A taxi driver provides service in two zones of a city. Fares picked up in zone A will have destination in zone A with probability 0.65 or in zone B with probability 0.35. Fares picked up in zone B will have destinations in zone A with probability 0.45 or in zone B with probability 0.55. The driver's expected profit for a trip entirely in zone A is $7; for a trip entirely in zone B is $8; and for a trip that involves both zones is $12. Find the taxi drivers average profit per trip.

The taxi drivers average profit per trip is $9.50.

Step-by-step explanation:

The taxi driver provides services in Zone A and Zone B.

Let = destination is in Zone A and = destination is in Zone B.

Given:

The probabilities are:

The Expected profit are:

If the trip is entirely in Zone A the expected profit is, E (A - A) = $7.

If the trip is entirely in Zone B the expected profit is, E (B - B) = $8.

If the trip involves both the zones the expected profit is,

E (A - B) = E (B - A) = $12.

Determine the expected profit earned in Zone A as follows:

Determine the expected profit earned in Zone B as follows:

The total expected profit is:

Thus, the taxi drivers average profit per trip is $9.50.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

idk

step-by-step explanation:

i don't know