4percent of all cars produced on a monday are deficient, while 3 percent of friday's production is deficient; on all other days, only 1 percent is defective. let a car chosen randomly from the production be deficient. what is the probability that it was produced on monday? what is the solution if 12 percent of the cars are produced on friday and the remaining production is distributed equally over the other 4 working days?

Let the total numbers of cars be ten then the total cars produced in one week would be 50

Probability of deficient cars on Monday = 4/10

Probability of deficient cars on Tuesday =1/10

Probability of deficient cars on Wednesday = 1/10

Probability of deficient cars on Thursday = 1/10

Probability of deficient cars on Friday = 3/10

Probability ( deficient car)= 10/50 or 1/5

Probability that it produced on Monday 4/10*1/5= 4/50= 2/25= 8/100= 0.08 or 8%

Probability of deficient cars on Monday = 4/10* 12/100= 48/1000= 0.048or 4.8%

Probability of deficient cars on Tuesday =1/10 *22/100= 22/1000= 0.022 or 2.2%

Probability of deficient cars on Wednesday = 1/10*22/100=22/1000= 0.022 or 2.2%

Probability of deficient cars on Thursday = 1/10*22/100=22/1000= 0.022 or 2.2%

Probability of deficient cars on Friday = 3/10*22/100= 66/1000= 0.066 or 6.6%

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.

im alright what do you need

step-by-step explanation: