VCAA Mathematical Methods Data analysis, probability and statistics

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.

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 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.

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

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

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.

Exclusion bias: Athletes with unexplained respiratory symptoms who do not seek medical intervention, or who do not identify themselves as athletes to their doctor, will not be represented in the sample.

The sample proportion
$\hat{p} = \frac{25}{71}$
Hence the 95% confidence interval is given by
$95\% \text{ CI} = \left( \frac{25}{71} - 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}}, \ \frac{25}{71} + 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}} \right)$
$= (0.2410, 0.4632)$

The margin of error is given by
$E = \frac{0.4632 - 0.2410}{2} = 0.1111$

The margin of error would decrease if the sample size is increased.

The margin of error would increase if the confidence level is increased.

Since the sample proportion is less than 0.5 (0.5 is the sample proportion yielding the largest margin of error), a decrease in sample proportion would move the margin of error away from being maximum. Hence the margin of error would decrease.

We require
$0.04 \ge 1.96\sqrt{\frac{0.5(1-0.5)}{n}}$
$\Rightarrow \left(\frac{0.04}{1.96}\right)^2 \ge \frac{0.25}{n}$
$\Rightarrow n \ge 0.25\left(\frac{1.96}{0.04}\right)^2$
$= 600.25$
Since $n$ must be an integer the minimum value of the sample size is $n = 601$.

Yes. The claimed American proportion is outside the confidence interval for Australian athletes, so it is possible to conclude that the proportions are different at the 95% level of confidence.

Given $\mu = 591$
Using GDC
z-score associated with 65th percentile
$= 0.38532$

$0.38532 = \frac{593 - 591}{\sigma}$

$\sigma = 5.1905$

Using GDC
z-score associated with 20% rejection region $= -0.841621$

To determine the smallest volume that will be accepted ($x$)
$-0.841621 = \frac{x - 591}{5.1905}$

$x = 587$ mL

Analytical procedure:

$\sigma=\sqrt{\dfrac{p(1-p)}{n}}$ (formula book)

$0.04=\sqrt{\dfrac{p(1-p)}{120}}$

$0.0016=\dfrac{p(1-p)}{120}$

$0.192=p(1-p)$

$0.192=p-p^2$

$p^2-p+0.192=0$

Use GDC equation solver or graph $y=0.04$ and the
square root function to find intersections:

$p=0.25921$

$=26\%$

or

$p=0.7408$

$=74\%$

Both population proportions (26% and 74%) are less than
80%.

Therefore, the bus route proposal would not be discussed
at a council meeting.