i. Explanation Given information: The probability distribution function is E x = μ . Calculation: The probability distribution function is E x = ∫ a b x . f x d x . Use the identity a − b 2 = a 2 + b 2 − 2 a b ii. To determine To prove: that E x 2 − 2 μ x + μ 2 = E x 2 − E 2 μ x + E μ 2 . iii. To determine To prove: that E x 2 − E 2 μ x + E μ 2 = E x 2 − 2 E μ E x + E x 2 . iv. To determine To prove: that E x 2 − 2 E μ E x + E x 2 = E x 2 − 2 E x 2 + E x 2 . v. To determine To prove: that E x 2 − 2 E x 2 + E x 2 = E x 2 − μ 2 .

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Author: BITTINGER

Publisher: PEARSON C

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Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 5.5, Problem 64E

To determine

To prove: that Ex−μ2=Ex2−2μx+μ2 .

ii.

To determine

To prove: that Ex2−2μx+μ2=Ex2−E2μx+Eμ2 .

iii.

To determine

To prove: that Ex2−E2μx+Eμ2=Ex2−2EμEx+Ex2 .

iv.

To determine

To prove: that Ex2−2EμEx+Ex2=Ex2−2Ex2+Ex2 .

To determine

To prove: that Ex2−2Ex2+Ex2=Ex2−μ2 .

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Chapter 5.5, Problem 63E

Chapter 5.5, Problem 65E

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Chapter 5 Solutions

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Ch. 5.1 - In Exercises 1-14, is the price, in dollars per...Ch. 5.1 - In Exercises 1-14, is the price, in dollars per...Ch. 5.1 - In Exercises 1-14, D(x) is the price, in dollars...Ch. 5.1 - In Exercises 1-14, is the price, in dollars per...Ch. 5.1 - In Exercises 1-14, D(x) is the price, in dollars...Ch. 5.1 - In Exercises 1-14, D(x) is the price, in dollars...Ch. 5.1 - In Exercises 1-14, D(x) is the price, in dollars...Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

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To prove: that E x − μ 2 = E x 2 − 2 μ x + μ 2 .

Solution Summary: The author concludes that the probability distribution function is E(x)=mu .

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To prove: that Ex−μ2=Ex2−2μx+μ2 .

To prove: that Ex2−2μx+μ2=Ex2−E2μx+Eμ2 .

To prove: that Ex2−E2μx+Eμ2=Ex2−2EμEx+Ex2 .

To prove: that Ex2−2EμEx+Ex2=Ex2−2Ex2+Ex2 .

To prove: that Ex2−2Ex2+Ex2=Ex2−μ2 .

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Chapter 5 Solutions