Pam210_PS2_answer - PAM 210 Spring 2008 Professor Owens...

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PAM 210 Spring 2008 Professor Owens Problem Set 2 The Suggested Solution 1) breathing rates X ~ N(12, 2.3) a) P(9.7<=X<=14.3)=P((9.7-12)/2.3<=Z<=(14.3-12)/2.3)=P(- 1<=Z<=1)=0.8413-0.1587=0.6826 (or, based on the empirical rule of a normal distribution, 68% of the observations fall between one standard deviation of the mean). b) P(7.4<=X<=16.6)=P((7.4-12)/2.3<=Z<=(16.6-12)/2.3)=P(- 2<=Z<=2)=0.9772-0.0228=0.9544 (or, based on the empirical rule of a normal distribution, 95% of the observations fall between two standard deviations of the mean). c) P(9.7<=X<=16.6)=P((9.7-12)/2.3<=Z<=(16.6-12)/2.3)=P(- 1<=Z<=2)=0.9772-0.1587=0.8185. d) P(X<=5.1 or X>=18.9)=P(X<=5.1)+P(X>=18.9)=P(Z<=(5.1- 12)/2.3)+P(Z>=(18.9-12)/2.3)=P(Z<=-3)+P(Z>=3)=0.0013*2=0.0026 2) The mean of the amounts of chloroform present in water sources is 34 per liter, and the standard deviation is 53 per liter. Think about the shape of a normal distribution. It is bell-shaped, unimodal, and symmetric density curve. If the chloroform amounts has a normal distribution, its density curve should also be
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Pam210_PS2_answer - PAM 210 Spring 2008 Professor Owens...

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