Question

A bottler of drinking water fills plastic bottles with a mean volume of 999 mL and standard deviation 4 mL. The fill volumes are normally distributed.

What proportion of bottles have volumes between 991 mL and 997 mL? (anyone also know the command and way to solve this on the TI-84 calculator to do this?)

Answer 1

Let X be the random variable denoting the fill volume of drinking water.

X ~ N(999, 4) i.e. (X - 999)/4 ~ N(0,1)

Thus, P(991 < X < 997) = P[(991 - 999)/4 < (X - 999)/4 < (997 - 999)/4] = P[-2 < (X - 999)/4 < -0.5] = (-0.5) - (-2) = 0.3085 - 0.0228 = 0.2857

So, 28.57% of bottles have volumes between 991 mL and 997 mL.

In, TI-84 calculator, the cdf values corresponding to -0.5 and -2 of standard normal distribution need to be computed.