Assignment 2 Exercise 2.5 [PDF]

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Members: PHAM NGOC MINH THUY – BAFNIU19034 PHAM DANG KHOA – BAFNIU19015 Course: Quantitative Methods for Finance ASSIGNMENT 2 Exercise 2.5 The systolic blood pressure of male adults is normally distributed with a mean of 127.7 and a standard deviation of 19.2. (a) Specify an interval in which the blood pressures of approximately 68% of the adult male population fall. According to the Empirical Rule: 68.27%:   1  127.7  19.2 = [108.5 ; 146.9] (b) Specify an interval in which the blood pressures of approximately 95% of the adult male population fall. 95.45%:   2  127.7  (2 x 19.2) = [89.3 ; 166.1] (c) Specify an interval in which the blood pressures of approximately 99.7% of the adult male population fall. 99.73%:   3  127.7  (3 x 19.2) = [70.1 ; 185.3]

Exercise 2.6 Suppose that the amount of time that a certain battery functions is a normal random variable with mean 400 hours and standard deviation 50 hours. Suppose that an individual owns two such batteries, one of which is to be used as a spare to replace the other when it fails. (a) What is the probability that the total life of the batteries will exceed 760 hours? X: the amount of time of a certain battery   = 400 ;  = 50 X1 + X2: the amount of time of two batteries   = 400 + 400 = 800  2 = 502 + 502 = 5000   = 50√ 2 760−800  P (X1 + X2 > 760) = P (Z > ) = P(Z > -0.565) = P(Z  0.565) = 0.7123 50 √ 2 (b) What is the probability that the second battery will outlive the first by at least 25 hours? X2 – X1: the amount of time that the second battery will outlive the first   = 400 - 400 = 0   = 50√ 2 25−0  P (X2 – X1 > 25) = P (Z > ) = P(Z > 0.353) = 1 - P(Z  0.353) = 1 – 0.6368 = 0.3632 50 √ 2

(c) What is the probability that the longer-lasting battery will outlive the other by at least 25 hours? P (| X1 - X2 | > 25) = 2P (X2 – X1 > 25) = 2 x 0.3632 = 0.7264

Exercise 2.7 The time it takes to develop a photographic print is a random variable with mean 18 seconds and standard deviation 1 second. Approximate the probability that the total amount of time that it takes to process 100 prints is (a) more than 1710 seconds X: the amount of time to develop a photographic print   = 18 ;  = 1 100X: the amount of time to develop 100 prints   = 18 x 100 = 1800  2 = 12 x 100 = 100   = 10 1710−1800  P (100X > 1710) = P (Z > ) = P (Z > -9) = P (Z  9) = 10 (b) between 1690 and 1710 seconds 1690−1800 1710−1800 P ( 1690 < 100X < 1710 ) = P (

) = P (Z > 0) = 0.5 12000 √ 30 (b) be between 23000 and 27000 23 000.30−750 000 27 000.30−750 000  P(