Explanation Given information: The median of a probability distribution to be that value a such that ∫ a ∞ p ( x ) d x = ∫ − ∞ a p ( x ) d x = 1 2 Formula used: A probability function p ( z ) is symmetric about the line z = a , then f ( a + z ) = f ( a − z ) , ∀ z . Mean, μ = ∫ − ∞ ∞ x p ( x ) d x Proof: Given condition is ∫ a ∞ p ( x ) d x = ∫ − ∞ a p ( x ) d x = 1 2 . Further simplify the above condition, we get ∫ a ∞ p ( x ) d x + ∫ − ∞ a p ( x ) d x = 1 ∫ − ∞ ∞ p ( x ) d x = 1 The probability function is symmetric about the line x = m . (Given) ⇒ p ( m + x ) = p ( m − x ) , ∀ x Mean is given by μ = ∫ − ∞ ∞ x p ( x ) d x = ∫ − ∞ ∞ ( m + ( x − m ) ) p ( x ) d x = ∫ − ∞ ∞ m p ( x ) d x +

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Chapter 9.1, Problem 25E

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To show thatm is both mean and median if a probability function is symmetric about the line x=m.

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Chapter 9.1, Problem 24E

Chapter 9.1, Problem 26E

Chapter 9 Solutions

CALCULUS (CLOTH)

