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

Answer:

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

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

$\sigma= \sqrt{Var(X)}= \sqrt{0.8996}= 0.933$

Step-by-step 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:

$\sigma= \sqrt{Var(X)}= \sqrt{0.8996}= 0.933$