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A probability density function on the interval [a,1] is given by 1/x2 and outside this interval the value of the function is zero.
The value of a is _________.
The value of a is _________.
Correct answer is '0.5'. Can you explain this answer?
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Verified Answer
A probability density function on the interval [a,1] is given by 1/x2 ...
We know that the sum of all the probabilities is 1.
Therefore on integrating
with limits a to 1, the result should be 1.
Hence a = 0.5
Therefore on integrating
with limits a to 1, the result should be 1.Hence a = 0.5
Integrate 1/x2 in the limits a to 1.. and equate it to 1.. solving we get a=0.5
Most Upvoted Answer
A probability density function on the interval [a,1] is given by 1/x2 ...
Probability Density Function
A probability density function (PDF) describes the probability distribution of a continuous random variable. In this case, the PDF is given by 1/x^2 on the interval [a,1], and outside this interval, the value of the function is zero.
Properties of a PDF
1. The integral of the PDF over its entire domain must be equal to 1.
2. The PDF must be non-negative for all values within its domain.
Finding the Value of 'a'
To find the value of 'a' in the given PDF, we need to satisfy the properties of a PDF.
1. Integral of the PDF over its entire domain must be equal to 1:
∫[a,1] (1/x^2) dx = 1
2. The PDF must be non-negative for all values within its domain:
1/x^2 ≥ 0
Integral Calculation
We can calculate the integral of the PDF over the interval [a,1] to find the value of 'a'.
∫[a,1] (1/x^2) dx = [-1/x] [a,1]
= [1/a] - [1/1]
= 1/a - 1
To satisfy the property of the PDF, this integral must be equal to 1. Therefore, we have:
1/a - 1 = 1
1/a = 2
a = 1/2
Conclusion
The value of 'a' in the given PDF is 1/2 or 0.5. This satisfies the properties of a PDF, as the integral over the interval [a,1] equals 1 and the PDF is non-negative for all values within its domain.
A probability density function (PDF) describes the probability distribution of a continuous random variable. In this case, the PDF is given by 1/x^2 on the interval [a,1], and outside this interval, the value of the function is zero.
Properties of a PDF
1. The integral of the PDF over its entire domain must be equal to 1.
2. The PDF must be non-negative for all values within its domain.
Finding the Value of 'a'
To find the value of 'a' in the given PDF, we need to satisfy the properties of a PDF.
1. Integral of the PDF over its entire domain must be equal to 1:
∫[a,1] (1/x^2) dx = 1
2. The PDF must be non-negative for all values within its domain:
1/x^2 ≥ 0
Integral Calculation
We can calculate the integral of the PDF over the interval [a,1] to find the value of 'a'.
∫[a,1] (1/x^2) dx = [-1/x] [a,1]
= [1/a] - [1/1]
= 1/a - 1
To satisfy the property of the PDF, this integral must be equal to 1. Therefore, we have:
1/a - 1 = 1
1/a = 2
a = 1/2
Conclusion
The value of 'a' in the given PDF is 1/2 or 0.5. This satisfies the properties of a PDF, as the integral over the interval [a,1] equals 1 and the PDF is non-negative for all values within its domain.
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A probability density function on the interval [a,1] is given by 1/x2 and outside this interval the value of the function is zero.The value of a is _________.Correct answer is '0.5'. Can you explain this answer?
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