Mathematics 9709/63 — October/November 2014

Walkthrough

Underweight packets are those with weight less than $250\text{ g}$, and $1\%$ of packets are underweight, so the area under the normal curve to the left of $250$ is $0.01$. Since the mean is $260$, the value $250$ is below the mean, so the corresponding $z$-value must be negative. Using normal distribution tables or a calculator for the lower $1\%$ tail gives $z = -2.326$.

The standardisation formula is $z = \frac{x - \mu}{\sigma}$. Substitute $x = 250$, $\mu = 260$, and $z = -2.326$:

$$\frac{250 - 260}{\sigma} = -2.326$$

This becomes

$$\frac{-10}{\sigma} = -2.326$$

Multiplying both sides by $\sigma$ and dividing by $-2.326$ gives

$$\sigma = \frac{10}{2.326} = 4.30\text{ g} \quad (3\text{ s.f.})$$

Key Takeaways

This question tests the core skill of converting a probability condition from a normal distribution into a standardised $z$-score. It also shows how to use the inverse normal distribution to find the boundary value when a tail probability is known, and then to solve for an unknown mean or standard deviation.

Common Mistakes

A common mistake is using the positive value $2.326$ instead of $-2.326$. The boundary $250$ is below the mean $260$, so the $z$-score must be negative.

Another common mistake is failing to standardise correctly and writing $250 - 260 = -2.326$ without dividing by $\sigma$. The $z$-score is not a raw weight; it is measured in standard deviations from the mean.

Rounding too early can also cause error. The mark scheme uses $z = -2.326$ to obtain $\sigma = 4.30$.

Things to Be Careful About

The phrase "$1\%$ are underweight" means $P(X < 250) = 0.01$, which is a lower-tail probability. This determines the sign of the $z$-value.

Because the distribution is already given as normal, do not apply a continuity correction. Continuity corrections are only used when approximating a discrete distribution, such as the binomial, by a normal distribution.

Include the unit grams in the final answer: $\sigma = 4.30\text{ g}$.