A machine fills tins of paint so that the volume, V litres, in a randomly chosen tin is modelled by V N(5, 0.1^2). What is the probability that a randomly chosen tin contains less than 4.85 litres of paint?

Answer: A. Standardising V N(5,0.1^2), so =5 and =0.1. Standardise using Z=V- /: z=4.85-5/0.1=-0.15/0.1=-1.5 Reading the table P(V<4.85)=P(Z<-1.5)= (-1.5) By symmetry of the standard normal curve, (-1.5)=1- (1.5). From the standard normal table, (1.5…

The Normal Distribution: Question 6

Syllabus 5.5

Multiple choice AS 1 mark

A machine fills tins of paint so that the volume, $V$ litres, in a randomly chosen tin is modelled by $V\sim N(5, 0.1^2)$ .

What is the probability that a randomly chosen tin contains less than $4.85$ litres of paint?

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Standardising

$V\sim N(5,0.1^2)$ , so $\mu=5$ and $\sigma=0.1$ . Standardise using $Z=\dfrac{V-\mu}{\sigma}$ :

$z=\frac{4.85-5}{0.1}=\frac{-0.15}{0.1}=-1.5$

Reading the table

$P(V<4.85)=P(Z<-1.5)=\Phi(-1.5)$

By symmetry of the standard normal curve, $\Phi(-1.5)=1-\Phi(1.5)$ . From the standard normal table, $\Phi(1.5)=0.9332$ , so

$\Phi(-1.5)=1-0.9332=0.0668$

Why the other options are wrong

$0.9332=\Phi(1.5)$ is $P(V>4.85)$ , the complement of what is asked.
$0.5000=\Phi(0)$ would only be correct if $4.85$ litres were the mean.
$0.1587=\Phi(-1)$ comes from using an incorrect standard deviation of $0.15$ (giving $z=-1$ ) instead of the correct $\sigma=0.1$ .

Final answer

$P(V<4.85)=\boxed{0.0668}$ , so the correct option is A.