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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?
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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="" />
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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).
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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?
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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?.
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