Mathematics 9709/61 — May/June 2014

Walkthrough

We are told the petrol consumption has a normal distribution with mean 24 km/litre and standard deviation 4.7 km/litre, and we want the probability that a randomly chosen car has consumption between 21.6 and 28.7 km/litre.

Because every normal distribution with different mean or standard deviation has its own curve, we first convert to the standard normal variable $Z$ using

$$z = \frac{x - \mu}{\sigma}$$

For the lower value $x = 21.6$:

$$z_1 = \frac{21.6 - 24}{4.7} = -0.5106$$

For the upper value $x = 28.7$:

$$z_2 = \frac{28.7 - 24}{4.7} = 1$$

The probability we want is therefore $P(-0.5106 < Z < 1)$.

The standard normal table gives cumulative probabilities $P(Z < z) = \Phi(z)$. Reading the table gives $\Phi(1) = 0.8413$, the area under the curve to the left of $z = 1$. For the negative value we use symmetry: the area to the left of $-0.5106$ equals the area to the right of $0.5106$, which is $1 - \Phi(0.5106)$. From the table $\Phi(0.5106) \approx 0.6953$, so

$$\Phi(-0.5106) = 1 - 0.6953 = 0.3047$$

The probability that $Z$ lies between $-0.5106$ and $1$ is the area between them, found by subtracting the smaller cumulative area from the larger:

$$0.8413 - 0.3047 = 0.5366$$

So the required probability is $0.537$ to three significant figures.

Key Takeaways

This question tests the central method of the normal distribution unit: standardising a normal variable and using the standard normal table. The key ideas are: (1) convert any normal variable to the standard normal so one table works for all problems; (2) use the symmetry $\Phi(-z) = 1 - \Phi(z)$ to handle negative Z-scores; (3) recognise that the probability of being between two values is the difference of the two cumulative probabilities. It also reinforces that no continuity correction is needed for a directly normal distribution.

Common Mistakes

• Applying a continuity correction. The mark scheme explicitly says "no cc" — a continuity correction only belongs to normal approximations of a binomial distribution.
• Treating the standard deviation as a variance, e.g. squaring $\sigma$ or taking a square root. We are already given $\sigma = 4.7$, so the mark scheme warns "no sq rt".
• Forgetting to turn $\Phi(-0.5106)$ into $1 - \Phi(0.5106)$ and instead looking up a negative Z-value as if it were positive.
• Subtracting in the wrong direction, which would give a negative probability.
• Rounding too early and getting a final answer that is visibly off.

Things to Be Careful About

• Read the table to a sensible precision: the mark scheme accepts one rounding to $\Phi(0.841)$ or $\Phi(0.695)$.
• Keep the correct sign on $z_1 = -0.5106$; a sign error produces a probability greater than 1.
• Because the normal distribution is continuous, $P(21.6 < X < 28.7) = P(21.6 \le X \le 28.7)$; the endpoints contribute zero probability.
• Round only at the end: $0.5366$ rounds to $0.537$ (3 s.f.).