Suppose the scores of an entrance test are normally distributed. Also, suppose the test has a mean of m and a standard deviation of s. You have set the cut off as 90 marks for the test. What is the probability of a student passing the test?
E.g. For a mean of 100 and standard deviation of 10, this probability is 0.84.
However, What is the formula or method to arrive at this number?
This is a very easy, standard question related to the normal distribution. We have a normally distributed variable $X \sim N(100,10)$. Then $$ P(X>90) = 1- P(X<90) = 1- \Phi \left( \frac{90-100}{10} \right) \approx 0.841~344 $$ It was necessary to normalize the value inside the cumulative density function $\Phi$ because it is calculated for the $N(0,1)$-case.