SCSA Mathematics Methods Continuous random variables and the normal distribution

The expected value of a uniformly distributed continuous random variable $X$ is midway between the maximum and minimum values, so the probability density function is

$f(x)=\begin{cases} \dfrac{1}{6}, & 3\le x\le 9 \\ 0, & \text{otherwise} \end{cases}$

The variance of $X$ is given by

$\begin{aligned} \operatorname{Var}(X)&=\int_{3}^{9} (x-6)^2 f(x)\,dx\\ &=\dfrac{1}{6}\int_{3}^{9} (x-6)^2\,dx\\ &=\dfrac{1}{6}\left[\dfrac{(x-6)^3}{3}\right]_{3}^{9}\\ &=\dfrac{1}{6}\left(9-(-9)\right)\\ &=3 \end{aligned}$

From the question $np=\dfrac{1}{2}$ and $np(1-p)=\dfrac{5}{12}$. It follows that

$\begin{aligned} \dfrac{1}{2}(1-p)&=\dfrac{5}{12}\\ \Rightarrow 1-p&=\dfrac{5}{6}\\ \Rightarrow p&=\dfrac{1}{6} \end{aligned}$

And

$\begin{aligned} \dfrac{n}{6}&=\dfrac{1}{2}\\ \Rightarrow n&=3 \end{aligned}$

Hence,

$\begin{aligned} P(W=1)&=\binom{3}{1}\left(\dfrac{1}{6}\right)^1\left(\dfrac{5}{6}\right)^2\\ &=3\times\dfrac{1}{6}\times\dfrac{25}{36}\\ &=\dfrac{25}{72} \end{aligned}$

The area under the curve must be equal to 1. Hence
$4 \times 0.1 + 0.5 \times 2 \times (k - 0.1) = 1$
$\Rightarrow k + 0.3 = 1$
$\Rightarrow k = 0.7$

The probability density function for $B$ is given by
$f(b) = \begin{cases} 0.1, & \bold{1 \le b < 3} \\ \bold{0.3b - 0.8}, & 3 \le b \le 5 \\ 0, & \text{otherwise} \end{cases}$

$E(B) = \int_{1}^{3} 0.1b \ db + \int_{3}^{5} b(0.3b - 0.8) \ db$
$= 3.8$

Therefore, the expected pickup time is 3:48 pm.

$P(A \le 3) = \frac{10(3) - 3^2 - 9}{16}$
$= 0.75$

$P(3 \le A \le 4) = P(A \le 4) - P(A \le 3)$
$= \frac{10(4) - 4^2 - 9}{16} - \frac{10(3) - 3^2 - 9}{16}$
$= \frac{15}{16} - \frac{12}{16}$
$= \frac{3}{16}$
$= 0.1875$

The probability density function is given by
$p(a) = \frac{d}{da}\left( \frac{10a - a^2 - 9}{16} \right)$
$= \frac{5}{8} - \frac{a}{8}$

for $1 \le a \le 5$ (0 otherwise). Hence the expected value is given by
$E(A) = \int_{1}^{5} a\left( \frac{5}{8} - \frac{a}{8} \right) da$
$= \frac{7}{3} \left( = 2\frac{1}{3} \right)$

Therefore, the expected pickup time is 2:20 pm.

We are given that $X \sim \text{N}(50, 10^2)$. Hence

$\text{P}(X \ge 64) = 0.0808$

and so, we expect 8.08% of Australian 15-year-old students to obtain a score of at least 64.

Solving for $\text{P}(X \ge x) = 0.01$ yields $x = 73.26\dots$. Hence the minimum score required for a student to be in the top 1% is 73.27 (or 74 if reporting only integer scores).

$\text{P}(43 \le X \le 57) = 0.5161$

The sample proportion $\hat{p}$ of students classified as 'average' has a mean of

$\mu = p = 0.5161$

and standard deviation

$\sigma = \sqrt{\frac{0.5161(1 - 0.5161)}{50}} = 0.0707$

Hence, we use the approximate distribution $\hat{p} \sim \text{N}(0.5161, 0.0707^2)$. The probability that $\hat{p}$ is no more than 0.46 given by

$\text{P}(\hat{p} < 0.46) = 0.2137$

Let $Z \sim \text{N}(0, 1)$. Given that

$\text{P}(Y \le 82) = \text{P}(Z \le -1.2) = 0.1151$

and

$\text{P}(Y \ge 130) = \text{P}(Z \ge 2) = 0.0228$

It follows that

$-1.2 = \frac{82 - \mu}{\sigma}$ and $2 = \frac{130 - \mu}{\sigma}$

where $\mu$ and $\sigma$ are the mean and standard deviation of $Y$ respectively. Solving these two equations yields

$\mu = 100$

$\sigma = 15$

Given that $Y = aX + b$ it follows from the means and standard deviations of $X$ and $Y$ that

$100 = 50a + b$

$15 = 10a$

Hence $a = 1.5$ and $b = 25$.