IB Maths AA SL · Paper 2

1. Enter the values in one list and the frequencies in another, then run one-variable statistics on the GDC.
2. There are $60$ students, so the median is the mean of the $30$th and $31$st values, both of which are $3$.
3. (iii) Use the population standard deviation, ${\sigma }_x$.

1. Read the five-number summary from the same GDC output: minimum, lower quartile, median, upper quartile, maximum.
2. Draw a box from $2$ to $4$ with a line at $3$, and whiskers out to $0$ and $7$.

Another way. Compare $\mathrm{P}(G\mid H)=\frac{30}{70}=\frac{3}{7}$ with $\mathrm{P}(G)=0.6$; they differ, so the events are not independent.

1. Draw a line from $B$ parallel to $DC$ meeting $AD$ at a point we will call $E$. Then $EBCD$ is a rectangle, so $ED=BC$ and $EB=40$.
2. In the right-angled triangle $BCD$: $BC=40\tan 10^{\circ }$.
3. Angle $EBD=10^{\circ }$ (alternate angles with $DC$), so angle $ABE=32^{\circ }-10^{\circ }=22^{\circ }$.
4. In the right-angled triangle $ABE$: $AE=40\tan 22^{\circ }$.
5. Add the two pieces: $AD=AE+ED$.

Another way. Find $BD=\frac{40}{\cos 10^{\circ }}=40.617\dots$, note that in triangle $ABD$ the angle at $D$ is $80^{\circ }$ and at $A$ is $68^{\circ }$, then use the sine rule: $AD=\frac{BD\sin 32^{\circ }}{\sin 68^{\circ }}$.

1. A closed cylinder has two circular ends plus the curved side.
2. Substitute your expression for $h$ from (a) and simplify.

1. Graph $y=2\pi x^2+\frac{1000}{x}$ for $x>0$ and use the GDC's minimum feature.
2. The $x$-coordinate of the minimum is $r$; the $y$-coordinate is the least area.

1. Distance is the integral of speed, $|v|$, not of $v$.
2. The velocity changes sign at $t=1.857\dots$, so integrating $v$ alone would let the two pieces cancel.

1. Acceleration is $\frac{\mathrm{d}v}{\mathrm{d}t}$. Differentiate, or graph the derivative on the GDC.
2. Solve $a=19.3$ for $t\ge 0$.

1. For $\mathrm{B}(n,p)$ the mean is $np$ and the standard deviation is $\sqrt{np(1-p)}$.
2. Set the mean equal to three times the standard deviation, square both sides and solve, using $p\ne 0$.

1. Use $\mathrm{P}(A\mid B)=\frac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}$. Since $9>6$, the event $X=9$ already sits inside $X>6$.
2. $X>6$ means $X\ge 7$, so the denominator is $1-\mathrm{P}(X\le 6)$.

1. Graph $y=x^2+99x$ and $y=750(1-0.6^x)$ and find where the first overtakes the second, or tabulate both for whole $n$.
2. $n$ counts terms, so round the crossing point up to the next whole number.

1. Let $X\sim \mathrm{N}(173,21^2)$ and use normal cdf for (i) and (ii).
2. (iii) Conditional probability: the event in (ii) lies inside $X>180$, so divide (ii) by $\mathrm{P}(X>180)$, which is $1$ minus (i).

1. Add the two given probabilities to get $\mathrm{P}(Y<180)$.
2. Use inverse normal on the standard normal to get the $z$-value for each boundary.
3. Write $z=\frac{x-\mu }{\sigma }$ for $170$ and $180$ and solve the pair of equations.

1. Use the complement: at least one over $180$ g is everything except both under $180$ g.
2. The two potatoes come from separate bags, so multiply the two probabilities.

Another way. Add the three separate cases: red only over, purple only over, and both over, using $0.369441\dots$ and $0.14099$.

1. Graph on the GDC over the given domain and describe it.
2. The curve starts at the origin, rises to a single maximum, then falls back to the $x$-axis at $x=\frac{\pi }{2}$.
3. The hump leans to the right because ${\mathrm{e}}^x$ grows, so the maximum is past the midpoint $x=0.785$.

1. The horizontal line is $y=1.27189\dots$. Find where it meets the curve again, to the right of $P$.
2. Between the two meeting points the curve lies above the line, so integrate curve minus line.

1. The curve is above the $x$-axis on the domain, so the area from $0$ to $k$ is just the integral.
2. Set it equal to twice (d) and solve for $k$ on the GDC.