Normally distributed honey fills: an underweight probability, the raised mean for a 1% underweight rate, the mean and standard deviation for two tail limits, and the jars needed to repay a new machine

Jars of honey are labelled \(500\) grams. A machine fills the jars, and the masses of honey in the jars are normally distributed with mean \(508\) grams and standard deviation \(4\) grams.

1. Find the probability that a jar chosen at random contains less than \(500\) grams of honey.[2]

To reduce the proportion of jars containing less than \(500\) grams to \(1\%\), the mean of the machine is raised and the standard deviation stays the same.

2. Show that the new mean is \(509.3\) grams, correct to \(1\) decimal place.[3]

A more precise filling machine is being considered. It must fill only \(1\%\) of jars with less than \(500\) grams and only \(1\%\) of jars with more than \(512\) grams.

3. Find the mean and the standard deviation that meet these requirements.[3]

The precise machine costs \(3960\) dollars, and the honey is sold at \(20\) dollars per kilogram.

4. The original machine runs with a mean of \(509.3\) grams. Compared with this, find the number of jars that the precise machine must fill to save enough honey to repay its cost.[3]