The following problem and solution appears in Judgment under uncertainty: Heuristics and Biases; Page 24
The mean IQ of the population of eighth graders in a city is known to be 100. You have selected a random sample of 50 children for a study of educational achievements. The first child tested has an IQ of 150. What do you expect the mean IQ to be for the whole sample? The correct answer is 101.
Question: how is it 101? How to arrive at that number?
Based on the given information we know that the population mean $\mu$ is equal to $100$, i.e $\mu=100$.
Then we extract a sample $n=50$, where the first child has IQ, i.e. $X_{1}=150$. And based on that we want to know what is the expected mean $\bar{X}_{50}$.
We start by taking the expected value of the sample mean $\bar{X}_{50}$
$$\mathbb{E}[\bar{X}_{50}]= \mathbb{E}[\frac{X_{1}+X_{2}+...+X_{50}}{50}]=\mathbb{E}[\frac{X_{1}}{50}+\frac{X_{2}+X_{3}+...+X_{50}}{50}]$$
however $X_{1}$ is known so it can get outside of the expected value
$$=\frac{X_{1}}{50}+\mathbb{E}[\frac{X_{2}+X_{3}+...+X_{50}}{50}]$$
and inside the expected value we have to construct the sample mean but this time for only the $49$ values.
$$=\frac{X_{1}}{50}+\mathbb{E}[\frac{X_{2}+X_{3}+...+X_{50}}{49}]\frac{49}{50}$$
We also, know that the sample mean is an unbiased estimator of the population mean i.e. $\mathbb{E}[\bar{X}]=\mu$. Hence,
$$=\frac{X_{1}}{50}+100\frac{49}{50}=\frac{150}{50}+\frac{4900}{50}=101$$
