Explanation Formula used: The mean, median and standard deviation for the continuous random variable are, (1) Mean μ = ∫ − ∞ ∞ x f ( x ) d x . (2) If m is a median, then m must satisfy P ( X ≤ m ) = 1 2 , where P ( X ≤ m ) is solved by finding cumulative distribution function F ( x ) = ∫ − ∞ x f ( t ) d t . (3) Standard deviation σ = V ( X ) where variance V ( X ) = ∫ − ∞ ∞ x 2 f ( x ) d x − μ 2 . Calculation: The probability density function f ( x ) = { 0.36 e − 0.36 x if x ≥ 0 0 otherwise . Use online calculator to compute the mean value on the interval ( − ∞ , ∞ ) by using the above formula. μ = ∫ − ∞ ∞ x f ( x ) d x = ∫ − ∞ 0 ( 0 ) d x + ∫ 0 ∞ 0.36 x e − 0.36 x d x = 0 + ∫ 0 ∞ 0.36 x e − 0.36 x d x ( ∵ x < 0 ) ≈ 2.778 Thus, the value of a mean is 2.778. Use online calculator to compute the mean value on the interval [ 0 , 100 ] by using the above formula. μ = ∫ − ∞ ∞ x f ( x ) d x = ∫ − ∞ 0 ( 0 ) d x + ∫ 0 100 0.36 x e − 0.36 x d x + ∫ 100 ∞ 0.36 x e − 0.36 x d x ≈ 0 + 2.778 + 0 ≈ 2.778 Here, the value of ∫ − ∞ ∞ 0.36 x e − 0.36 x d x = ∫ 0 100 0.36 x e − 0.36 x d x . On the interval ( − ∞ , 0 ] , the probability density function is zero. So, it is immaterial to integrate over ( − ∞ , 0 ] . On further, the probability density function is defined only the interval [ 0 , ∞ ) . Split the interval [ 0 , ∞ ) into two subintervals as [ 0 , 100 ] and [ 100 , ∞ ) . Then, the value of the integral over [ 100 , ∞ ) becomes zero and the value of the integral of x f ( x ) over [ 0 , 100 ] will be the same as the value of the integral of x f ( x ) obtained on the interval ( − ∞ , ∞ ) . Therefore, it is reasonable to replace the interval ( − ∞ , ∞ ) by [ 0 , 100 ] . Compute the median with probability density function. Use the definition of the cumulative distribution function to obtain F ( x ) as follows. For x < 0 , F ( x ) = P ( X < 0 ) = ∫ − ∞ x 0 d t [ ∵ f ( x ) = 0 for x < 0 ] = 0 Hence, the cumulative distribution function for x < 0 is F ( x ) = 0 . For x ≥ 0 , F ( x ) = P ( X ≤ x ) = ∫ − ∞ x f ( t ) d t = ∫ − ∞ 0 ( 0 ) d t + ∫ 0 x 0.36 e − 0.36 t d t = 0 + ∫ 0 x 0.36 e − 0.36 t d t ( ∵ x < 0 ) On further simplification gives, F ( x ) = 0.36 [ e − 0.36 t − 0.36 ] 0 x = − 1 [ e − 0.36 ( x ) − e − 0.36 ( 0 ) ] = − 1 [ e − 0

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Chapter 11.4, Problem 64E

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To approximate: The mean, median and standard deviation of the probability density function and explain the reason for replacing the interval (−∞,∞) by [0,100] and verify the obtained result with the exponential probability density function.

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Chapter 11.4, Problem 63E

Chapter 11.4, Problem 65E

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

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Ch. 11.1 - Evaluate the following, if it converges: 3dx(x1)2.Ch. 11.1 - Prob. 2MP Ch. 11.1 - Prob. 3MP Ch. 11.1 - Prob. 4MP Ch. 11.1 - Prob. 5MP Ch. 11.1 - Prob. 6MP Ch. 11.1 - Prob. 1ED Ch. 11.1 - Prob. 2ED Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E

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To approximate: The mean, median and standard deviation of the probability density function and explain the reason for replacing the interval ( − ∞ , ∞ ) by [ 0 , 100 ] and verify the obtained result with the exponential probability density function.

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To approximate: The mean, median and standard deviation of the probability density function and explain the reason for replacing the interval (−∞,∞) by [0,100] and verify the obtained result with the exponential probability density function.

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