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Statistics 270 - Lecture 15. Percentile for Normal Distributions The 100p th percentile of the N( ,  2 ) distribution is  +  (p)  Where  (p) is.

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Presentation on theme: "Statistics 270 - Lecture 15. Percentile for Normal Distributions The 100p th percentile of the N( ,  2 ) distribution is  +  (p)  Where  (p) is."— Presentation transcript:

1 Statistics 270 - Lecture 15

2 Percentile for Normal Distributions The 100p th percentile of the N( ,  2 ) distribution is  +  (p)  Where  (p) is the 100p th percentile of the standard normal Proof:

3 Why is this useful?

4 Example Find the 97.50 percentile of the N(10,5) distribution

5 Example Verbal SAT scores follow approximately a N(430, 100) distribution What is the interval that encompasses the middle 50% of scores

6 Normal Approximation to the Binomial If X has a Binomial(n, p) distribution and n is large, then the calculations for this distribution become cumbersome (e.g., P(X<100) ) If X has a binomial distribution with parameters n and p, with np 5 and n(1-p) 5, then X has approximately a distribution

7 Normal Approximation to the Binomial Recall, Can use Poisson approximation when Can use Normal approximation when

8 Continuity Correction

9 Example Suppose that 25% of all licensed drivers in a particular province do not have insurance Let X be the number of uninsured drivers in a random sample of size n = 50

10 Example Suppose that 25% of all licensed drivers in a particular province do not have insurance Let X be the number of uninsured drivers in a random sample of size n = 50


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