Mathematics 9709/63 — May/June 2019

Walkthrough

We are told that the crossing time $X$ is normally distributed with mean $\mu = 85$ and standard deviation $\sigma = 6.8$. The question asks for the probability that a randomly chosen crossing takes between 79 and 91 minutes.

The normal table gives probabilities for the standard normal distribution $Z \sim N(0,1)$, so we first convert the endpoints 79 and 91 into $z$-scores:

$$z_1 = \frac{79 - 85}{6.8} = -0.8824, \quad z_2 = \frac{91 - 85}{6.8} = 0.8824$$

Because the two $z$-scores are equal and opposite, the required area is symmetric about the mean. From the table, $\Phi(0.8824) = 0.8111$, so the area above $0.8824$ is $1 - 0.8111 = 0.1889$. By symmetry the area below $-0.8824$ is also $0.1889$. Therefore the middle area is:

$$0.8111 - 0.1889 = 0.6222$$

which rounds to $0.622$.

Key Takeaways

• A normal probability is an area under the normal curve.
• Standardisation converts any normal distribution to $N(0,1)$ using $Z = \frac{X - \mu}{\sigma}$.
• Symmetry can simplify probabilities between equal and opposite $z$-values.

Common Mistakes

• Forgetting to standardise before using the normal table.
• Using the variance $6.8^2$ instead of the standard deviation $6.8$ in the $z$-score formula.
• Applying a continuity correction; this is not a binomial approximation, so no continuity correction is needed.

Things to Be Careful About

• Keep the sign of the lower $z$-score negative.
• Use the cumulative probability $\Phi(z)$ correctly: $\mathrm{P}(Z < -0.8824) = 1 - \Phi(0.8824)$.
• Round the final probability to 3 significant figures as shown in the mark scheme.

Walkthrough

We need the time $t$ such that 96% of crossings take longer than $t$. In probability notation, $\mathrm{P}(X > t) = 0.96$.

Since the total probability is 1, this is equivalent to $\mathrm{P}(X < t) = 0.04$. This tells us that $t$ is below the mean, so the corresponding $z$-value must be negative. Looking up the lower-tail probability 0.04 in the standard normal table gives $z = -1.751$.

Now use the standardisation formula:

$$-1.751 = \frac{t - 85}{6.8}$$

Multiply both sides by 6.8:

$$t - 85 = 6.8 \times (-1.751) = -11.9068$$

Therefore:

$$t = 85 - 11.9068 = 73.0932$$

So $t = 73.1$ minutes, correct to 3 significant figures.

Key Takeaways

• A statement like '96% are longer than $t$' means a right-tail probability $\mathrm{P}(X > t) = 0.96$.
• Convert a right-tail probability to a lower-tail probability before using the table.
• Inverse normal problems require solving the standardisation equation for the unknown value.

Common Mistakes

• Using $z = +1.751$ instead of $-1.751$; this would give a time above the mean, which contradicts '96% are longer'.
• Using the variance instead of the standard deviation in the equation.
• Looking up 0.96 directly as the lower-tail probability without realising that 0.96 is the upper-tail probability.

Things to Be Careful About

• The mark scheme accepts $\pm 1.751$ seen, but the sign in the equation must be negative to get $t$ below the mean.
• Keep the standard deviation as 6.8, not $6.8^2$.
• Round the final answer to 3 significant figures: 73.1 minutes.