IB Maths AI HL · Paper 2

1. Work out the expected prize by multiplying each prize by its probability and adding.
2. A game is fair when the expected prize equals the cost of playing, so compare with the 1-dollar cost (or find the expected gain).

1. She did not reject $H_0$, so the possible mistake is keeping a null hypothesis that is actually false: a Type II error.
2. Collecting a bigger sample (or raising the significance level) makes this error less likely.

1. Each month $200$ from Kelley plus $200$ matched goes in, so the payment is $400$.
2. Use the GDC finance solver with $40\times 12=480$ monthly periods and solve for the future value.

1. Find the value after $10$ years ($120$ periods).
2. Subtract the $50000$ withdrawal to get the new starting value.
3. Run the solver again for the remaining $30$ years ($360$ periods) from that present value, still paying $400$ a month.
4. Subtract this new total from the answer to (b).

1. Use the solver with the $30$-year set-up from (d), set the future value to $2000000$ and solve for the monthly payment.
2. Take away the current $400$ a month and round up, because the target is a minimum.

1. $b$ comes from the period: $b=\frac{2\pi }{12.5}$ in radians (or $\frac{360}{12.5}$ in degrees).
2. $d$ is the midpoint of the minimum and maximum depths; $|a|$ is half their difference.
3. A plain cosine starts at a maximum, but this model starts at a minimum, so $a$ is negative.

1. Time to reach the port is distance divided by speed.
2. Check the depth at that time: it is below $53$ feet, so the ship must wait.
3. Find the next time after that when $w(t)=53$ with the depth rising, which is two periods after the answer to (c).
4. Subtract to get the wait.

Another way. Graph $w(t)$ and the line $w=53$ on the GDC and read off the first intersection after $t=26.19$.

1. The hands meet after the total of all these times, which is an infinite geometric series with first term $50$ and $r=\frac{1}{12}$.
2. Subtract the $54$ minutes and convert the fraction of a minute into seconds.

1. Enter the values and frequencies as two lists on the GDC and run one-variable statistics ($n=168$).
2. The best estimate of the population variance uses the sample standard deviation $s_{n-1}$; square it.

1. Nine hours contain $36$ intervals of $15$ minutes, so the daily count follows $X\sim \mathrm{Po}(2\times 36)=\mathrm{Po}(72)$.
2. (i) Use the Poisson probability function on the GDC.
3. (ii) At least $60$ means $1-P(X\le 59)$.

1. Over a school day, under $H_0$ the count follows $\mathrm{Po}(72)$. The test is lower-tailed, so look for the largest $c$ with $P(X\le c)\le 0.05$.
2. (i) Try values near the answer on the GDC.
3. (ii) Compare the observed $59$ with the critical region and conclude in context.

1. Set the two position vectors equal and write one equation per component.
2. Solve any two of them simultaneously, then check the third.
3. Substitute back into either line.

1. Use the direction vectors only, not the position vectors.
2. Apply $\cos \theta =\frac{\mathbf{u}\cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$; if the angle is obtuse, take it from $180^{\circ }$.

Another way. Take the absolute value of the dot product at the start, which gives the acute angle straight away.

1. Set the equation of $l_2$ equal to the coordinates of $B$ and solve one component for $\mu$.
2. Check that the same $\mu$ works in the other two components.

1. Write a general point on $l_1$ and form the vector $\overrightarrow{CB}$ from it to $B$.
2. At the closest point, $\overrightarrow{CB}$ is perpendicular to $l_1$, so its dot product with the direction of $l_1$ is zero.
3. Expand, solve for $\lambda$ and substitute back into $l_1$.

Another way. Write $|\overrightarrow{CB}{|}^2=14{\lambda }^2+104\lambda +712$ and minimise it, by differentiating or with the GDC.

Another way. Use the quotient rule with a constant numerator; the derivative of the denominator is $-0.34068{\mathrm{e}}^{-0.051t}$.