(a) Explanation Given: The normal distribution is I = ∫ − ∞ ∞ e − x 2 d x . The integral is I 2 = ∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y = ∫ − ∞ ∞ ∫ − ∞ ∞ e − x 2 − y 2 d x d y Formula used: If f is a polar rectangle R given by 0 ≤ a ≤ r ≤ b , α ≤ θ ≤ β , where 0 ≤ β − α ≤ 2 π , then, ∬ R f ( x , y ) d A = ∫ α β ∫ a b f ( r cos θ , r sin θ ) r d r d θ (1) If g ( x ) is the function of x and h ( y ) is the function of y then, ∫ a b ∫ c d g ( x ) h ( y ) d y d x = ∫ a b g ( x ) d x ∫ c d h ( y ) d y (2) Calculation: From the given region, it is observed that r varies from 0 to ∞ and θ varies from 0 to 2 π (b) To determine To evaluate: The given integrals

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 13.3, Problem 71E

(a)

To determine

To show: The value of I is π .

(b)

To determine

To evaluate: The given integrals

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Chapter 13.3, Problem 70E

Chapter 13.3, Problem 72E

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The value of I is π .

Solution Summary: The author explains that the normal distribution is I=displaystyle underset-infty