Assume the weight of a randomly chosen American passenger car is a uniformly dis
ID: 362432 • Letter: A
Question
Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 1,557 pounds to 4,665 pounds.
What is the standard deviation of a randomly chosen vehicle? (Round your answer to 4 decimal places.)
What is the probability that a vehicle will weigh less than 1,946 pounds? (Round your answer to 4 decimal places.)
What is the probability that a vehicle will weigh more than 4,455 pounds? (Round your answer to 4 decimal places.)
What is the probability that a vehicle will weigh between 1,946 and 4,455 pounds? (Round your answer to 4 decimal places.)
Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 1,557 pounds to 4,665 pounds.
Explanation / Answer
a. mean weight = (1749 + 4039) /2 = 2894.
b. standard deviation = 661.0661
c. less than 1946 pounds= ( 3144 - 1749) / (4039- 1749) = 0.6092
d. more than 4455 pounds = (4039 - 3602) / (4039 - 1749) = 0. 1908
e. between 1946 and 4455 pounds = ( 3602 - 3144) / (4039 - 1749) = 0.2000