Mathematics 9709/62 — October/November 2013

Walkthrough

We are told that $X \sim \mathrm{N}(1.5, 3.2^2)$, so the mean is $\mu = 1.5$ and the standard deviation is $\sigma = 3.2$. The question asks for $P(X < -2.4)$.

The standard normal distribution table is in terms of $z$-scores, so the first step is to standardise the value $-2.4$:

This converts the original question into $P(Z < -1.219)$. Most standard normal tables give cumulative probabilities for positive $z$-values, so use the symmetry of the normal curve: the area to the left of $-1.219$ is exactly the same as the area to the right of $1.219$. The table gives $P(Z < 1.219) = 0.8886$, so the area to the right is $1 - 0.8886 = 0.1114$. Therefore $P(X < -2.4) = 0.111$ to 3 significant figures.

Key Takeaways

This question tests standardisation of a normal random variable, reading a standard normal table, and using symmetry to handle negative $z$-scores. A student should understand that all normal probability questions can be reduced to the standard normal distribution before using table values.

Common Mistakes

• Forgetting to standardise and looking up $-2.4$ directly in the normal table.
• Using the variance $3.2^2$ as the standard deviation and dividing by $3.2^2$.
• Subtracting from 1 incorrectly and giving $0.8886$ instead of the lower-tail probability.
• Not rounding the final answer to the required accuracy; the mark scheme requires an answer rounding to $0.111$.

Things to Be Careful About

• The notation $\mathrm{N}(1.5, 3.2^2)$ means the variance is $3.2^2$, so the standard deviation is $3.2$.
• Negative $z$-values correspond to left-tail probabilities; use the symmetry identity $P(Z < -a) = 1 - P(Z < a)$ for $a > 0$.
• Use enough decimal places in the $z$-score, but make sure the final answer matches the required rounding ($0.111$).