Shower-cleaner liquid is sold in spray bottles. The volume of liquid in a bottle may be modelled by a normal distribution with mean 955 ml and a standard deviation of 5 ml. Determine the probability that the volume in a particular bottle is:

a) at most 960 ml

Let X be the volume of liquid in a particular bottle, then we want to find P(X <= 960). We first need to transform from X to Z (to the standard normal distribution) so that we can use the tables.

Z = (X - mu)/sigma, where mu is the mean and sigma is the standard deviation.

Therefore:

P(X <= 960) = P(Z <= (960 - mu)/sigma) = P(Z <= (960 - 955)/5) = P(Z <= 1)

If we look this up in the cumulative table, we will find that P(Z<=1) = 0.84134.

b) more than 946ml

We want to find P(X > 946), and again we need to transform this to the standard normal distribution.

P(X > 946) = P(Z > (946 - 955)/5) = P(Z > - 1.8) = P(Z <= 1.8)

If we look this up in the cumulative table, we find P(Z <= 1.8) = 0.96407.