigma = \sqrt{420} = 20.4939$. Your Z-score will be $$\frac{160-150}{20.4939} = 0.48795$$So, $\mathbb{P}(Z > 0.48795) \approx 0.3121$
What you did wrong, and I did aswell, was using the fact that $\text{Var}(aX) = a^2\text{Var}(X)$, and this is true, but this is only scaling the stochast $X$, but that is not what we are doing here. We are using the rule that you can add variances when stochasts are independent. And that is the case here I think (it isn't described in the question but it feels right).