The Cozy Car Company created a new model of car especially for Japan and Vietnam, which they start shipping to both countries at
the same time.The company had shipped 222 of that model of car to Japan by the end of the first month, and the cumulative total of cars they've shipped triples each month.The company had shipped 555 of that model of car to Vietnam by the end of the first month, and the cumulative total of cars they've shipped increases by 111111 cars each month.In which month will the cumulative number of cars of that model shipped to Japan first exceed the cumulative total shipped to Vietnam?
1 answer:
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Answer:
(A) 0.15625
(B) 0.1875
(C) Can't be computed
Step-by-step explanation:
We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.
Let X = Amount of time taken by student to complete a statistics quiz
So, X ~ U(32 , 64)
The PDF of uniform distribution is given by;
f(X) = , a < X < b where a = 32 and b = 64
The CDF of Uniform distribution is P(X <= x) =
(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)
P(X > 59) = 1 - P(X <= 59) = 1 - = 1 -
=
= 0.15625
(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)
P(X <= 43) = =
= 0.34375
P(X < 37) = =
= 0.15625
P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875
(C) Probability that student complete the quiz in exactly 44.74 minutes
= P(X = 44.74)
The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.
Answer:
0.00
Step-by-step explanation:
If the national average score on a standardized test is 1010, and the standard deviation is 200, where scores are normally distributed, to calculate the probability that a test taker scores at least 1600 on the test, we should first to calculate the z-score related to 1600. This z-score is , then, we are seeking P(Z > 2.95), where Z is normally distributed with mean 0 and standard deviation 1. Therefore, P(Z > 2.95) = 0.00