# Thread: The probability distribution, how to solve it?

1. ## The probability distribution, how to solve it?

The probability distribution of the random variable X that describes the number of cars sold by a dealer is weekly

X 0 1 2 3 4 5 6 7 8
P [X = k] p 0.04 0.04 0.11 0.30 0.23 0.10 0.05 0.03

a) What is the average number of cars sold per week?, What is the most likely?
b)What is the probability that the number of cars sold in a week is more than 2 but less than 5?, What is the maximum number of cars sold weekly with a probability of 50%?,
c) If the middle of a week the number of cars sold is less than 3,
What is the probability that at week's end the average overcome?

a) the average it will be 0 * 0.04 +0.04 * 1. .0,03 * 8 = 4.11
most likely from the histogram I read that the value 1/2 is for 4
b) add the probability histogram between 2 and 5 (excluding Fridays: 0.1 0.11 0.30 = 0.51)?
I do not know how to check the max number of cars with a probability of 50%?
c) I have no idea

2. ## Re: The probability distribution, how to solve it?

b) Find the probability of P(3)+P(4)

c) Find $\displaystyle \displaystyle P(X=4.11|X<3) = \frac{P(X=4.11\cap X<3)}{P(X<3)}$

3. ## Re: The probability distribution, how to solve it?

Thank You for help I will try now to solve this and copy my results.

What is the maximum number of cars sold weekly with a probability of 50%?,

a)
$\displaystyle average=0*0,04+1*0,04+2*0,1+3*0,11+4*0,3+5*0,23+6* 0,1+7*0,05+8*0,03=4,11$

b)
$\displaystyle P(B)=P(3)+P(4)=0,11+0,3=0,41$

c)
$\displaystyle P(X<3)= \0,04+0,04+0,1=0,18$

$\displaystyle P(X=4,11)=0,5$

how to do this?

$\displaystyle P(X=4,11|X<3)= \frac{0,5*XXX}{0,04+0,04+0,1}=$