VCAA Mathematical Methods Data analysis, probability and statistics

$\frac{1}{32} = 0.03125$

$\frac{13}{16} = 0.8125$

$\text{Pr}(X \ge 2 | X < 5) = \frac{\text{Pr}(2 \le X \le 4)}{\text{Pr}(X \le 4)} = \frac{0.78125}{0.96875} = 0.806$

$\text{E}(X) = \frac{5}{2} = 2.5$, $\text{sd}(X) = \frac{\sqrt{5}}{2}$

$1$

$a = -0.8 = -\frac{4}{5}$, $b = 3.4 = \frac{17}{5}$, $c = -2.78\dot{3} = -\frac{167}{60}$

$r = -1, s = 3$

Discrete, countable

$(0.208, 0.592)$

$100$

Let $X$ be a binomial random variable representing the number of casts that caught a fish.
$p$ corresponds to a probability of a catch (success).

Blue pond:
$X \sim B\left(3, \frac{2}{3}\right)$
Red pond:
$X \sim B\left(3, \frac{1}{3}\right)$

In the blue pond, 30 points will require three successes in three casts:

In the red pond, 30 points will require at least two successes in three casts:
i.e. $X \ge 2$

In the blue pond, the required probability is $\frac{8}{27}$
In red pond, the required probability is $\frac{7}{27}$

The claim is correct. Probability of winning if casting in the blue pond is $\frac{1}{27}$ more than the probability of winning using the red pond.

$p_1 =$ proportion of Town A who prefer to drink tea
$p_2 =$ proportion of Town B who prefer to drink tea

The sample proportions are:
$\hat{p}_1 = \frac{111}{216}$
$\hat{p}_2 = \frac{150}{257}$

Using the 99% confidence interval for the difference of two proportions
$\left(\frac{111}{216} - \frac{150}{257} - 2.576\sqrt{\frac{\frac{111}{216}(1-\frac{111}{216})}{216} + \frac{\frac{150}{257}(1-\frac{150}{257})}{257}}, \frac{111}{216} - \frac{150}{257} + 2.576\sqrt{\frac{\frac{111}{216}(1-\frac{111}{216})}{216} + \frac{\frac{150}{257}(1-\frac{150}{257})}{257}}\right)$

$= (-0.188, 0.048)$

This interval contains zero; therefore, there is no evidence in the data to say that the two proportions are different, i.e. preference to drink tea does not depend on where the person lives.

Given that $p(10)=0.0135$

$0.0135=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{10-18}{\sigma}\right)^2}$

$0.0135=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{-8}{\sigma}\right)^2}$

Using a GDC:

Two solutions are obtained for the std deviation

$\sigma=4.0002$ and $28.4020$

Reject 28.4020 as it is not a possible standard deviation
because for example three standard deviations less than
the mean would produce a negative speed or three above
would result in an impossible speed on an e-scooter
($>100\ \text{km/h}$).

Use a GDC to determine $P(X\ge 23)$ with $N\sim(18,4.00)$,

$P(X\ge 23)=0.10566$

The number of riders is:

$0.10566\times 75=7.9245$

The total fines obtained:

$7.9245\times 143=1133.20$

The expected total fines is about ＄1133, which is less
than the suggested ＄1500, so the suggestion is not
reasonable.

$\therefore k = \frac{1}{64}$

$E(T) = \frac{1}{64} \int_0^4 \left(16t^2 - t^4\right) dt$

$= \frac{1}{64} \left[ \frac{16t^3}{3} - \frac{t^5}{5} \right]_0^4$

$= \frac{1}{64} \left( \frac{1024}{3} - \frac{1024}{5} - 0 \right)$

$= \frac{1}{64} \times \frac{2048}{15}$

$= \frac{64}{30} = \frac{32}{15} = 2\frac{2}{15}$

The sample proportion of heads is given by

$\hat{p} = \frac{30}{50} = 0.6$

Hence, the 90% confidence interval is

$0.6 - 1.645\sqrt{\frac{0.6 \times 0.4}{50}} \le p \le 0.6 + 1.645\sqrt{\frac{0.6 \times 0.4}{50}}$

$0.4860 \le p \le 0.7140$

$X \sim \text{Bin}(20, 0.9)$

The expected value of $X$ is given by

$E(X) = 20 \times 0.9$

$= 18$

The variance of $X$ is given by

$Var(X) = 20 \times 0.9 \times 0.1$

$= 1.8$

If three confidence intervals did not contain the true proportion, then 17 did contain the true proportion.

