IB Maths AI SL · Paper 2

The prices fall steadily as the distance rises, so the ranks are exactly reversed.

(i) Enter the pairs into the calculator.

(ii) The value of $r$ is close to $-1$.

(i), (ii) The regression line of $y$ on $x$.

(iii) $b$ is the price when $x=0$. It represents the typical price (about $3.19 million) of an apartment in the centre of the city.

(i) Substitute $x=19.6$.

(ii) The distance 19.6 km is inside the range of the data (7 to 20.9 km), so this is interpolation, and the correlation is strong ($r\approx -0.979$).

$H_1$: the mean apartment price in $A$ is different from the mean price in $B$.

Enter the two samples into a two-sample $t$-test with the variances pooled, two-tailed.

Since $0.0224<0.05$, reject $H_0$. There is sufficient evidence that the mean price in location $A$ is different from the mean price in location $B$.

Xavie has assumed that the prices in each location come from normal distributions.

64 is the mean and the normal distribution is symmetric.

Use the normal CDF on the calculator.

(i) Sketch: a bell-shaped curve centred on 64 with the area to the left of a value $k$ (below the mean) shaded.

(ii) Use the inverse normal with area 0.3, mean 64 and standard deviation 12.

Each package is below $k$ with probability 0.3. Let $X\sim B(5,0.3)$.

The charge is $4.50 plus $2.00 for each kilogram above 1.

Substitute $x=5.3$.

Set the charge equal to $7.20.

Solve $\frac{40000}{x^2}=400$ for $x>0$.

(i) At $x=50$.

(ii) The income is the number sold times the price.

(i) Profit is the number sold times the profit per smoothie.

(ii) Differentiate $P=40000x^{-1}-800000x^{-2}$.

(iii) Set the derivative to zero and multiply by $x^3$.

(iv) Substitute $x=40$ in the model for $n$.

Since this is the only turning point and $P\to 0$ as $x\to \infty$, this is the maximum profit.

Switch $A$ fails with probability 0.1 and does not fail with 0.9. Switch $B$ fails with probability 0.3 and does not fail with 0.7.

Both fail.

No switch fails, so both do not fail.

Substitute $x=2$ into $h^{\prime }(x)=0.6x^2-1.8x$.

(i) The shaded region is under the curve between $x=0$ and $x=4$.

(ii) Evaluate.

The volume is the cross-sectional area times the width of 1.2 m.

Subtract the two gradient equations to get $2a=1.2$, so $a=0.6$. Then $b=-3.6$ from $6a+b=0$. Substitute into the first equation to find $c$.

At $x=2$ the original design has $h(2)=1.6-3.6+3=1$ and $h^{\prime }(2)=-1.2$. The new design has $h(2)=0.6(4)-3.6(2)+5.8=1$ and $h^{\prime }(2)=-1.2$. The two parts have the same height and the same gradient at $x=2$, so they join without a corner and the path is smooth.