Assume the weight of a randomly chosen American passenger car is a uniformly dis
ID: 3359155 • Letter: A
Question
Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 1,516 pounds to 4,661 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 1,753 pounds? (Round your answer to 4 decimal places.)
What is the probability that a vehicle will weigh more than 3,666 pounds? (Round your answer to 4 decimal places.)
What is the probability that a vehicle will weigh between 1,753 and 3,666 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 1,516 pounds to 4,661 pounds.
Explanation / Answer
Solution:- given information: ranging from 1,516 pounds to 4,661 pounds
a) mean = (a+b)/2
= (1,516+4,661)/2 = 3088.5
=> Mean weight = 3089
b) Standard deviation = (b-a)/sqrt12
= (4661-1516)/sqrt12
= 3145/3.4641
=>Standard deviation = 907.8837
c) P(X < x) = (x - a)/(b - a)
P(X < 1753) = (1753 - 1516)/3145
= 0.0754
d) P(X > x) = 1 - P(X < x)
= 1 - P(X < 3666)
= 1 - [(3666 - 1516)/3145]
= 1 - 0.6836
= 0.3164
e) P(c < x < d) = (d - c)/(b - a)
= (3666 - 1753)/3145
= 0.6083