Question

In a nation wide survey of adults, one variable measured was how many days vacationers spent driving on the road on their longest trip. Consider the following (partial) probability distribution for the random variable X= the number of days for the longest car trip.

Value of X; 4 5 6 7 8

Probability:o.10 0.20 0.25 ? ??

a) Suppose the probability of 7 days is twice as likely as the probability of 8 days. What are the two missing probabilities to complete the distribution for X?

b) What is the probability that vacationers spent 4 or 5 days driving on the road?

c) Find the Probability that vacationers spent 4 or 5 days on the road, P(4 days or 5 days).

d) Calculate the expected number of days for the longest car trip done last summer, E(X) and Interpret

Answer 1

a) Let a and b denote the probabilities corresponding to values of X =7 and X= 8 respectively. It is given that a=2b. Using the property of total probability = 1,

P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)=1

0.10+0.20+0.25+a+b= 1

0.10+0.20+0.25+2b+b = 1

0.55 + 3b = 1

3b = 0.45

b = 0.15

And a = 2 b = 2 (0.15) = 0.30

Hence, P(X=7) = 0.30 and P(X=8) = 0.15

b) P(4 or 5 days)= P(X=4) + P(X=5)

= 0.10 + 0.20

= 0.30

Hence, Probability that vacationers spent 4 or 5 days on the road = 0.30

c) P(4 days or 5 days) = 0.30

d) Expected number of days = E(X)

  

= 4(0.10) + 5(0.20) + 6 (0.25) + 7( 0.30) + 8 (0.15)

= 6.2

The above obtained expectation is nothing but the average number of days spent for the longest car trip done last summer; i.e. on average, a person spends about 6 days for the longest car trip.