Mathematics 9709/61 — October/November 2023

Walkthrough

The question gives a normal population with mean 410 and variance 400. Since the variance is 400, the standard deviation is $\sqrt{400} = 20$. We take a random sample of 36 values. The sample mean $\bar{X}$ is itself a random variable. Because the population is normal, $\bar{X}$ is also normally distributed, with the same mean $\mu = 410$ but a smaller variance $\sigma^2/n = 400/36$. This is why a sample mean is more tightly concentrated around the population mean than individual values.

To find $P(\bar{X} < 405)$, we standardise $\bar{X}$ to a standard normal variable $Z$. The standard error of the mean is $\sigma/\sqrt{n} = 20/\sqrt{36} = 20/6 = 10/3$. So:

We then need $P(Z < -1.5)$. Since the standard normal table usually gives $P(Z < z)$ for positive $z$, we use symmetry: $P(Z < -1.5) = 1 - P(Z < 1.5) = 1 - \Phi(1.5)$. From the table, $\Phi(1.5) = 0.9332$, so the probability is $1 - 0.9332 = 0.0668$.

Key Takeaways

• The sample mean of a normal population is normally distributed with mean $\mu$ and variance $\sigma^2/n$.
• The standard deviation of the sample mean (standard error) is $\sigma/\sqrt{n}$.
• Standardising uses $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$.
• Use symmetry of the normal distribution to handle negative $z$-values.

Common Mistakes

• Forgetting to divide the standard deviation by $\sqrt{n}$ — using $\sigma = 20$ instead of $\sigma/\sqrt{n} = 10/3$.
• Confusing variance and standard deviation: the variance is 400, so $\sigma = 20$.
• Forgetting to subtract from 1 when the $z$-value is negative, or reading the wrong tail.
• Using the wrong formula for the variance of the sample mean.

Things to Be Careful About

• The distribution given is $\text{N}(410, 400)$: the second parameter is the variance, not the standard deviation.
• The sample size is 36, so $\sqrt{36} = 6$.
• The mark scheme allows the totals method: the sum of 36 values has distribution $\text{N}(36 \times 410, 36 \times 400) = \text{N}(14760, 14400)$, and $P(\text{sum} < 36 \times 405) = P(\text{sum} < 14580)$, then standardise $\frac{14580 - 14760}{\sqrt{14400}} = \frac{-180}{120} = -1.5$. This gives the same result. Do not mix the two methods.