QCAA Specialist Mathematics Statistical inference

As $n \ge 30$, the sample mean distribution can be assumed to be normal.
$\mu_{\bar{x}} = 64\,800, \sigma_{\bar{x}} = \frac{4500}{\sqrt{360}}$

$P(64\,000 \le \bar{X} \le 65\,000) = 0.8$

$\bar{x} = \frac{65\,811 + 64\,589}{2} = ＄65\,200$

$\mu$ lies within the confidence interval.

The website's claim is reasonable.

$\bar{x} = \frac{2321.4 + 2423.6}{2} = 2372.5 \text{ hours}$

Approximate confidence interval for $\mu$ is
$\left( \bar{x} - z \frac{\sigma}{\sqrt{n}}, \bar{x} + z \frac{\sigma}{\sqrt{n}} \right)$
Using the upper value of the 95% confidence interval
$2372.5 + 1.96 \frac{125}{\sqrt{n}} = 2423.6$

Solving equation: $n \approx 22.98$

Sample size used was 23.

Yes, the assumption is required.

Because the sample size is less than 30, the distribution of the sample mean cannot be assumed to be normal unless the population is normally distributed.

Using the property of a PDF
$\int_{-\infty}^{\infty} p(x) dx = 1$
Using $n=5$ in the given PDF
$\int_0^\infty \frac{k^5 t^4}{4!} e^{-\frac{t}{3}} dt = 1$

Solving the equation: $k = \frac{1}{3}$
Mean of distribution for waiting time until 5th call, $\mu$
$E(X) = \int_{-\infty}^{\infty} x p(x) dx$
$\mu = \int_0^\infty t \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = \int_0^\infty \frac{\left(\frac{1}{3}\right)^5 t^5}{4!} e^{-\frac{t}{3}} dt$
$= 15 \text{ minutes}$

Variance of distribution for 5th call
$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x) dx$
$= \int_0^\infty (t-15)^2 \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = 45 \text{ minutes}^2$
$\therefore \sigma = \sqrt{45} \text{ minutes}$

Consider the distribution of the sample mean of the waiting time until the 5th phone call is received, $\bar{T}$.
As the sample size is large, the distribution of $\bar{T}$ can be considered normal.
$\mu_{\bar{T}} = 15$ and $\sigma_{\bar{T}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{45}}{\sqrt{80}} = 0.75$

Using normal cdf on GDC: $P(\bar{T} > 16) \approx 0.09$