Answer to Question #301506 in Statistics and Probability for insect

Question

The amount of coffee dispensed in a dispensing machine is normally distributed with a mean of 190 ml and a standard deviation of 10 ml. What is the z-value that will give an area of 0.975 on the right side of the normal curve? 

Answer 1

$\mu=190\\\sigma=10$

Let the random variable $X$ represent the amount of coffee dispensed.

We determine a value $c$ such that, $p(X\gt c)=0.975$

So,

$p(X\gt c)=p({X-\mu\over \sigma}\gt{c-\mu\over\sigma} )=p(Z\gt {c-190\over10})=0.975$

This probability is equivalent to,

$p(Z\lt {c-190\over10})=1-0.975=0.025\implies \phi({c-190\over10})=0.025$

Now,

$({c-190\over10})=Z_{0.025}=-1.96\implies c=170.4$

Thus, the Z value that leaves an area of 0.975 to the right side of the normal curve is -1.96