In a normal distribution where $\mu=180$ and $\sigma=36$, what value of the random variable $x$ has $98.3 \%$ of the area to its right?

Step 1 ：The problem is asking for the value of \(x\) such that the area to the right of \(x\) under the normal distribution curve is 98.3%. This is equivalent to finding the 1.7% percentile point (since 100% - 98.3% = 1.7%) of the normal distribution with mean \(\mu = 180\) and standard deviation \(\sigma = 36\).

Step 2 ：The percentile point can be found using the inverse of the cumulative distribution function (CDF), also known as the quantile function.

Step 3 ：By calculating, we get \(x = 103.67741916928256\)

Step 4 ：This means that approximately 98.3% of the area under the normal distribution curve is to the right of this value.

Step 5 ：Final Answer: The value of the random variable \(x\) that has 98.3% of the area to its right in a normal distribution where \(\mu=180\) and \(\sigma=36\) is approximately \(\boxed{103.68}\).