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Unformatted text preview: 7.4. We know that, is distributed as . Therefore, and the value of the random variable associated with S 2 =21.96 is Therefore, Thus, the probability that when s 2 21.96 is about 0.0577. (JMP) 7.25. is a random variable with 9 degrees of freedom since Z i s are random samples from the standard normal distribution. So P( c)=0.9=P() or 1 P()=0.1. Then using the JMP, we can calculate, so c=14.68. (JMP) 7.28. Let X i be the length of duration for each heat lamp. We want to find P(3100). We can use the Central Limit Theorem here because the sample size of 30 is sufficiently large. Therefore ~ N(30*100=3000, 10*sqrt(30)=54.77). So P(3100) = P(Z(31003000)/54.77) = P(Z1.83) = 1 P(Z1.83) = .0336. 7.30. Since ~ , we have = = = .9947....
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 Spring '10
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