At each turn in a game a player score is denoted by x is found as follows players throw a fair six sided die once and if the number showing on the die is 4,5,6 that number is the value of X if the number shows an 1,2,3 then the thrower throws the die again and x is the number showing on the second throw probability distribution of x is shown in the following terms

There are two scenarios for $X=6$:

1. The first roll is a 6. The probability of this is $P(\text{first roll is 6}) = \frac{1}{6}$.
2. The first roll is 1, 2, or 3 (probability $\frac{3}{6} = \frac{1}{2}$), AND the second roll is a 6 (probability $\frac{1}{6}$). The probability of this sequence is $P(\text{first roll is 1,2,3}) \times P(\text{second roll is 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

To combine, use a common denominator of 12:

To subtract, use a common denominator of 48:

Note: The problem asks to show that $Var(X) = 118/48$. Based on the given probability distribution and game rules, the calculated variance is $\mathbf{113/48}$. There appears to be a typo in the problem statement's expected variance value.

From part (b):
Population mean, $\mu = E(X) = \frac{17}{4} = 4.25$
Population variance, $\sigma^2 = Var(X) = \frac{113}{48}$
Population standard deviation, $\sigma = \sqrt{\frac{113}{48}} \approx 1.534336$

For a large sample size ($n=200$), the sample mean $\bar{X}$ is approximately normally distributed with:
Mean of sample means, $\mu_{\bar{X}} = \mu = 4.25$
Standard deviation of sample means (standard error), $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$

We want to estimate $P(\bar{X} > 4)$.

We need to find $P(Z > -2.3043)$.
Using a standard normal distribution table or calculator: