The Normal Distribution: Question 4

Worked solution

Setting up two equations

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

Upper boundary: 350 ml

$5\%$ of cans contain more than $350$ ml, so $P(V>350)=0.05$, i.e. $P(Z>z_1)=0.05$, so $\Phi(z_1)=0.95$.

From the standard normal (percentage points) table, $z_1=1.6449$. Since $350$ lies above the mean $\mu$: $\frac{350-\mu}{\sigma}=1.6449 \implies 350-\mu=1.6449\sigma \qquad \text{...(1)}$

Lower boundary: 306 ml

$2.5\%$ of cans contain less than $306$ ml, so $P(V<306)=0.025$. By symmetry, the corresponding upper-tail probability is also $0.025$, with $\Phi(1.9600)=0.975$, so $z_2=1.9600$. Because $306$ lies below the mean, the standardised value is negative: $\frac{306-\mu}{\sigma}=-1.9600 \implies \mu-306=1.9600\sigma \qquad \text{...(2)}$

Part (b): Solving for μ and σ

From (1): $\mu=350-1.6449\sigma$ From (2): $\mu=306+1.9600\sigma$

Setting these equal: $350-1.6449\sigma=306+1.9600\sigma$ $350-306=1.9600\sigma+1.6449\sigma$ $44=3.6049\sigma$ $\sigma=\frac{44}{3.6049}=12.2056\ldots$

So $\sigma=12.2$ ml (3 s.f.).

Substituting back into (2): $\mu=306+1.9600\times12.2056=306+23.923=329.923$

So $\mu=330$ ml (3 s.f.).

Check using equation (1): $\mu=350-1.6449\times12.2056=350-20.077=329.923$, which agrees with the value found from equation (2). ✓

Final answers

• (a) $350-\mu=\boxed{1.6449\sigma}$ and $\mu-306=\boxed{1.9600\sigma}$
• (b) $\mu=\boxed{330}$ ml (3 s.f.), $\sigma=\boxed{12.2}$ ml (3 s.f.)