Only 35% of the drivers in a particular city wear seat belts. suppose that 20 drivers are stopped at random what is the probability that exactly four are wearing a seatbelt? (round your answer to 4 decimal places)

Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.

The drivers stopped are assumed to be random and independent.

These conditions are suitable for modelling using he binomial distribution, where

where n=number of drivers stopped (sample size = 20)

x=number of drivers wearing seatbelts (4)

p=probability that a driver wears seatbelts (0.35), and

C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)

So the probability of finding 4 drivers wearing seatbelts out of a sample of 20

P(4;20;0.35)

=C(20,4)*(0.35)^4*(0.65)^16

= 4845*0.0150061*0.0010153

= 0.07382

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

$3720

step-by-step explanation:

taxable income = $31000

as per the 2018-2019 us tax bracket, the tax rate is 12% for an individual whose taxable income is between $9,526 to $38,700.

so, the tax to be paid by him = 12% of 31000

= 0.12 * 31000     [12% converted to decimal as 0.12]

= $3720