Question

A survey of cars on a certain stretch of highway during morning commute hours showed that 70% had only one occupant, 15% had 2, 10% had 3, 3% had 4, and 2% had 5. Let Xrepresent the number of occupants in a randomly chosen car.

a. Find the probability mass function of X.

b. Find P(X ≤ 2).

c. Find P(X > 3).

d. Find μX.

e. Find σX

Answer 1

a) X 1 2 3 4 5

P(X) 0.7 0.15 0.10 0.03 0.02

b)
`P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85`

c)
`P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05`

d)
`E(X) = \sum_(i=1)^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52`

e)
`E(X^2) = \sum_(i=1)^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18`

`Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996`

`\sigma= √(Var(X))= √(0.8996)= 0.933`

Explanation:

Part a

From the information given we define the probability distribution like this:

X 1 2 3 4 5

P(X) 0.7 0.15 0.10 0.03 0.02

And we see that the sum of the probabilities is 1 so then we have a probability distribution

Part b

We want to find this probability:

`P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85`

Part c

We want to find this probability
`P(X>3)`

And for this case we can use the complement rule and we got:

`P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05`

Part d

We can find the expected value with this formula:

`E(X) = \sum_(i=1)^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52`

Part e

For this case we need to find first the second moment given by:

`E(X^2) = \sum_(i=1)^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18`

And we can find the variance with the following formula:

`Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996`

And we can find the deviation taking the square root of the variance:

`\sigma= √(Var(X))= √(0.8996)= 0.933`