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E

Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Prob. 32E Ch. 9.2 - Prob. 1PQ Ch. 9.2 - Prob. 2PQ Ch. 9.2 - Prob. 3PQ Ch. 9.2 - Prob. 4PQ Ch. 9.2 - Prob. 5PQ Ch. 9.2 - Prob. 1E Ch. 9.2 - Prob. 2E Ch. 9.2 - Prob. 3E Ch. 9.2 - Prob. 4E Ch. 9.2 - Prob. 5E Ch. 9.2 - Prob. 6E Ch. 9.2 - Prob. 7E Ch. 9.2 - Prob. 8E Ch. 9.2 - Prob. 9E Ch. 9.2 - Prob. 10E Ch. 9.2 - Prob. 11E Ch. 9.2 - Prob. 12E Ch. 9.2 - Prob. 13E Ch. 9.2 - Prob. 14E Ch. 9.2 - Prob. 15E Ch. 9.2 - Prob. 16E Ch. 9.2 - Prob. 17E Ch. 9.2 - Prob. 18E Ch. 9.2 - Prob. 19E Ch. 9.2 - Prob. 20E Ch. 9.2 - Prob. 21E Ch. 9.2 - Prob. 22E Ch. 9.2 - Prob. 23E Ch. 9.2 - Prob. 24E Ch. 9.2 - Prob. 25E Ch. 9.2 - Prob. 26E Ch. 9.2 - Prob. 27E Ch. 9.2 - Prob. 28E Ch. 9.2 - Prob. 29E Ch. 9.2 - Prob. 30E Ch. 9.2 - Prob. 31E Ch. 9.2 - Prob. 32E Ch. 9.2 - Prob. 33E Ch. 9.2 - Prob. 34E Ch. 9.2 - Prob. 35E Ch. 9.2 - Prob. 36E Ch. 9.2 - Prob. 37E Ch. 9.2 - Prob. 38E Ch. 9.2 - Prob. 39E Ch. 9.2 - Prob. 40E Ch. 9.2 - Prob. 41E Ch. 9.2 - Prob. 42E Ch. 9.2 - Prob. 43E Ch. 9.2 - Prob. 44E Ch. 9.2 - Prob. 45E Ch. 9.2 - Prob. 46E Ch. 9.2 - Prob. 47E Ch. 9.2 - Prob. 48E Ch. 9.2 - Prob. 49E Ch. 9.2 - Prob. 50E Ch. 9.2 - Prob. 51E Ch. 9.2 - Prob. 52E Ch. 9.2 - Prob. 53E Ch. 9.2 - Prob. 54E Ch. 9.2 - Prob. 55E Ch. 9.2 - Prob. 56E Ch. 9.2 - Prob. 57E Ch. 9.2 - Prob. 58E Ch. 9.2 - Prob. 59E Ch. 9.2 - Prob. 60E Ch. 9.2 - Prob. 61E Ch. 9.2 - Prob. 62E Ch. 9.2 - Prob. 63E Ch. 9.2 - Prob. 64E Ch. 9.2 - Prob. 65E Ch. 9.3 - Prob. 1PQ Ch. 9.3 - Prob. 2PQ Ch. 9.3 - Prob. 3PQ Ch. 9.3 - Prob. 4PQ Ch. 9.3 - Prob. 5PQ Ch. 9.3 - Prob. 1E Ch. 9.3 - Prob. 2E Ch. 9.3 - Prob. 3E Ch. 9.3 - Prob. 4E Ch. 9.3 - Prob. 5E Ch. 9.3 - Prob. 6E Ch. 9.3 - Prob. 7E Ch. 9.3 - Prob. 8E Ch. 9.3 - Prob. 9E Ch. 9.3 - Prob. 10E Ch. 9.3 - Prob. 11E Ch. 9.3 - Prob. 12E Ch. 9.3 - Prob. 13E Ch. 9.3 - Prob. 14E Ch. 9.3 - Prob. 15E Ch. 9.3 - Prob. 16E Ch. 9.3 - Prob. 17E Ch. 9.3 - Prob. 18E Ch. 9.3 - Prob. 19E Ch. 9.3 - Prob. 20E Ch. 9.3 - Prob. 21E Ch. 9.3 - Prob. 22E Ch. 9.3 - Prob. 23E Ch. 9.3 - Prob. 24E Ch. 9.3 - Prob. 25E Ch. 9.3 - Prob. 26E Ch. 9.3 - Prob. 27E Ch. 9.3 - Prob. 28E Ch. 9.3 - Prob. 29E Ch. 9.3 - Prob. 30E Ch. 9.4 - Prob. 1PQ Ch. 9.4 - Prob. 2PQ Ch. 9.4 - Prob. 3PQ Ch. 9.4 - Prob. 4PQ Ch. 9.4 - Prob. 5PQ Ch. 9.4 - Prob. 6PQ Ch. 9.4 - Prob. 1E Ch. 9.4 - Prob. 2E Ch. 9.4 - Prob. 3E Ch. 9.4 - Prob. 4E Ch. 9.4 - Prob. 5E Ch. 9.4 - Prob. 6E Ch. 9.4 - Prob. 7E Ch. 9.4 - Prob. 8E Ch. 9.4 - Prob. 9E Ch. 9.4 - Prob. 10E Ch. 9.4 - Prob. 11E Ch. 9.4 - Prob. 12E Ch. 9.4 - Prob. 13E Ch. 9.4 - Prob. 14E Ch. 9.4 - Prob. 15E Ch. 9.4 - Prob. 16E Ch. 9.4 - Prob. 17E Ch. 9.4 - Prob. 18E Ch. 9.4 - Prob. 19E Ch. 9.4 - Prob. 20E Ch. 9.4 - Prob. 21E Ch. 9.4 - Prob. 22E Ch. 9.4 - Prob. 23E Ch. 9.4 - Prob. 24E Ch. 9.4 - Prob. 25E Ch. 9.4 - Prob. 26E Ch. 9.4 - Prob. 27E Ch. 9.4 - Prob. 28E Ch. 9.4 - Prob. 29E Ch. 9.4 - Prob. 30E Ch. 9.4 - Prob. 31E Ch. 9.4 - Prob. 32E Ch. 9.4 - Prob. 33E Ch. 9.4 - Prob. 34E Ch. 9.4 - Prob. 35E Ch. 9.4 - Prob. 36E Ch. 9.4 - Prob. 37E Ch. 9.4 - Prob. 38E Ch. 9.4 - Prob. 39E Ch. 9.4 - Prob. 40E Ch. 9.4 - Prob. 41E Ch. 9.4 - Prob. 42E Ch. 9.4 - Prob. 43E Ch. 9.4 - Prob. 44E Ch. 9.4 - Prob. 45E Ch. 9.4 - Prob. 46E Ch. 9.4 - Prob. 47E Ch. 9.4 - Prob. 48E Ch. 9.4 - Prob. 49E Ch. 9.4 - Prob. 50E Ch. 9.4 - Prob. 51E Ch. 9 - Prob. 1CRE Ch. 9 - Prob. 2CRE Ch. 9 - Prob. 3CRE Ch. 9 - Prob. 4CRE Ch. 9 - Prob. 5CRE Ch. 9 - Prob. 6CRE Ch. 9 - Prob. 7CRE Ch. 9 - Prob. 8CRE Ch. 9 - Prob. 9CRE Ch. 9 - Prob. 10CRE Ch. 9 - Prob. 11CRE Ch. 9 - Prob. 12CRE Ch. 9 - Prob. 13CRE Ch. 9 - Prob. 14CRE Ch. 9 - Prob. 15CRE Ch. 9 - Prob. 16CRE Ch. 9 - Prob. 17CRE Ch. 9 - Prob. 18CRE Ch. 9 - Prob. 19CRE Ch. 9 - Prob. 20CRE Ch. 9 - Prob. 21CRE Ch. 9 - Prob. 22CRE Ch. 9 - Prob. 23CRE Ch. 9 - Prob. 24CRE Ch. 9 - Prob. 25CRE Ch. 9 - Prob. 26CRE Ch. 9 - Prob. 27CRE Ch. 9 - Prob. 28CRE

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To show thatm is both mean and median if a probability function is symmetric about the line x = m .

Solution Summary: The author shows thatm is both mean and median if a probability function is symmetric about the line x=m.

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To show thatm is both mean and median if a probability function is symmetric about the line x=m.

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Chapter 9 Solutions