Assume the weight of a randomly chosen American passenger ca

Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 2,003 pounds to 4,906 pounds.

What is the mean weight of a randomly chosen vehicle? (Round your answer to the nearest whole number.)

What is the standard deviation of a randomly chosen vehicle? (Round your answer to 4 decimal places.)(c

What is the probability that a vehicle will weigh less than 3,060 pounds? (Round your answer to 4 decimal places

What is the probability that a vehicle will weigh more than 3,914 pounds? (Round your answer to 4 decimal places.)

What is the probability that a vehicle will weigh between 3,060 and 3,914 pounds? (Round your answer to 4 decimal places.)

Solution

A)

mean = (a+ b)/2 = (2003 + 4906)/2 = 3454.5 lbs [ANSWER]

**************

b)

standard deviation = sqrt((b-a)^2 / 12) = sqrt((4906-2003)^2/12) = 838.0239157 [ANSWER]

****************

c)

P(x<3060) = (3060-2003)/(4906-2003) = 0.364106097 [ANSWER]

***************

d)

P(x>3914) = (4906 - 3914)/(4906-2003) = 0.341715467 [ANSWER]

***************

e)

P(3060<x<3914) = (3914-3060)/(4906-2003) = 0.294178436 [ANSWER]