$\text{P}(X = 17) = 0.1901$

$(0.04 + 0.16) \div 2 = 0.1$

$\hat{p} = 0.1$

$0.06 = 2 \times \sqrt{\frac{0.1 \times 0.9}{n}}$

$0.03 = \sqrt{\frac{0.1 \times 0.9}{n}}$

$(0.03)^2 = \frac{0.1 \times 0.9}{n}$

$(0.03)^2 = \frac{9}{100n}$

$n = 100$

Confidence interval width is halved (reduced or decreased by a factor of 2; altered by a factor of $\frac{1}{2}$).

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,

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

$\int_1^a 2x - 2 dx = 0.36$
$x^2 - 2x \Big|_1^a = 0.36$
$(a^2 - 2a) - (1 - 2) = 0.36$
$a^2 - 2a + 1 = 0.36$
$a^2 - 2a + 0.64 = 0$
$\therefore a = \frac{2\pm\sqrt{4-4\times1\times0.64}}{2}$
$\therefore a = \frac{2 \pm \sqrt{1.44}}{2}$
$\therefore a = \frac{2 \pm 1.2}{2}$
$\therefore a = 1.6$ or $0.4$
Given $1 \le x \le 2$
$\therefore a = 1.6$

The confidence interval corresponds to $(\hat{p}-E, \hat{p}+E)$, where E is the margin of error about $\hat{p}$.
$\hat{p} + E = \frac{7}{10}$
$\hat{p} - E = \frac{3}{10}$
subtracting:
$\therefore 2E = \frac{7}{10} - \frac{3}{10} = \frac{4}{10}$
$\therefore E = \frac{2}{10} = \frac{1}{5}$

$\hat{p} + E = \frac{7}{10}$
$\hat{p} + \frac{2}{10} = \frac{7}{10}$
$\hat{p} = \frac{5}{10} = \frac{1}{2}$
Upper CI value = $\hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$\frac{7}{10} = \frac{1}{2} + 2\sqrt{\frac{\frac{1}{2}\left(1-\frac{1}{2}\right)}{n}}$
$\frac{1}{5} = 2\sqrt{\frac{\frac{1}{2}\left(1-\frac{1}{2}\right)}{n}}$
$\frac{1}{5} = 2\sqrt{\frac{\frac{1}{4}}{n}}$
$\frac{1}{5} = 2\frac{\sqrt{\frac{1}{4}}}{\sqrt{n}}$
$\frac{1}{5} = 2\frac{\frac{1}{2}}{\sqrt{n}}$
$\frac{1}{5} = \frac{1}{\sqrt{n}}$
$\frac{1}{10} = \sqrt{\frac{1}{4n}}$
$\frac{1}{100} = \frac{1}{4n}$
$100 = 4n$
$n = 25$
25 people were surveyed.

$X \sim \text{Bi}\left(4, \frac{3}{5}\right)$

$\Pr(X \geq 3)$

$= \Pr(X = 3) + \Pr(X = 4)$

$= \binom{4}{3}\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right) + \left(\frac{3}{5}\right)^4$

$= \frac{297}{625}$

$\Pr(X = 2 | X \geq 1) = \frac{\Pr(X = 2) \cap \Pr(X \geq 1)}{\Pr(X \geq 1)}$

$= \frac{\Pr(X = 2)}{1 - \Pr(X = 0)}$

$= \frac{6 \times \left(\frac{3}{5}\right)^2 \times \left(\frac{2}{5}\right)^2}{\left(\frac{5}{5}\right)^4 - \left(\frac{2}{5}\right)^4}$

$= \frac{6^3}{5^4 - 2^4}$

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.