Calculate the upper quartile of car speeds and the number of cars exceeding 35 km/h using the provided cumulative frequency data.

The total number of cars is 200. The upper quartile (Q3) corresponds to the $75^{\text{th}}$ percentile of the data.

$$\text{Upper quartile position} = 0.75 \times 200 = 150$$

From the cumulative frequency table (from the previous step):

• At $v \le 45$, cumulative frequency is 112.
• At $v \le 50$, cumulative frequency is 170. The $150^{\text{th}}$ car's speed lies between 45 km/h and 50 km/h. We estimate this value by interpolating between these points. The range of cumulative frequency is $170 - 112 = 58$. The range of speed is $50 - 45 = 5$. We need to find the speed corresponding to a cumulative frequency of 150. This is $150 - 112 = 38$ units above the cumulative frequency at 45 km/h.

$$\text{Estimated upper quartile speed} = 45 + \left(\frac{38}{58}\right) \times (50 - 45)$$ $$= 45 + \left(\frac{38}{58}\right) \times 5$$ $$= 45 + \frac{190}{58}$$ $$\approx 45 + 3.276$$ $$\approx 48.3 \text{ km/h}$$

First, find the cumulative frequency at 35 km/h from the diagram.
From the cumulative frequency table:

• At $v \le 20$, cumulative frequency is 16.
• At $v \le 40$, cumulative frequency is 50. The speed 35 km/h lies between 20 km/h and 40 km/h. We estimate the cumulative frequency by interpolating. The range of speed is $40 - 20 = 20$. The range of cumulative frequency is $50 - 16 = 34$. We need to find the cumulative frequency at 35 km/h. This is $35 - 20 = 15$ units above 20 km/h.

$$\text{Estimated cumulative frequency at } 35 \text{ km/h} = 16 + \left(\frac{15}{20}\right) \times (50 - 16)$$ $$= 16 + 0.75 \times 34$$ $$= 16 + 25.5$$ $$= 41.5$$

This means approximately 41.5 cars have a speed less than or equal to 35 km/h. The number of cars with a speed greater than 35 km/h is the total number of cars minus the number of cars with speed less than or equal to 35 km/h.

$$\text{Number of cars with speed } > 35 \text{ km/h} = 200 - 41.5 = 158.5$$

Since the number of cars must be an integer, we round this to the nearest whole number.

$$\approx 159 \text{ cars}$$

First, determine the number of cars with a speed greater than 50 km/h.
From the cumulative frequency table, the cumulative frequency for $v \le 50$ is 170.
So, the number of cars with speed $> 50$ km/h is $200 - 170 = 30$.
The probability that the first car chosen has a speed greater than 50 km/h is:

$$P(\text{1st car} > 50) = \frac{\text{Number of cars with speed } > 50}{\text{Total number of cars}} = \frac{30}{200}$$

After choosing one such car, there are 29 cars left with a speed greater than 50 km/h, and 199 total cars remaining.
The probability that the second car chosen also has a speed greater than 50 km/h (given the first did) is:

$$P(\text{2nd car} > 50 \mid \text{1st car} > 50) = \frac{29}{199}$$

The probability that both cars have a speed greater than 50 km/h is the product of these probabilities:

$$P(\text{both } > 50) = \frac{30}{200} \times \frac{29}{199}$$ $$P(\text{both } > 50) = \frac{870}{39800}$$ $$P(\text{both } > 50) = \frac{87}{3980}$$ $$P(\text{both } > 50) \approx 0.021859...$$

Rounding to 3 significant figures:

$$P(\text{both } > 50) \approx 0.0219$$

b)

iii)

a) The upper quartile is $\boxed{48.3 \text{ km/h}}$. (Estimated from interpolation)

b) The number of cars with a speed greater than 35 km/h is $\boxed{159 \text{ cars}}$. (Estimated from interpolation)

c) The probability that they both have a speed greater than 50 km/h is $\boxed{0.0219}$.