Q.The marks scored by a large batch of students in an examination are normally distributed with a mean of 60 and a standard deviation of 10. Find the percentage of students scoring more than 75 marks. (Area between Z = 0 and Z = 1.50 is 0.4332.)

Any normally distributed variable $X$ with mean $\mu$ and standard deviation $\sigma$ can be converted to the standard normal variate $Z = \dfrac{X-\mu}{\sigma}$, which always has mean $0$ and standard deviation $1$. This lets a single, universal standard normal table — listing the area under the curve between $Z=0$ and any positive $Z$ — answer probability questions for every normal distribution, regardless of its own $\mu$ and $\sigma$. Because the curve is symmetric, the table area for $-z$ equals the area for $+z$, and the total area on either side of the mean is $0.5$. Reading a probability from a business problem is a three-step process: con …