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(Z≥1.83) = 1 – P(Z≤1.83) = .0336. 7.30. Since ~ , we have = = = .9947....
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 Spring '10
 BARNES
 Central Limit Theorem, Normal Distribution, Standard Deviation, Variance, Probability theory, finite population correction

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