Mathematics Exam > Mathematics Questions > Consider the continuous random variable with ...
Start Learning for Free
Consider the continuous random variable with probability density function
f(t) = 1 + t for -1 ≤ t ≤ 0
= 1 - t for 0 ≤ t ≤ 1
The standard deviation of the random variables is
f(t) = 1 + t for -1 ≤ t ≤ 0
= 1 - t for 0 ≤ t ≤ 1
The standard deviation of the random variables is
- a)1/√3
- b)1/√6
- c)1/3
- d)1/6
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Consider the continuous random variable with probability density funct...
The probability density function (PDF) given is:
f(t) = 1/t for -1 < t="" />< />
To determine the cumulative distribution function (CDF) F(t) for this continuous random variable, we need to integrate the PDF over the given range:
F(t) = ∫[from -∞ to t] f(u) du
For -1 < t="" />< 1,="" we="" />
F(t) = ∫[from -∞ to t] (1/u) du
To evaluate this integral, we need to split it into two parts:
F(t) = ∫[from -∞ to 0] (1/u) du + ∫[from 0 to t] (1/u) du
The first integral from -∞ to 0 is undefined because the PDF is not defined for negative values. Therefore, we can ignore this part.
For the second integral from 0 to t, we have:
F(t) = ∫[from 0 to t] (1/u) du
Now, to evaluate this integral, we can use the natural logarithm function:
F(t) = ln|u| [from 0 to t]
F(t) = ln|t| - ln|0|
Since ln|0| is undefined, we cannot evaluate it. However, as t approaches 0 from the positive side, ln|t| approaches -∞. Therefore, we can write:
F(t) = ln|t| for 0 ≤ t < />
So, the cumulative distribution function (CDF) for the given continuous random variable is:
F(t) = ln|t| for 0 ≤ t < 1="" />
f(t) = 1/t for -1 < t="" />< />
To determine the cumulative distribution function (CDF) F(t) for this continuous random variable, we need to integrate the PDF over the given range:
F(t) = ∫[from -∞ to t] f(u) du
For -1 < t="" />< 1,="" we="" />
F(t) = ∫[from -∞ to t] (1/u) du
To evaluate this integral, we need to split it into two parts:
F(t) = ∫[from -∞ to 0] (1/u) du + ∫[from 0 to t] (1/u) du
The first integral from -∞ to 0 is undefined because the PDF is not defined for negative values. Therefore, we can ignore this part.
For the second integral from 0 to t, we have:
F(t) = ∫[from 0 to t] (1/u) du
Now, to evaluate this integral, we can use the natural logarithm function:
F(t) = ln|u| [from 0 to t]
F(t) = ln|t| - ln|0|
Since ln|0| is undefined, we cannot evaluate it. However, as t approaches 0 from the positive side, ln|t| approaches -∞. Therefore, we can write:
F(t) = ln|t| for 0 ≤ t < />
So, the cumulative distribution function (CDF) for the given continuous random variable is:
F(t) = ln|t| for 0 ≤ t < 1="" />
Free Test
FREE
| Start Free Test |
Community Answer
Consider the continuous random variable with probability density funct...
Var 
T being the random variable of f(t).
T being the random variable of f(t).
|
Explore Courses for Mathematics exam
|
|
Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer?
Question Description
Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer?.
Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer?.
Solutions for Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
Here you can find the meaning of Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider the continuous random variable with probability density functionf(t) =1 + t for -1≤ t≤ 0= 1 - t for 0≤ t≤ 1The standard deviation of the random variables isa)1/√3b)1/√6c)1/3d)1/6Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Mathematics tests.
|
|
Explore Courses for Mathematics exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup