Mathematics 9709/63 — May/June 2022

Walkthrough

We are told that the number of characters in one email follows a normal distribution with mean 1250 and standard deviation 480. When we take a random sample of 100 emails, the sample mean is also normally distributed. Its mean is the same as the population mean, 1250, but its standard deviation is smaller: the population standard deviation divided by the square root of the sample size, 480/sqrt(100) = 48. This is why we use 48 in the denominator, not 480.

We want the probability that the sample mean exceeds 1300. To use the standard normal table, convert 1300 to a z-score: subtract the mean 1250 and divide by the standard deviation of the sample mean, 48. This gives z ≈ 1.042. Since we want "more than 1300", we need the area to the right of z = 1.042, which is 1 - Phi(1.042). Looking up 1.042 in the normal table gives about 0.8513, so the required probability is about 0.1487, or 0.149 to three significant figures.

An equivalent method uses the total number of characters in the 100 emails. The total has mean 100 × 1250 = 125000 and standard deviation 480 × sqrt(100) = 4800. Because the total is a discrete count, a continuity correction may be used: P(mean > 1300) is the same as P(total > 130000), corrected to P(total > 130000.5). This also gives z ≈ 1.042 and the same probability.

Key Takeaways

• The sample mean of a normal population is normally distributed: $\bar{X} \sim N(\mu, \sigma^2/n)$.
• The standard deviation of the sample mean is $\sigma/\sqrt{n}$, often called the standard error.
• To find probabilities about the sample mean, standardise using the standard error, not the population standard deviation.
• "More than" means a one-tailed upper-tail probability, $1 - \Phi(z)$.
• A continuity correction may be appropriate when working with the total of discrete counts, but it makes negligible difference here.

Common Mistakes

• Using 480 as the standard deviation of the sample mean instead of 48. This gives a much smaller z-score and a wrong probability.
• Using the variance $480^2$ instead of $480^2/100$ when standardising.
• Forgetting that the sample size is 100 and not using $\sqrt{100}$ in the denominator.
• Using a two-tailed probability for "more than 1300"; this question requires the upper tail only.
• Omitting the working for the z-score. The mark scheme awards M1 for the standardisation, so an unsupported final answer may not receive full credit.

Things to Be Careful About

• The normal distribution is continuous, so no continuity correction is needed when working directly with the sample mean. If using the total method, the mark scheme allows the continuity correction but also accepts omitting it.
• Keep at least three significant figures in intermediate values; rounding z too early could change the final answer.
• Check the direction of the inequality: "more than 1300" means $P(\bar{X} > 1300)$, so use $1 - \Phi(z)$.
• The mark scheme accepts either $z = \frac{1300 + \frac{1}{200} - 1250}{\frac{480}{10}}$ or $z = \frac{1300 - 1250}{\frac{480}{10}}$, both giving 1.042.