It is given that one in five cups of coffee contain more than 120 mg of caffeine.
It is also known that three in five cups contain more than 110 mg of caffeine.

Assume that the caffeine content of coffee is modelled by a normal distribution.
Find the mean and standard deviation of the caffeine content of coffee.

let \(X\) be the random variable “amount of caffeine content in coffee”

\({\text{P}}(X > 120) = 0.2,{\text{ P}}(X > 110) = 0.6\)     (M1)

\(( \Rightarrow {\text{P}}(X < 120) = 0.8,{\text{ P}}(X < 110) = 0.4)\)

Note:     Award M1 for at least one correct probability statement.

\(\frac{{120 - \mu }}{\sigma } = 0.84162 \ldots ,{\text{ }}\frac{{110 - \mu }}{\sigma } =  - 0.253347 \ldots \)     (M1)(A1)(A1)

Note:     Award M1 for attempt to find at least one appropriate \(z\)-value.

\(120 - \mu  = 0.84162\sigma ,{\text{ }}110 - \mu  =  - 0.253347\sigma \)

attempt to solve simultaneous equations     (M1)

\(\mu  = 112,{\text{ }}\sigma  = 9.13\)     A1

[6 marks]