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A Random variable X is uniformly distributed on the interval (-5, 15). Another random variableY = e -X/5 is formed. The value of E[Y ] is
- a)2
- b)0.667
- c)1.387
- d)2.967
Correct answer is option 'B'. Can you explain this answer?
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A Random variable X is uniformly distributed on the interval (-5, 15)....
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A Random variable X is uniformly distributed on the interval (-5, 15)....
Given information:
- Random variable X is uniformly distributed on the interval (-5, 15)
- Random variable Y = e^(-X/5)
To find: Expected value of Y, i.e., E[Y]
Solution:
The formula for expected value of a continuous random variable is:
E[X] = ∫xf(x)dx, where f(x) is the probability density function of X
Here, X is uniformly distributed on the interval (-5, 15). So, the probability density function of X is:
f(x) = 1/(15 - (-5)) = 1/20, for -5 ≤ x ≤ 15
= 0, for otherwise
So, the expected value of X can be calculated as:
E[X] = ∫xf(x)dx
= ∫(-5)^15(x/20)dx
= [x^2/40](-5)^15
= (15^2 - (-5)^2)/(2*20)
= 5
Now, we need to find the expected value of Y, which is given by:
E[Y] = ∫yf(y)dy, where f(y) is the probability density function of Y
Here, Y = e^(-X/5), so we can find the probability density function of Y using the transformation method:
f(y) = f(x)/|dy/dx|
= (1/20)/(5e^(-x/5))
= 1/(100ye^(ln(y)/5)), for 0 < y="" ≤="" />
= 0, for otherwise
So, the expected value of Y can be calculated as:
E[Y] = ∫yf(y)dy
= ∫0^e^3(y/(100ye^(ln(y)/5)))dy
= (1/100)∫0^e^3e^(4ln(y)/5)dy
= (1/100)∫0^e^3y^(4/5)dy
= (1/100)[(5/9)y^(9/5)]0^e^3
= (1/100)[(5/9)(e^3)^(9/5) - (5/9)(0)^(9/5)]
= (1/100)(5/9)(e^3)^(9/5)
= (1/100)(5/9)(e^3)
= 5e^(-3)/9
= 0.6667 (approx.)
Therefore, the value of E[Y] is 0.667, which is option (B).
- Random variable X is uniformly distributed on the interval (-5, 15)
- Random variable Y = e^(-X/5)
To find: Expected value of Y, i.e., E[Y]
Solution:
The formula for expected value of a continuous random variable is:
E[X] = ∫xf(x)dx, where f(x) is the probability density function of X
Here, X is uniformly distributed on the interval (-5, 15). So, the probability density function of X is:
f(x) = 1/(15 - (-5)) = 1/20, for -5 ≤ x ≤ 15
= 0, for otherwise
So, the expected value of X can be calculated as:
E[X] = ∫xf(x)dx
= ∫(-5)^15(x/20)dx
= [x^2/40](-5)^15
= (15^2 - (-5)^2)/(2*20)
= 5
Now, we need to find the expected value of Y, which is given by:
E[Y] = ∫yf(y)dy, where f(y) is the probability density function of Y
Here, Y = e^(-X/5), so we can find the probability density function of Y using the transformation method:
f(y) = f(x)/|dy/dx|
= (1/20)/(5e^(-x/5))
= 1/(100ye^(ln(y)/5)), for 0 < y="" ≤="" />
= 0, for otherwise
So, the expected value of Y can be calculated as:
E[Y] = ∫yf(y)dy
= ∫0^e^3(y/(100ye^(ln(y)/5)))dy
= (1/100)∫0^e^3e^(4ln(y)/5)dy
= (1/100)∫0^e^3y^(4/5)dy
= (1/100)[(5/9)y^(9/5)]0^e^3
= (1/100)[(5/9)(e^3)^(9/5) - (5/9)(0)^(9/5)]
= (1/100)(5/9)(e^3)^(9/5)
= (1/100)(5/9)(e^3)
= 5e^(-3)/9
= 0.6667 (approx.)
Therefore, the value of E[Y] is 0.667, which is option (B).
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A Random variable X is uniformly distributed on the interval (-5, 15). Another random variableY = e -X/5is formed. The value of E[Y ] isa)2b)0.667c)1.387d)2.967Correct answer is option 'B'. Can you explain this answer?
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