Explanation Given: The probability density function is f ( x ) = 1 2 e − x 2 on the interval [ 0 , ∞ ) . The median of X is defined to be the number m such that P ( X ≤ m ) = 1 2 . Concept used: The probability that an observed value of the random variable X lies in the interval [ a , b ] can be expressed as below. P ( a ≤ X ≤ b ) = ∫ a b f ( x ) d x ...... ( 1 ) Here, f ( x ) is the probability density function. Calculation: It is given that the probability density function is f ( x ) = 1 2 e − x 2 on the interval [ 0 , ∞ ) . The median of X is defined to be the number m such that P ( X ≤ m ) and that can be written as follows: P ( X ≤ m ) = ∫ 0 m f ( x ) d x Substitute 1 2 e − x 2 for f ( x ) in the above expression to obtain the probability as follows: P ( X ≤ m ) = ∫ 0 m ( 1 2 e − x 2 ) d x = 1 2 [ − 2 e − x 2 ] 0 m = − [ e − m 2<

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Chapter 10.2, Problem 34E

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The median of the random variable X with the probability density function f(x)=12e−x2 on the interval [0,∞).

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Chapter 10.2, Problem 33E

Chapter 10.2, Problem 35E

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The median of the random variable X with the probability density function f ( x ) = 1 2 e − x 2 on the interval [ 0 , ∞ ) .

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The median of the random variable X with the probability density function f(x)=12e−x2 on the interval [0,∞).

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Chapter 10 Solutions