Mathematics 9709/62 — May/June 2020

Walkthrough

We need the probability that the total weight of 6 plums is below 220 g. Each plum has mass $N(40.4, 5.2^2)$, and the plums are chosen independently.

The total weight $T$ is the sum of 6 independent normal random variables. A key fact is that the sum of independent normal variables is also normal. Its mean is the sum of the means, so $6 \times 40.4 = 242.4$. Its variance is the sum of the variances, so $6 \times 5.2^2 = 162.24$. Notice that we do not multiply the standard deviation by 6; we multiply the variance by 6, then take the square root if needed.

Now that we know $T \sim N(242.4, 162.24)$, we standardise the value 220. The $z$-score is

The negative sign shows that 220 is below the mean. The normal table usually gives probabilities for positive $z$, so we use symmetry:

This gives approximately 0.0393, or 3.93%.

Key Takeaways

• The sum of independent normal random variables is normal.
• For a sum of $n$ independent variables each with mean $\mu$ and variance $\sigma^2$, the total has mean $n\mu$ and variance $n\sigma^2$.
• To find a normal probability, standardise using $z = \frac{x - \mu}{\sigma}$, then use the standard normal table.
• For negative $z$-values, use the symmetry $\Phi(-z) = 1 - \Phi(z)$.

Common Mistakes

• Using $6 \times 5.2 = 31.2$ as the standard deviation of the total. The correct standard deviation is $\sqrt{6 \times 5.2^2}$, not $6 \times 5.2$.
• Forgetting to square $5.2$ before multiplying by 6.
• Treating the total as if it were a single plum rather than a sum of 6 plums.
• Reading the normal table incorrectly for a negative $z$-value; remember to subtract from 1.

Things to Be Careful About

• The variance of a sum of independent variables is the sum of their variances; the standard deviation is the square root of that sum.
• Since the normal distribution is continuous, $T < 220$ and $T \leq 220$ give the same probability, so no continuity correction is needed here.
• The final answer must be given as a percentage: $3.93\%$.