I got this question as home work but i am not able to understand how I will update distribution using Bayes' rule. I want to a direction to proceed?
Let X be the score of a randomly selected student in this class on the quiz. Let Y denote the mean score. If we knew Y , suppose that a reasonable model for X would be that X is normal with mean Y and variance 10.
When I write the quiz, I do not have a very good idea of what the mean Y will be. I think the mean is equally likely to be anywhere between 60 and 90, i.e.,
Suppose that I randomly select a quiz, grade it, and it turns out to receive a 100. Using Bayes rule, how should I update my distribution for Y ? That is, what is fY |X(y|X = 100)? (Simplify as much as possible, but you may leave your answer in terms of the standard normal cumulative distribution function Φ if you wish.)
According to Bayes Rule we have
$$f_{Y|X}(y|x) = \frac{f_{X|Y}(x|y) f_Y(y)}{f_X(x)}$$
In a Bayesian context, the left hand side is known as the posterior distribution of $Y$ given the observed data $X$ (in your case $X=x = 100$), and the prior distribution $f_Y$.
You have given the formulae for $f_{X|Y}, \, f_Y$, however we do not know the formula for $f_X$. This, however, is okay, because $x = 100$ is known, $f_X(100)$ is a constant. i.e. we have
$$f_{Y|X}(y|100) \propto f_{X|Y}(100|y) f_Y(y)$$
Further, we know that $f_{Y|X}(y|100)$ is a probability density function, so we know that the normalizing constant has to be $\int f_{X|Y}(100|y) f_Y(y)dy$.
Often you will not have to compute the normalizing constant, however in this case the form of the prior distribution $f_Y(y)$ will require you to compute this; but as hinted in the question, this can be done as a standard integral of a Normal distribution, and hence returned in terms of $\Phi$.
Hopefully that is enough of a hint.