Wooden lawn edging is supplied in 1.8 m length rolls. The actual length, \(X\) metres, of a roll may be modelled by a normal distribution with mean 1.81 and standard deviation 0.08 .
Determine the probability that a randomly selected roll has length:
less than 1.90 m ;
greater than 1.85 m ;
between 1.81 m and 1.85 m .
Plastic lawn edging is supplied in 9 m length rolls. The actual length, \(Y\) metres, of a roll may be modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).
An analysis of a batch of rolls, selected at random, showed that
$$\mathrm { P } ( Y < 9.25 ) = 0.88$$
Use this probability to find the value of \(z\) such that
$$9.25 - \mu = z \times \sigma$$
where \(z\) is a value of \(Z \sim \mathrm {~N} ( 0,1 )\).
Given also that
$$\mathrm { P } ( Y > 8.75 ) = 0.975$$
find values for \(\mu\) and \(\sigma\).