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Quiz on continuous random variable NAME 1. Let X be a number chosen at random be

ID: 3065384 • Letter: Q

Question

Quiz on continuous random variable NAME 1. Let X be a number chosen at random between 8 and 10 uniformly. Determine the probabilities that X is between 8.5 and 9.5 and X is between 8.5 and 8.7. What is the expectation of your choice? 2. Old Faithful erupts every 91 minutes. You arrive there at random and only wait for 20 minutes. What is the probability you will see it erupt? 3. The weights of adult are normally distributed with a mean of 173 lb and a standard deviation of 29 Ib. a. What is the probability that one random selected adult will weigh more than 1901b? b. What is the probability that 25 randomly selected adult will have a average weight of more than 190 lb? Suppose the reaction times of teenage drivers are normally distributed with a mean of 0.53 seconds and a standard deviation of 0.11 seconds. 4. a. What is the probability that a teenage driver chosen at random will have a reaction b. Find the probability that a teenage driver chosen at random will have a reaction c. What is the probability that a teenage driver chosen at random will have a reaction time less than 0.65 seconds? time between 0.4 and 0.6 seconds. time greater than 0.8 seconds?

Explanation / Answer

1) P(8.5 < X < 9.5) = (9.5 - 8.5)/(10 - 8) = 0.5

P(8.5 < X < 8.7) = (8.7 - 8.5)/(10 - 8) = 0.2/2 = 0.1

3) a) P(X > 190) = P((X - mean)/sd > (190 - mean)/sd)

                           = P(Z > (190 - 173)/29)

                           = P(Z > 0.59)

                           = 1 - P(Z < 0.59)

                           = 1 - 0.7224 = 0.2776

b) P(X > 190) = P((X - mean)/(sd/sqrt(n)) > (190 - mean)/(sd/sqrt(n)))

                           = P(Z > (190 - 173)/(29/sqrt(25)))

                           = P(Z > 2.93)

                           = 1 - P(Z < 2.93)

                           = 1 - 0.9983

                           = 0.0017

4)a) P(X < 0.65)

       = P((X - mean)/sd < (0.65 - mean)/sd)

       = P(Z < (0.65 - 0.53)/0.11)

       = P(Z < 1.09)

       = 0.8621

b) P(0.4 < X < 0.6)

= P((0.4 - mean)/sd < (X - mean)/sd < (0.6 - mean)/sd)

= P((0.4 - 0.53)/0.11 < Z < (0.6 - 0.53)/0.11)

= P(-1.18 < Z < 0.64)

= P(Z < 0.64) - P(Z < -1.18)

= 0.7389 - 0.1190

= 0.6199

c) P(X > 0.8)

       = P((X - mean)/sd > (0.8 - mean)/sd)

       = P(Z > (0.8 - 0.53)/0.11)

       = P(Z > 2.45)

       = 1 - P(Z < 2.45)

       = 1 - 0.9929

       = 0.0071