Originally Posted by

**CaptainBlack** This bit allows you to determin the standard deviation. 0.9332 is the

probability of observing a z of less than 1.5 from a standard normally

distributed random variable. So the z score corresponding to a mark

of 85 is:

$\displaystyle z = \frac{85-70}{\sigma} = 1.5$

so:

$\displaystyle \sigma=10$

The probability p that at least 1 score more than 80 is 1 minus the probability

that all score less than 80. So if Q(80) is the probability that one score less

than 80, then:

$\displaystyle

p = 1-[Q(80)]^4

$

RonL